0.3 Fractions

Aligned outcomes:

SLO 6.NS

The Number System. Divide fractions by fractions; compute fluently with

SLO 6.EE

Expressions and Equations. Write, read and evaluate expressions in which

SLO 7.NS

The Number System. Add, subtract, multiply and divide rational numbers,

SLO 7.EE

Expressions and Equations. Apply properties of operations to add, subtract,

SLO 8.NS

The Number System. Know that numbers which are not rational are irrational;

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:
  • explain what a fraction means, both as a part of a whole and as a division;
  • build equivalent fractions and simplify a fraction to lowest terms;
  • convert between improper fractions and mixed numbers, and place the sign of a negative fraction correctly;
  • multiply and divide fractions, and explain why dividing by a fraction means multiplying by its reciprocal;
  • add and subtract fractions with like and unlike denominators using the least common denominator;
  • simplify complex fractions and apply the order of operations to expressions with fractions in them.

The integers from Section 0.2 count things that come in whole pieces. Most of what you measure does not. Two and a half hours of practice, three quarters of a tank of gas, a recipe scaled down to five sixths — none of those can be written with an integer, and rounding them off throws away information you may need later.

Fractions handle those in-between amounts exactly. That word matters. Section 0.4 covers decimals, which describe the same quantities in a different notation but sometimes only approximately: one third is \(0.333\dots\) forever, so any decimal you actually write down is a little bit wrong. A fraction is never a little bit wrong. That is why algebra keeps answers in fraction form and converts to decimals at the very end, if at all.

Here is the thing to watch for as you read. Most fraction mistakes are not arithmetic mistakes. They happen when a rule from one operation leaks into a different operation. Multiplication really does go straight across; addition does not, and never has. Cancelling works on factors; it does not work on terms. This section is careful to tell you not just what each rule says, but exactly where it stops.

0.3.1 Visualizing and Simplifying Fractions

Cut a pan of cornbread into 8 equal pieces and take 3 of them. You have taken \(\dfrac{3}{8}\) of the pan. The bottom number says how many equal pieces the whole was cut into, and the top number says how many of those pieces you are talking about.

Definition 0.3.1: Fraction

A fraction is a number written \(\dfrac{a}{b}\), where \(b \neq 0\). The denominator \(b\) is the number of equal parts the whole has been divided into, and the numerator \(a\) is how many of those parts are being counted.

The word equal is doing real work in that definition. If you hack the pan into 8 pieces of wildly different sizes and grab 3, you do not have \(\dfrac{3}{8}\) of the pan. Every part-of-a-whole reading of a fraction assumes the pieces are the same size.

Two conditions in the definition are worth pulling out on their own, because both come back later. First, the denominator can never be zero: \(\dfrac{5}{0}\) asks you to cut a whole into zero equal parts, which describes nothing at all. That is the same undefined division you met in Section 0.2, and spotting when a variable denominator could turn into zero is a job you will do over and over in Chapter 3.

Second, a fraction is also a division. The expression \(\dfrac{a}{b}\) means \(a \div b\). That is not a separate meaning bolted on; it is the same idea coming at you from the other direction. Split 3 pans of cornbread evenly among 8 people and each person gets \(3 \div 8 = \dfrac{3}{8}\) of a pan, which is exactly what the picture already showed.

FractionAs parts of a wholeAs a divisionValue
\(\dfrac{3}{8}\)3 of 8 equal pieces\(3 \div 8\)\(0.375\)
\(\dfrac{6}{6}\)all 6 of 6 equal pieces\(6 \div 6\)\(1\)
\(\dfrac{0}{9}\)none of 9 equal pieces\(0 \div 9\)\(0\)
\(\dfrac{11}{4}\)11 quarter-pieces, more than one whole\(11 \div 4\)\(2.75\)
\(\dfrac{7}{0}\)0 equal pieces, which is meaningless\(7 \div 0\)undefined

The second row generalizes into a rule you will use constantly. Any nonzero number divided by itself is 1, so a fraction whose numerator and denominator match is one whole:

$$\frac{a}{a} = 1 \qquad (a \neq 0)$$

That small fact is the engine behind everything else in this subsection.

A fraction names how many of a whole's equal pieces are counted A pan of cornbread drawn as a single bar divided into eight equal pieces. A brace underneath the whole bar is labelled cut into 8 equal pieces, which is the denominator. Three of the pieces at the left end are then shaded and labelled 3 of them taken, which is the numerator. Below, the fraction three eighths is written out, with the top number labelled numerator, how many are counted, and the bottom number labelled denominator, equal pieces in the whole. A closing line notes that because a fraction is also a division, three eighths is the same number as three divided by eight. cut into 8 equal pieces — the denominator 3 of them taken — the numerator 3 8 numerator — how many are counted denominator — equal pieces in the whole a fraction is also a division, so 3/8 is the same number as 3 ÷ 8

Definition 0.3.1 — A fraction: the denominator names the equal pieces, the numerator counts them.

Equivalent fractions

Now slice each of those 8 cornbread pieces in half. There are 16 pieces on the table, and the 3 you took have become 6. Nothing about how much cornbread you have changed — only how finely it was cut. So

$$\frac{3}{8} = \frac{6}{16}$$

Fractions that name the same amount are called equivalent, and every fraction has infinitely many equivalent forms. You generate them by cutting the pieces finer or by grouping them coarser, and the rule below does both.

Definition 0.3.2: Equivalent Fractions Property

If \(a\), \(b\), and \(c\) are numbers with \(b \neq 0\) and \(c \neq 0\), then

$$\frac{a}{b} = \frac{a \cdot c}{b \cdot c} \qquad \text{and} \qquad \frac{a}{b} = \frac{a \div c}{b \div c}$$
Slicing does not feed more people

Cutting a pizza into 16 slices instead of 8 does not give you more pizza. It gives you more pieces, each one half as big. Equivalent fractions are the same amount described with a different piece size.

Read left to right, that property builds a fraction up to a bigger denominator. Read right to left, it reduces one. Both directions are allowed for the same reason: multiplying the numerator and denominator by the same \(c\) is really multiplying by \(\dfrac{c}{c}\), which is 1, and multiplying by 1 cannot change what a number is worth.

Start withMultiply top and bottom byEquivalent fraction
\(\dfrac{2}{5}\)3\(\dfrac{6}{15}\)
\(\dfrac{2}{5}\)4\(\dfrac{8}{20}\)
\(\dfrac{2}{5}\)9\(\dfrac{18}{45}\)
\(\dfrac{7}{6}\)5\(\dfrac{35}{30}\)
\(\dfrac{1}{4}\)12\(\dfrac{12}{48}\)

The requirement that \(c \neq 0\) is not fussiness. Multiplying top and bottom by 0 would turn every fraction into \(\dfrac{0}{0}\), which is undefined and tells you nothing about what you started with.

Slicing every piece in half renames the fraction without changing the amount A pan drawn as one bar cut into eight equal pieces, with the three pieces at the left end shaded and labelled three eighths. A dividing line then appears inside every piece, splitting the bar into sixteen pieces; the shaded region does not move or change width, but it now covers six of the sixteen pieces and is labelled six sixteenths. A bracket under the shaded region is labelled the same amount of pan, both times, because the shaded width is identical before and after. The closing line shows three eighths equals three times two over eight times two, which is six sixteenths. 3/8 now 16 pieces — the shaded part covers 6 of them the same amount, both times 3/8 = (3 × 2) / (8 × 2) = 6/16 multiplying top and bottom by the same number is multiplying by 1

Definition 0.3.2 — Equivalent Fractions Property: finer pieces, the same amount.

Simplifying to lowest terms

Definition 0.3.3: Lowest terms

A fraction is in lowest terms when its numerator and denominator share no common factor other than 1.

To simplify a fraction, divide the numerator and the denominator by their greatest common factor. If the greatest common factor does not jump out at you, divide by any common factor you do see and repeat — you land in the same place, it just takes more steps.

Lowest terms is the point where no common factor is left to group by A bar divided into twenty-four equal pieces with the first eighteen shaded, labelled eighteen twenty-fourths. Heavier dividing lines then appear every six pieces, gathering the bar into four equal groups, three of which are entirely shaded. The shaded width does not change. The reading underneath becomes three quarters, and a closing line notes that three and four share no common factor but one, so there is nothing left to group by and the fraction is in lowest terms. 18/24 6 6 6 6 gathered in sixes, the greatest common factor of 18 and 24 18/24 = 3 groups shaded of 4 = 3/4 3 and 4 share no common factor but 1, so there is nothing left to group by

Definition 0.3.3 — Lowest terms: the point where no common factor is left to group by.

Example 0.3.1: Simplifying to lowest terms

Simplify \(\dfrac{42}{70}\).

Solution

Step 1 — Look for common factors. Both 42 and 70 are even, so 2 divides both. Both are also divisible by 7. Putting those together, 14 divides both, and nothing larger does.

Step 2 — Divide the top and the bottom by 14.

$$\frac{42}{70} = \frac{42 \div 14}{70 \div 14} = \frac{3}{5}$$

Step 3 — Check that you are done. The numbers 3 and 5 share no factor but 1, so the fraction is in lowest terms.

Answer: \(\dfrac{3}{5}\).

FractionGreatest common factorLowest terms
\(\dfrac{18}{24}\)6\(\dfrac{3}{4}\)
\(\dfrac{42}{70}\)14\(\dfrac{3}{5}\)
\(\dfrac{27}{45}\)9\(\dfrac{3}{5}\)
\(\dfrac{16}{40}\)8\(\dfrac{2}{5}\)
\(\dfrac{30}{6}\)6\(\dfrac{5}{1} = 5\)
\(\dfrac{13}{20}\)1already lowest

Two of those rows are worth a comment. The last one shows that simplifying does not always change the fraction: when the greatest common factor is 1, you are already finished. The row above it shows a fraction simplifying all the way down to a whole number, which happens whenever the denominator divides the numerator evenly.

Cancel factors, not terms

Factors cancel, terms do not

\(\dfrac{3 \cdot 5}{3 \cdot 7}\) really is \(\dfrac{5}{7}\), but \(\dfrac{3 + 5}{3 + 7}\) is not. In the first the 3 is being multiplied, so it is a factor. In the second it is being added, so it is a term. Only factors cancel.

This is the most common fraction error in all of algebra, so it is worth stating exactly. Simplifying divides the whole numerator and the whole denominator by a common factor. A factor is something being multiplied. If a piece of the numerator is being added instead, it is a term, and it cannot be cancelled.

Check the bad one with actual numbers: \(\dfrac{3+5}{3+7} = \dfrac{8}{10} = \dfrac{4}{5}\), which is nowhere near \(\dfrac{5}{7}\). The two 3s look identical on the page, but in one expression the 3 is a factor and in the other it is a term.

ExpressionCan the 4s cancel?WhySimplified
\(\dfrac{4 \cdot 9}{4 \cdot 11}\)yes4 is a factor of both\(\dfrac{9}{11}\)
\(\dfrac{4 + 9}{4 + 11}\)no4 is a term, not a factor\(\dfrac{13}{15}\)
\(\dfrac{4x}{4y}\)yes4 multiplies each\(\dfrac{x}{y}\)
\(\dfrac{4 + x}{4}\)nothe numerator is a sum\(\dfrac{4+x}{4}\)

Here is the habit that protects you. Before cancelling anything, ask whether the numerator and the denominator are each written as a single product. If either one has a \(+\) or a \(-\) sitting at the top level, nothing inside it can be cancelled until you factor it first, which is a skill Chapter 8 builds properly.

Improper fractions and mixed numbers

Definition 0.3.4: Proper and improper fractions

A proper fraction has a numerator smaller than its denominator, so its value is less than 1. An improper fraction has a numerator greater than or equal to its denominator, so its value is 1 or more.

The one-whole mark is what sorts a proper fraction from an improper one Two bars of quarter-pieces are drawn against a common scale with a marked line at one whole. The upper bar shades three quarters and stops short of that line, and is labelled a proper fraction because its numerator three is smaller than its denominator four, so its value is less than one. The lower bar shades eleven quarters and runs past the line, using two full wholes and three more quarters, and is labelled an improper fraction because its numerator eleven is greater than its denominator four, so its value is one or more. one whole 3/4 — stops short of one whole a proper fraction numerator 3 is smaller than denominator 4 two wholes 11/4 — runs past one whole, filling two wholes and three more quarters an improper fraction — numerator 11 is greater than denominator 4

Definition 0.3.4 — The one-whole mark is what sorts a proper fraction from an improper one.

Definition 0.3.5: Mixed number

A mixed number is a whole number written beside a proper fraction, and it means their sum: \(2\frac{3}{4}\) means \(2 + \frac{3}{4}\).

Where the quarters are hiding

Each whole holds 4 quarters, so 5 wholes hold 20 of them. That is all the "multiply by the denominator" step is doing — counting the quarter-pieces buried inside the whole numbers before adding the 3 loose ones.

Neither form is more correct than the other, and they are useful in different places. Mixed numbers read the way people talk, which is what you would say at a lumber yard. Improper fractions are much easier to compute with, which is why algebra almost always converts to them first.

To go from improper to mixed, divide the numerator by the denominator. The quotient is the whole-number part. The remainder is the new numerator, and the denominator stays put. To go the other way, multiply the whole number by the denominator, then add the numerator. The denominator does not change.

That second procedure is not a recipe you have to take on faith. Since \(5\frac{3}{4}\) means \(5 + \frac{3}{4}\), and \(5 = \frac{20}{4}\), the sum is \(\frac{20}{4} + \frac{3}{4} = \frac{23}{4}\).

A mixed number is the same pieces gathered into whole units plus a remainder Eleven shaded quarter-pieces sit in a single row, labelled eleven quarters. Heavier dividers then appear after the fourth and the eighth piece, gathering the row into two full wholes of four quarters each, with three loose quarters left over. Braces label the two full wholes and the three loose quarters. The reading underneath is eleven quarters equals two and three quarters, and a closing line notes that writing a whole number beside a fraction means their sum, so two and three quarters is two plus three quarters, not two times three quarters. 11 quarter-pieces — 11/4 4 quarters = 1 whole 4 quarters = 1 whole 3 loose quarters 11/4 = 2 and 3/4 a whole number written beside a fraction means their SUM: 2 + 3/4, never 2 × 3/4

Definition 0.3.5 — A mixed number is the same pieces gathered into wholes plus a remainder.

Example 0.3.2: Converting both directions

Write \(\dfrac{23}{4}\) as a mixed number, then convert your answer back to an improper fraction.

Solution

Step 1 — Divide the numerator by the denominator. How many times does 4 go into 23?

$$23 \div 4 = 5 \text{ with a remainder of } 3$$

Step 2 — Assemble the mixed number. The quotient 5 is the whole-number part, the remainder 3 is the new numerator, and the denominator stays 4.

$$\frac{23}{4} = 5\frac{3}{4}$$

Step 3 — Convert back. Multiply the whole number by the denominator, then add the numerator.

$$5 \cdot 4 + 3 = 23$$

Keeping the same denominator gives \(\dfrac{23}{4}\), which is what we started with.

Answer: \(5\frac{3}{4}\), and converting back gives \(\dfrac{23}{4}\).

Improper fractionThe divisionMixed number
\(\dfrac{23}{4}\)\(23 \div 4 = 5\) remainder \(3\)\(5\dfrac{3}{4}\)
\(\dfrac{17}{5}\)\(17 \div 5 = 3\) remainder \(2\)\(3\dfrac{2}{5}\)
\(\dfrac{31}{8}\)\(31 \div 8 = 3\) remainder \(7\)\(3\dfrac{7}{8}\)
\(\dfrac{9}{9}\)\(9 \div 9 = 1\) remainder \(0\)\(1\)
\(\dfrac{40}{6}\)simplify to \(\dfrac{20}{3}\), then \(20 \div 3 = 6\) remainder \(2\)\(6\dfrac{2}{3}\)

One warning about mixed-number notation. Writing a whole number next to a fraction means addition, but writing a number next to a variable means multiplication, as Section 0.1 established. So \(2\frac{3}{4}\) is \(2 + \frac{3}{4}\), while \(2x\) is \(2 \cdot x\). The two notations look parallel and mean opposite things, which is one more reason algebra prefers improper fractions — \(\frac{11}{4}\) carries no such ambiguity.

Negative fractions and where the sign goes

A fraction can be negative, and the sign can be written in three different places. All three mean the same number.

$$-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b} \qquad (b \neq 0)$$

The sign rules from Section 0.2 are what make that true. A fraction is a division, and a division with exactly one negative sign among its parts comes out negative. It makes no difference whether that sign is sitting on the numerator, on the denominator, or out in front.

Where the sign isExampleValue
Out in front\(-\dfrac{3}{7}\)negative
On the numerator\(\dfrac{-3}{7}\)negative
On the denominator\(\dfrac{3}{-7}\)negative
On both\(\dfrac{-3}{-7} = \dfrac{3}{7}\)positive

That last row is the like-signs-give-a-positive rule from Section 0.2, and it is the fraction version of \(-(-6) = 6\).

Keep the sign out in front

\(-\dfrac{2}{3}\), \(\dfrac{-2}{3}\), and \(\dfrac{2}{-3}\) are all the same number, but only the first is easy to keep track of. A minus sign buried in a denominator gets lost partway through a long problem, and once it is lost you will not find it by rereading.

Make moving the sign to the front an automatic habit. When you simplify \(\dfrac{-24}{36}\), for instance, the greatest common factor of 24 and 36 is 12, so the fraction reduces to \(\dfrac{-2}{3}\), and you write that as \(-\dfrac{2}{3}\).

Try It Now 0.3.1

Simplify \(\dfrac{54}{72}\). Then write \(\dfrac{19}{5}\) as a mixed number, and rewrite \(\dfrac{6}{-13}\) with the sign in front.

Solution

Part 1 — Simplify \(\dfrac{54}{72}\). Both numbers are divisible by 18, and nothing larger divides both.

$$\frac{54}{72} = \frac{54 \div 18}{72 \div 18} = \frac{3}{4}$$

Part 2 — Write \(\dfrac{19}{5}\) as a mixed number. Divide 19 by 5.

$$19 \div 5 = 3 \text{ with a remainder of } 4$$

The quotient is the whole number, the remainder is the new numerator, and the denominator stays 5.

$$\frac{19}{5} = 3\frac{4}{5}$$

Part 3 — Move the sign on \(\dfrac{6}{-13}\). One negative sign among the parts makes the fraction negative, and the standard place to write it is out front.

$$\frac{6}{-13} = -\frac{6}{13}$$

Answer: \(\dfrac{3}{4}\); \(3\dfrac{4}{5}\); \(-\dfrac{6}{13}\).

0.3.2 Multiplying and Dividing Fractions

Definition 0.3.6: Fraction multiplication

If \(a\), \(b\), \(c\), and \(d\) are numbers with \(b \neq 0\) and \(d \neq 0\), then

$$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$

Multiply the numerators, multiply the denominators, then simplify. That really is the whole rule:

$$\frac{2}{5} \cdot \frac{3}{7} = \frac{2 \cdot 3}{5 \cdot 7} = \frac{6}{35}$$

Here is the picture behind it. You have \(\frac{3}{7}\) of a pan of cornbread, and you want \(\frac{2}{5}\) of that. Cutting the pan into 7 strips and then cutting each strip into 5 pieces makes \(5 \cdot 7 = 35\) small pieces in the whole pan, and you are keeping \(2 \cdot 3 = 6\) of them. The denominators multiply because the two rounds of cutting compound, and the numerators multiply for the same reason. Notice also that the word "of" is signaling multiplication here, exactly as the translation table in Section 0.1 promised.

Signs work the way Section 0.2 said: like signs give a positive product, unlike signs give a negative one. A whole number joins in by writing it over 1, since \(n = \dfrac{n}{1}\).

ProductStraight acrossSimplified
\(\dfrac{2}{5} \cdot \dfrac{3}{7}\)\(\dfrac{6}{35}\)\(\dfrac{6}{35}\)
\(-\dfrac{3}{4} \cdot \dfrac{8}{9}\)\(-\dfrac{24}{36}\)\(-\dfrac{2}{3}\)
\(\dfrac{5}{6} \cdot 12\)\(\dfrac{60}{6}\)\(10\)
\(-\dfrac{2}{3} \cdot \left(-\dfrac{9}{10}\right)\)\(\dfrac{18}{30}\)\(\dfrac{3}{5}\)
Two rounds of cutting compound, which is why both numerators and both denominators multiply A square pan is cut into seven vertical strips and three of them are shaded, showing three sevenths. The pan is then cut again into five horizontal bands, so the whole pan now holds thirty-five small pieces. Taking two of the five bands from within the shaded strips leaves six small pieces doubly covered. A reading underneath shows two fifths times three sevenths equals two times three over five times seven, which is six thirty-fifths, and notes that the denominators multiply because the two rounds of cutting compound. 7 strips, 3 of them shaded that is 3/7 of the pan cut again into 5 bands 5 × 7 = 35 small pieces in the whole pan keep 2 of those 5 bands, inside the shaded strips 2 × 3 = 6 small pieces are kept 2/5 × 3/7 = (2 × 3) / (5 × 7) = 6/35 the denominators multiply because the two rounds of cutting compound

Definition 0.3.6 — Two rounds of cutting compound, so both numerators and both denominators multiply.

Simplify before you multiply

Every product in that table needed simplifying afterward, and the numbers you had to simplify were bigger than the ones you started with. There is a better order to work in: cancel the common factors first, then multiply.

That is allowed because the Equivalent Fractions Property does not care which numerator sits above which denominator. Once the product is written as one fraction, \(\dfrac{a \cdot c}{b \cdot d}\), any factor on top can be cancelled against any factor on the bottom.

Two ways through the same product. Both columns compute \(\dfrac{14}{15} \cdot \dfrac{25}{21}\):

Multiply firstSimplify first
Step 1. \(\dfrac{14 \cdot 25}{15 \cdot 21} = \dfrac{350}{315}\)Step 1. 14 and 21 share 7, becoming 2 and 3.
Step 2. Divide by 5 to get \(\dfrac{70}{63}\).Step 2. 25 and 15 share 5, becoming 5 and 3.
Step 3. Divide by 7 to get \(\dfrac{10}{9}\).Step 3. \(\dfrac{2}{3} \cdot \dfrac{5}{3} = \dfrac{10}{9}\)
Biggest number handled: 350.Biggest number handled: 25.

Same answer, but the largest product the right-hand column ever formed was \(2 \cdot 5 = 10\), against \(14 \cdot 25 = 350\) on the left. The bigger the numbers get, the more that gap matters. It matters most once the fractions contain variables: a variable factor cancels cleanly, while a variable product has to be expanded and then factored all over again.

Example 0.3.3: Cancelling before multiplying

Multiply \(\dfrac{14}{15} \cdot \dfrac{25}{21}\) by simplifying first.

Solution

Step 1 — Write it as one fraction. Every number here is a factor, so any of them can cancel against any other.

$$\frac{14 \cdot 25}{15 \cdot 21}$$

Step 2 — Cancel the 7. The 14 on top and the 21 on the bottom are both multiples of 7, so they become 2 and 3.

$$\frac{2 \cdot 25}{15 \cdot 3}$$

Step 3 — Cancel the 5. The 25 on top and the 15 on the bottom are both multiples of 5, so they become 5 and 3.

$$\frac{2 \cdot 5}{3 \cdot 3}$$

Step 4 — Multiply what is left.

$$\frac{10}{9}$$

Answer: \(\dfrac{10}{9}\).

One caution follows straight from Section 0.3.1: this cancelling is still cancelling factors. It is allowed here only because multiplication is the sole operation in sight. The moment a sum shows up in a numerator, the same-looking move becomes wrong.

Reciprocals

Definition 0.3.7: Reciprocal

The reciprocal of a nonzero number is the number you multiply it by to get 1. For a fraction, swap the numerator and the denominator.

$$\frac{5}{8} \cdot \frac{8}{5} = \frac{40}{40} = 1$$
NumberReciprocalTheir product
\(\dfrac{4}{7}\)\(\dfrac{7}{4}\)1
\(-\dfrac{5}{3}\)\(-\dfrac{3}{5}\)1
\(11\)\(\dfrac{1}{11}\)1
\(-\dfrac{1}{6}\)\(-6\)1
\(0\)none exists

Two facts in that table matter later on. A reciprocal keeps the sign of the original number, since flipping a fraction over does not change whether it is positive or negative, and \(-\frac{5}{3}\) times \(-\frac{3}{5}\) is \(+1\) by the like-signs rule. And zero has no reciprocal at all, because no number times 0 gives 1. That is the same fact as "you cannot divide by zero" wearing a different hat, and it is why every division rule below comes with a nonzero condition attached.

A reciprocal is the number that turns a product into one The fraction five eighths is written on the left. A curved arrow labelled turn it over leads to eight fifths on the right. Below, the two are multiplied together: five eighths times eight fifths gives forty over forty, which is one. Two closing notes record that turning a fraction over does not change its sign, so the reciprocal of negative five thirds is negative three fifths, and that zero has no reciprocal because no number times zero can give one. 5 8 turn it over 8 5 5 8 × 8 5 = 40 40 = 1 turning a fraction over keeps its sign, so the reciprocal of −5/3 is −3/5 0 has no reciprocal — no number times 0 can give 1

Definition 0.3.7 — A reciprocal is the number that turns a product into 1.

Dividing by a fraction

Definition 0.3.8: Fraction division

If \(b\), \(c\), and \(d\) are all nonzero, then

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$$

To divide by a fraction, multiply by its reciprocal instead. Only the second fraction gets flipped — flipping both, or flipping the first one, is a common slip that throws the answer off badly enough that a rough estimate will catch it.

Dividing counts how many of the divisor fit inside, which is what multiplying by the reciprocal does Three whole bars are drawn side by side. Each is then cut into four quarter-pieces, and the twelve pieces are numbered one through twelve to show how many quarters fit inside three wholes. The reading underneath is three divided by one quarter equals twelve, which is the same as three times four. A closing line gives the general rule that a over b divided by c over d equals a over b times d over c, and warns that only the second fraction is turned over. 1 whole 1 whole 1 whole each whole holds 4 quarters 1 2 3 4 5 6 7 8 9 10 11 12 3 ÷ 1/4 = 12, which is 3 × 4 a/b ÷ c/d = a/b × d/c only the SECOND fraction is turned over — flipping both, or the first, is the common slip

Definition 0.3.8 — Dividing counts how many fit inside, which is what the reciprocal does.

Example 0.3.4: Dividing by a fraction

Divide \(\dfrac{3}{4} \div \dfrac{2}{5}\).

Solution

Step 1 — Find the reciprocal of the divisor. The divisor is \(\dfrac{2}{5}\), so its reciprocal is \(\dfrac{5}{2}\).

Step 2 — Change the division to a multiplication.

$$\frac{3}{4} \cdot \frac{5}{2}$$

Step 3 — Check for common factors. The numbers 3 and 5 on top share nothing with 4 and 2 on the bottom, so there is nothing to cancel.

Step 4 — Multiply straight across.

$$\frac{15}{8}$$

Answer: \(\dfrac{15}{8}\).

DivisionRewritten as a multiplicationResult
\(\dfrac{3}{8} \div \dfrac{1}{4}\)\(\dfrac{3}{8} \cdot \dfrac{4}{1}\)\(\dfrac{3}{2}\)
\(-\dfrac{5}{9} \div \dfrac{10}{3}\)\(-\dfrac{5}{9} \cdot \dfrac{3}{10}\)\(-\dfrac{1}{6}\)
\(6 \div \dfrac{2}{5}\)\(\dfrac{6}{1} \cdot \dfrac{5}{2}\)\(15\)
\(\dfrac{7}{12} \div \left(-\dfrac{7}{12}\right)\)\(\dfrac{7}{12} \cdot \left(-\dfrac{12}{7}\right)\)\(-1\)

Why multiplying by the reciprocal works

Memorizing "flip and multiply" will get you through the exercises. Understanding it keeps you from using it in the wrong place, so here are two explanations — one you can picture, one you can prove.

Small pieces fit many times

How many quarters fit inside 3? Twelve, because each whole holds four of them. So \(3 \div \frac{1}{4} = 12\), which is just \(3 \cdot 4\). Dividing by something smaller than 1 gives a bigger answer, and multiplying by its reciprocal is the only move that does the same.

The picture version is about counting. Division asks how many copies of the divisor fit inside the dividend, and when the divisor is small, a lot of them fit. The reciprocal of a small number is a large number, so multiplying by it produces the same growth that dividing by a small piece does.

The proof version is short. Write the division as a fraction, then multiply the top and the bottom by the reciprocal of the divisor. That move is legal — it is the Equivalent Fractions Property, multiplying by a well-chosen form of 1.

$$\frac{a}{b} \div \frac{c}{d} \;=\; \frac{\;\frac{a}{b}\;}{\;\frac{c}{d}\;} \;=\; \frac{\;\frac{a}{b} \cdot \frac{d}{c}\;}{\;\frac{c}{d} \cdot \frac{d}{c}\;} \;=\; \frac{\;\frac{a}{b} \cdot \frac{d}{c}\;}{1} \;=\; \frac{a}{b} \cdot \frac{d}{c}$$

The reciprocal gets chosen for exactly one reason: it is the one multiplier that turns the denominator into 1 so it disappears. Nothing else about it is special. Hold onto that argument, because Section 0.3.4 uses the same move on complex fractions, and it is the reason both topics live in one section.

Try It Now 0.3.2

Multiply \(\dfrac{9}{16} \cdot \dfrac{4}{15}\), cancelling first. Then divide \(-\dfrac{3}{10} \div \dfrac{9}{20}\).

Solution

Part 1 — Multiply \(\dfrac{9}{16} \cdot \dfrac{4}{15}\). Write it as a single fraction and look for common factors.

$$\frac{9 \cdot 4}{16 \cdot 15}$$

The 9 and the 15 share a factor of 3, becoming 3 and 5. The 4 and the 16 share a factor of 4, becoming 1 and 4.

$$\frac{3 \cdot 1}{4 \cdot 5} = \frac{3}{20}$$

Part 2 — Divide \(-\dfrac{3}{10} \div \dfrac{9}{20}\). Multiply by the reciprocal of the second fraction.

$$-\frac{3}{10} \cdot \frac{20}{9}$$

The 3 and the 9 share a factor of 3, and the 20 and the 10 share a factor of 10.

$$-\frac{1}{1} \cdot \frac{2}{3} = -\frac{2}{3}$$

Unlike signs give a negative result, which is why the answer stays negative.

Answer: \(\dfrac{3}{20}\) and \(-\dfrac{2}{3}\).

0.3.3 Adding and Subtracting Fractions

Definition 0.3.9: Fraction addition and subtraction

If \(a\), \(b\), and \(c\) are numbers with \(c \neq 0\), then

$$\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c} \qquad \text{and} \qquad \frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$$

When the denominators already match, add or subtract the numerators and leave the denominator alone:

$$\frac{2}{9} + \frac{5}{9} = \frac{7}{9} \qquad\qquad \frac{11}{12} - \frac{5}{12} = \frac{6}{12} = \frac{1}{2}$$
Fractions are like terms in disguise

\(3x + 5x = 8x\) works because both terms count the same kind of thing. \(\frac{2}{9} + \frac{5}{9} = \frac{7}{9}\) works for exactly the same reason, with "ninths" playing the role of \(x\). Section 0.1 called those like terms.

The denominator does not change because it is not a quantity you are adding. It is the name of the unit. Two ninths plus five ninths is seven ninths for the same reason that 2 feet plus 5 feet is 7 feet and not 7 square feet. You count the objects; you do not count the noun.

Subtraction needs no separate treatment here either. The numerators are integers, so Section 0.2 already covers them, negative results included.

Like denominators add their counts and keep the denominator, because the denominator names the unit Three bars are stacked, each divided into nine equal pieces of the same size. The first has two pieces shaded and is labelled two ninths. The second has five pieces shaded and is labelled five ninths. The third shows the two shadings combined, seven pieces of the same nine, labelled seven ninths. All three bars have exactly nine pieces, so the piece size never changes. A closing line notes that you count the objects and do not count the noun: two ninths plus five ninths is seven ninths for the same reason that two feet plus five feet is seven feet. 2/9 + 5/9 = 7/9 still 9 pieces — the denominator names the unit, so it does not change you count the objects; you do not count the noun — just as 2 ft + 5 ft = 7 ft

Definition 0.3.9 — Like denominators add their counts; the denominator names the unit.

Why you never add the denominators

$$\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$

Adding the denominators instead would give \(\frac{2}{8}\), which is \(\frac{1}{4}\) — the same quarter you started with. That rule says adding a quarter to a quarter changes nothing at all, and two quarters of a pizza is plainly half a pizza. Any rule that returns a sum equal to one of the things being added is broken on its face.

SumCorrectDenominators added, which is wrongHow you can tell
\(\dfrac{1}{4} + \dfrac{1}{4}\)\(\dfrac{1}{2}\)\(\dfrac{2}{8} = \dfrac{1}{4}\)a sum cannot equal one of its addends
\(\dfrac{3}{5} + \dfrac{1}{5}\)\(\dfrac{4}{5}\)\(\dfrac{4}{10} = \dfrac{2}{5}\)a sum cannot be less than \(\dfrac{3}{5}\)
\(\dfrac{1}{2} + \dfrac{1}{2}\)\(1\)\(\dfrac{2}{4} = \dfrac{1}{2}\)two halves make a whole

The confusion is understandable, because multiplication does multiply the denominators. But multiplication is answering a different question — a part of a part — and parts of parts really are smaller. Addition combines two pieces of the same-size whole, and the size of that whole does not change just because you put pieces together.

The least common denominator

Unlike denominators cannot be added until you make them alike. Any common denominator gets the job done, but the smallest one keeps the numbers manageable.

Definition 0.3.10: Least common denominator

The least common denominator, or LCD, is the smallest positive number that every denominator in the problem divides into evenly.

There are two ways to find it. You can go by inspection, running through multiples of the larger denominator until you hit one that the smaller denominator also divides: for 6 and 8, try 8, then 16, then 24, and 24 works because it is \(6 \cdot 4\). Or you can use prime factorization, which keeps working when the numbers get ugly. Factor each denominator into primes, then take each prime the greatest number of times it shows up in any single factorization.

For 12 and 18, note that \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\). Take two 2s, which is the most in either, and two 3s for the same reason, giving \(2^2 \cdot 3^2 = 36\).

DenominatorsPrime factorizationsLCDNote
4 and 6\(2^2\), \(2 \cdot 3\)12not \(4 \cdot 6 = 24\)
12 and 18\(2^2 \cdot 3\), \(2 \cdot 3^2\)36
6 and 18\(2 \cdot 3\), \(2 \cdot 3^2\)18one denominator divides the other
5 and 9\(5\), \(3^2\)45no shared factor, so the LCD is the product
10 and 15\(2 \cdot 5\), \(3 \cdot 5\)30
14 and 21\(2 \cdot 7\), \(3 \cdot 7\)42

Two patterns in that table are worth memorizing. When the denominators share no factor, the LCD is just their product. When one denominator divides the other, the larger one is already the LCD.

The least common denominator is the first number both ladders of multiples land on Two rows of chips are laid out along a shared scale. The upper row holds the multiples of six: six, twelve, eighteen, twenty-four, thirty, thirty-six, forty-two and forty-eight. The lower row holds the multiples of eight: eight, sixteen, twenty-four, thirty-two, forty and forty-eight. The chips at twenty-four in both rows are highlighted and joined by a vertical connector, marking the first value the two rows share. The chips at forty-eight are also marked as shared but noted as not the least. The reading is that the least common denominator of six and eight is twenty-four. multiples of 6 6 12 18 30 36 42 multiples of 8 8 16 32 40 24 24 the first value both rows land on 48 48 also shared LCD of 6 and 8 = 24 48 works too, so any common denominator would do — 24 is just the smallest

Definition 0.3.10 — The LCD is the first value both ladders of multiples land on.

Adding with unlike denominators

The procedure has four steps: find the LCD, build each fraction up to it with the Equivalent Fractions Property, add the numerators, and simplify.

Example 0.3.5: Adding with unlike denominators

Add \(\dfrac{5}{12} + \dfrac{7}{18}\).

Solution

Step 1 — Find the LCD. Since \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\), take two 2s and two 3s.

$$\text{LCD} = 2^2 \cdot 3^2 = 36$$

Step 2 — Build the first fraction up to 36. Since \(12 \cdot 3 = 36\), multiply the top by 3 as well.

$$\frac{5}{12} = \frac{5 \cdot 3}{12 \cdot 3} = \frac{15}{36}$$

Step 3 — Build the second fraction up to 36. Since \(18 \cdot 2 = 36\), multiply the top by 2 as well.

$$\frac{7}{18} = \frac{7 \cdot 2}{18 \cdot 2} = \frac{14}{36}$$

Step 4 — Add the numerators. The denominators match now, so they stay put.

$$\frac{15}{36} + \frac{14}{36} = \frac{29}{36}$$

Step 5 — Simplify. 29 is prime and does not divide 36, so this is already in lowest terms.

Answer: \(\dfrac{29}{36}\).

Whatever you multiply a denominator by, you have to multiply its numerator by as well. Multiplying only the bottom changes the number instead of renaming it, and that mistake hides well — nothing about \(\frac{5}{36}\) looks wrong on the page, it simply is not equal to \(\frac{5}{12}\).

Example 0.3.6: A subtraction that comes out negative

Subtract \(\dfrac{7}{10} - \dfrac{5}{6}\).

Solution

Step 1 — Find the LCD. Since \(10 = 2 \cdot 5\) and \(6 = 2 \cdot 3\), the LCD is \(2 \cdot 3 \cdot 5 = 30\).

Step 2 — Build both fractions up to 30.

$$\frac{7}{10} = \frac{21}{30} \qquad \frac{5}{6} = \frac{25}{30}$$

Step 3 — Subtract the numerators.

$$\frac{21 - 25}{30} = \frac{-4}{30}$$

The numerator subtraction is ordinary integer work from Section 0.2: \(21 - 25 = -4\).

Step 4 — Simplify and move the sign out front. Both 4 and 30 are divisible by 2.

$$\frac{-4}{30} = -\frac{2}{15}$$

Answer: \(-\dfrac{2}{15}\).

You are not actually required to use the least common denominator. Running that same problem over 60 gives \(\frac{42}{60} - \frac{50}{60} = \frac{-8}{60} = -\frac{2}{15}\), which is the same answer reached through bigger numbers and one extra round of simplifying. The LCD is a convenience, not a law.

Signed fractions

Nothing new is required here. Convert each fraction to the common denominator, then let the integer rules from Section 0.2 handle the numerators. Writing negative fractions with the sign up in the numerator during this step keeps the bookkeeping where you can see it.

ProblemOver the LCDNumerator arithmeticResult
\(-\dfrac{3}{8} + \dfrac{5}{6}\)\(\dfrac{-9}{24} + \dfrac{20}{24}\)\(-9 + 20 = 11\)\(\dfrac{11}{24}\)
\(\dfrac{1}{3} - \dfrac{7}{9}\)\(\dfrac{3}{9} - \dfrac{7}{9}\)\(3 - 7 = -4\)\(-\dfrac{4}{9}\)
\(-\dfrac{2}{5} + \left(-\dfrac{1}{2}\right)\)\(\dfrac{-4}{10} + \dfrac{-5}{10}\)\(-4 + (-5) = -9\)\(-\dfrac{9}{10}\)
\(-\dfrac{5}{6} - \left(-\dfrac{1}{4}\right)\)\(\dfrac{-10}{12} - \dfrac{-3}{12}\)\(-10 + 3 = -7\)\(-\dfrac{7}{12}\)

That last row uses the rule from Section 0.2 that subtracting a negative is the same as adding, so \(-10 - (-3)\) becomes \(-10 + 3 = -7\). Rewriting the subtraction as an addition before you touch the numbers costs one extra line and prevents most of the sign errors that happen here.

Knowing which procedure you need

The two procedures are so different that the most valuable habit is checking which one the problem is asking for before you write anything down.

QuestionMultiplyingAdding
Do the denominators have to match?noyes
What happens to the denominatorsmultiply themkeep the common one
What happens to the numeratorsmultiply themadd them
Can you cancel across the two fractions?yes, before multiplyingno, never
Rough size of the answersmaller than both, if both are positive and properbigger than both, if both are positive

That fourth row is the one to watch. Cancelling across an addition is the fraction version of cancelling a term, and it is wrong for exactly the reason Section 0.3.1 gave.

Try It Now 0.3.3

Add \(\dfrac{7}{12} + \dfrac{5}{8}\). Then subtract \(-\dfrac{1}{6} - \dfrac{3}{4}\).

Solution

Part 1 — Add \(\dfrac{7}{12} + \dfrac{5}{8}\). Since \(12 = 2^2 \cdot 3\) and \(8 = 2^3\), take three 2s and one 3, so the LCD is 24.

$$\frac{7}{12} = \frac{14}{24} \qquad \frac{5}{8} = \frac{15}{24}$$

Add the numerators and keep the denominator:

$$\frac{14 + 15}{24} = \frac{29}{24}$$

Since 29 is prime, this is in lowest terms.

Part 2 — Subtract \(-\dfrac{1}{6} - \dfrac{3}{4}\). The LCD of 6 and 4 is 12.

$$\frac{-2}{12} - \frac{9}{12}$$

Now the numerators are ordinary integers, and subtracting 9 from \(-2\) moves further negative:

$$\frac{-2 - 9}{12} = \frac{-11}{12} = -\frac{11}{12}$$

Answer: \(\dfrac{29}{24}\) and \(-\dfrac{11}{12}\).

0.3.4 Complex Fractions and Order of Operations

Section 0.1 listed the fraction bar among the grouping symbols, right alongside parentheses and brackets. That was not a technicality. It changes the order you have to work in.

The fraction bar as a grouping symbol.

Simplify the entire numerator. Simplify the entire denominator. Then divide.

Treat the numerator and the denominator as though each one were wrapped in invisible parentheses. Written on one line with a division sign, such an expression would need those parentheses spelled out: \((5 + 3 \cdot 4) \div (2^3 - 6)\). The bar supplies the grouping for free, which is exactly why algebra prefers it to \(\div\).

Example 0.3.7: The bar groups the top and the bottom

Simplify \(\dfrac{5 + 3(4)}{2^3 - 6}\).

Solution

Step 1 — Simplify the numerator by itself. Multiplication comes before addition.

$$5 + 3(4) = 5 + 12 = 17$$

Step 2 — Simplify the denominator by itself. The exponent comes before the subtraction.

$$2^3 - 6 = 8 - 6 = 2$$

Step 3 — Now divide.

$$\frac{17}{2}$$

Since 17 and 2 share no factor, this is in lowest terms.

Answer: \(\dfrac{17}{2}\).

Here is a second one with signs in it. In \(\dfrac{6 - 10}{-2 - 2}\) the numerator works out to \(-4\) and the denominator also works out to \(-4\), so the value is \(\dfrac{-4}{-4} = 1\) by the like-signs rule from Section 0.2.

Complex fractions

Definition 0.3.11: Complex fraction

A complex fraction is a fraction whose numerator, denominator, or both are themselves fractions.

$$\frac{\;\dfrac{3}{4}\;}{\;\dfrac{9}{10}\;} \qquad\qquad \frac{\;\dfrac{2}{3} + \dfrac{1}{4}\;}{\;\dfrac{5}{6}\;} \qquad\qquad \frac{\;\dfrac{7}{8}\;}{\;4\;}$$

These look intimidating and are not. That main bar is a division sign, so a complex fraction is just a division problem written vertically, and Section 0.3.2 already told you what to do with a division problem.

The main bar of a complex fraction is the division that happens last A complex fraction is drawn with three quarters above a long heavy horizontal bar and nine tenths below it. The two short bars inside the small fractions are drawn thin, while the bar between them is drawn thick and in the accent colour. Labels identify the thin bars as the fractions themselves and the thick bar as the division that happens last. The whole thing is then rewritten on one line as three quarters divided by nine tenths, and a closing note says that a complex fraction is a division problem written vertically, so the reciprocal rule from earlier in the section applies unchanged. 3 4 9 10 a thin bar: this is a fraction a thin bar: this is a fraction the MAIN bar: the division that happens last so finish the top, finish the bottom, then divide = 3/4 ÷ 9/10 = 3/4 × 10/9 a complex fraction is a division written vertically, so multiply by the reciprocal

Definition 0.3.11 — The main bar of a complex fraction is the division that happens last.

Example 0.3.8: Simplifying a complex fraction

Simplify \(\dfrac{\;\frac{3}{4}\;}{\;\frac{9}{10}\;}\).

Solution

Step 1 — Read the main bar as a division.

$$\frac{3}{4} \div \frac{9}{10}$$

Step 2 — Multiply by the reciprocal of the divisor.

$$\frac{3}{4} \cdot \frac{10}{9}$$

Step 3 — Cancel before multiplying. The 3 and the 9 share a factor of 3, and the 10 and the 4 share a factor of 2.

$$\frac{1}{2} \cdot \frac{5}{3}$$

Step 4 — Multiply straight across.

$$\frac{5}{6}$$

Answer: \(\dfrac{5}{6}\).

When the numerator or the denominator has an operation inside it, the grouping rule comes first: finish the top, finish the bottom, and only then divide. Do not try to combine an addition in the numerator with the division in one move — the addition happens inside the grouping, and the division happens after. Skipping that order is the usual failure here, and it produces an answer that looks plausible but is not.

Example 0.3.9: A complex fraction with a sum on top

Simplify \(\dfrac{\;\frac{2}{3} + \frac{1}{4}\;}{\;\frac{5}{6}\;}\).

Solution

Step 1 — Simplify the numerator first. The LCD of 3 and 4 is 12.

$$\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$$

Step 2 — Rewrite the complex fraction. The top is now a single fraction.

$$\frac{\;\frac{11}{12}\;}{\;\frac{5}{6}\;}$$

Step 3 — Multiply by the reciprocal.

$$\frac{11}{12} \cdot \frac{6}{5}$$

Step 4 — Cancel and multiply. The 6 and the 12 share a factor of 6.

$$\frac{11}{2} \cdot \frac{1}{5} = \frac{11}{10}$$

Answer: \(\dfrac{11}{10}\), which is \(1\dfrac{1}{10}\) as a mixed number.

Complex fractionRewrittenSimplified
\(\dfrac{\;\frac{7}{8}\;}{\;4\;}\)\(\dfrac{7}{8} \cdot \dfrac{1}{4}\)\(\dfrac{7}{32}\)
\(\dfrac{\;6\;}{\;\frac{3}{5}\;}\)\(6 \cdot \dfrac{5}{3}\)\(10\)
\(\dfrac{\;-\frac{2}{9}\;}{\;\frac{4}{3}\;}\)\(-\dfrac{2}{9} \cdot \dfrac{3}{4}\)\(-\dfrac{1}{6}\)
\(\dfrac{\;\frac{1}{2} - \frac{1}{6}\;}{\;\frac{2}{3}\;}\)\(\dfrac{1}{3} \cdot \dfrac{3}{2}\)\(\dfrac{1}{2}\)

The second row makes a point about size worth repeating. Dividing 6 by a number smaller than 1 gives an answer bigger than 6. If you divide a positive number by a positive proper fraction and your answer comes out smaller than what you started with, you flipped the wrong fraction. (The size check needs both numbers positive — with negatives in play, compare magnitudes instead.)

Order of operations with fractions

The four-level order from Section 0.1 does not change at all. Fractions do not get their own rules; they just make each step take more care, the same way Section 0.2 said about negative numbers.

One fact deserves its own line before the example. An exponent on a fraction applies to the numerator and the denominator both, since \(\left(\frac{2}{5}\right)^2\) means \(\frac{2}{5} \cdot \frac{2}{5}\), which multiplies straight across to \(\frac{4}{25}\). The placement warnings from Section 0.1 still apply, so \(\left(\frac{2}{5}\right)^2\) and \(\frac{2^2}{5}\) are different numbers.

Example 0.3.10: Order of operations with fractions

Simplify \(\;\dfrac{3}{4} + \dfrac{1}{2}\left(\dfrac{2}{5}\right)^2\).

Solution

Step 1 — Exponents first. Square the numerator and the denominator both.

$$\left(\frac{2}{5}\right)^2 = \frac{4}{25}$$

Step 2 — Then multiplication. Cancel the 2 into the 4 before multiplying.

$$\frac{1}{2} \cdot \frac{4}{25} = \frac{2}{25}$$

Step 3 — Addition last, so find the LCD. Since \(4 = 2^2\) and \(25 = 5^2\) share no factor, the LCD is \(4 \cdot 25 = 100\).

$$\frac{3}{4} = \frac{75}{100} \qquad \frac{2}{25} = \frac{8}{100}$$

Step 4 — Add the numerators.

$$\frac{75 + 8}{100} = \frac{83}{100}$$

Answer: \(\dfrac{83}{100}\).

Evaluating expressions containing fractions

To evaluate an expression, substitute the given value and simplify, wrapping the substituted value in parentheses exactly as Sections 0.1 and 0.2 insisted. Those parentheses are not decoration. Without them, \(2x\) with \(x = -\frac{1}{2}\) invites the reading \(2 - \frac{1}{2}\), and the whole problem goes wrong on the first line.

Example 0.3.11: Evaluating at a negative value

Evaluate \(\;\dfrac{2x}{x+5}\;\) when \(x = -\dfrac{1}{2}\).

Solution

Step 1 — Substitute, with parentheses around the value.

$$\frac{2\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right) + 5}$$

Step 2 — Simplify the numerator. Unlike signs give a negative product.

$$2 \cdot \left(-\frac{1}{2}\right) = -1$$

Step 3 — Simplify the denominator. Write 5 as \(\frac{10}{2}\) so the denominators match.

$$-\frac{1}{2} + \frac{10}{2} = \frac{9}{2}$$

Step 4 — Divide. This is now a complex fraction, so multiply by the reciprocal.

$$-1 \cdot \frac{2}{9} = -\frac{2}{9}$$

Answer: \(-\dfrac{2}{9}\).

A zero denominator has no value

The expression \(\dfrac{2x}{x+5}\) means nothing when \(x = -5\), because the denominator becomes zero. Every fractional expression carries a hidden restriction like this one, and finding it is the first move in almost every Chapter 3 problem.

A value that makes a denominator zero is not merely awkward to work with. It is not in the expression's domain at all, so the expression has no value there — the same conclusion Section 0.2 reached about \(\frac{5}{0}\).

Fractions in a real setting

A crew is repainting a hallway. Materials for the whole job cost $48, and the crew has finished \(\frac{5}{8}\) of the hallway. To find what they have used so far, take \(\frac{5}{8}\) of 48, cancelling the 8 into the 48 before multiplying:

$$\frac{5}{8} \cdot 48 = \frac{5}{8} \cdot \frac{48}{1} = 5 \cdot 6 = 30$$

So they have used $30 of materials, and the remaining \(\frac{3}{8}\) of the job accounts for the other $18. Notice the check built into that: \(\frac{5}{8} + \frac{3}{8} = \frac{8}{8} = 1\), so the two pieces add to one whole job.

Now flip the question. Suppose you know that $30 of materials covered \(\frac{5}{8}\) of the hallway and you want the cost of the whole job. That is a division:

$$30 \div \frac{5}{8} = 30 \cdot \frac{8}{5} = 48$$

Same two numbers, opposite operation, because the unknown moved from the part to the whole. Deciding which of those two setups a situation calls for is the real skill, and Section 0.5 gives you the properties that let you rearrange relationships like this one on purpose instead of by guessing.

Try It Now 0.3.4

Simplify \(\dfrac{\;\frac{3}{4} + \frac{1}{2}\;}{\;\frac{5}{8}\;}\). Then evaluate \(\;\dfrac{4x}{x+3}\;\) when \(x = -\dfrac{1}{2}\).

Solution

Part 1 — Simplify the complex fraction. Start with the numerator, where the LCD of 4 and 2 is 4.

$$\frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$$

Now the main bar is a division, so multiply by the reciprocal.

$$\frac{5}{4} \div \frac{5}{8} = \frac{5}{4} \cdot \frac{8}{5}$$

Cancel the 5s, and cancel the 4 into the 8.

$$1 \cdot 2 = 2$$

Part 2 — Evaluate \(\dfrac{4x}{x+3}\) at \(x = -\dfrac{1}{2}\). Substitute with parentheses.

$$\frac{4\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right) + 3}$$

The numerator is \(4 \cdot \left(-\frac{1}{2}\right) = -2\). The denominator is \(-\frac{1}{2} + \frac{6}{2} = \frac{5}{2}\).

$$\frac{-2}{\;\frac{5}{2}\;} = -2 \cdot \frac{2}{5} = -\frac{4}{5}$$

Answer: \(2\) and \(-\dfrac{4}{5}\).

Problem Set 0.3

Problem 1. Simplify \(\dfrac{24}{36}\).

Solution

Step 1 — Break both numbers into prime factors: Prime-factoring makes the greatest common factor (GCF) easy to spot. $$24 = 2^3 \cdot 3, \qquad 36 = 2^2 \cdot 3^2$$ The shared factors are \(2^2\) and \(3^1\), so \(\text{GCF}(24,36) = 2^2\cdot 3 = 12\).

Step 2 — Apply the Equivalent Fractions Property: Dividing numerator and denominator by the same nonzero number (here, the GCF) produces an equivalent fraction in lowest terms — this is cancelling a common factor of the whole numerator and whole denominator, not cancelling a term. $$\frac{24}{36} = \frac{24 \div 12}{36 \div 12} = \frac{2}{3}$$

Answer: \(\dfrac{2}{3}\)

Problem 2. Simplify \(\dfrac{45}{60}\).

Solution

Step 1 — Find the greatest common factor of 45 and 60: \(45 = 3^2\cdot 5\) and \(60 = 2^2\cdot 3\cdot 5\), so the common factors are \(3\) and \(5\), giving \(\text{GCF}(45,60) = 3\cdot 5 = 15\).

Step 2 — Divide both parts by the GCF: By the Equivalent Fractions Property, dividing top and bottom by the same factor keeps the value unchanged while reaching lowest terms. $$\frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4}$$

Answer: \(\dfrac{3}{4}\)

Problem 3. Simplify \(\dfrac{-28}{63}\) and write the sign in front.

Solution

Step 1 — Simplify the magnitude first: Ignore the sign for a moment and reduce \(\dfrac{28}{63}\). Since \(28 = 2^2\cdot 7\) and \(63 = 3^2\cdot 7\), the only shared factor is \(7\), so \(\text{GCF}(28,63) = 7\).

Step 2 — Divide numerator and denominator by the GCF: Cancelling the common factor of \(7\) (a factor of both entire numbers, not just a piece of one) gives $$\frac{28}{63} = \frac{28 \div 7}{63 \div 7} = \frac{4}{9}$$

Step 3 — Move the sign out front: A fraction's negative sign belongs out in front of the fraction bar, not buried in the numerator or denominator. $$\frac{-28}{63} = -\frac{4}{9}$$

Answer: \(-\dfrac{4}{9}\)

Problem 4. Write three fractions equivalent to \(\dfrac{3}{7}\).

Solution

Step 1 — Recall the Equivalent Fractions Property: Multiplying (or dividing) the numerator and denominator of a fraction by the same nonzero number produces a new fraction with the same value.

Step 2 — Pick three different multipliers and apply them to \(\dfrac{3}{7}\): $$\frac{3}{7} = \frac{3 \cdot 2}{7 \cdot 2} = \frac{6}{14}, \qquad \frac{3}{7} = \frac{3 \cdot 3}{7 \cdot 3} = \frac{9}{21}, \qquad \frac{3}{7} = \frac{3 \cdot 4}{7 \cdot 4} = \frac{12}{28}$$

Answer: \(\dfrac{6}{14}\), \(\dfrac{9}{21}\), and \(\dfrac{12}{28}\) are all equivalent to \(\dfrac{3}{7}\) (many other correct choices exist).

Problem 5. Is \(\dfrac{9}{15}\) equal to \(\dfrac{3}{5}\)? Explain how you know.

Solution

Step 1 — Reduce \(\dfrac{9}{15}\) to lowest terms: Find the GCF of \(9\) and \(15\): both share a factor of \(3\), so \(\text{GCF}(9,15) = 3\). $$\frac{9}{15} = \frac{9 \div 3}{15 \div 3} = \frac{3}{5}$$

Step 2 — Compare the reduced form to the target fraction: Since dividing numerator and denominator of \(\dfrac{9}{15}\) by their GCF lands exactly on \(\dfrac{3}{5}\), the two fractions are the same number by the Equivalent Fractions Property, just written with different-sized pieces.

Answer: Yes — \(\dfrac{9}{15}\) equals \(\dfrac{3}{5}\), because dividing \(9\) and \(15\) by their greatest common factor, \(3\), reduces \(\dfrac{9}{15}\) exactly to \(\dfrac{3}{5}\).

Problem 6. Write \(\dfrac{29}{6}\) as a mixed number.

Solution

Step 1 — Divide the numerator by the denominator: \(29 \div 6\) gives a whole-number quotient and a remainder: since \(6 \cdot 4 = 24\) and \(29 - 24 = 5\), the quotient is \(4\) with remainder \(5\).

Step 2 — Write the mixed number: The quotient becomes the whole-number part, and the remainder is placed over the original denominator. $$\frac{29}{6} = 4\frac{5}{6}$$

Answer: \(4\dfrac{5}{6}\)

Problem 7. Write \(7\frac{1}{3}\) as an improper fraction.

Solution

Step 1 — Multiply the whole number by the denominator: For \(7\frac{1}{3}\), multiply the whole part \(7\) by the denominator \(3\): \(7 \cdot 3 = 21\).

Step 2 — Add the numerator, keeping the same denominator: Add the original numerator \(1\) to get the new numerator, and keep the denominator \(3\) unchanged. $$7\frac{1}{3} = \frac{7 \cdot 3 + 1}{3} = \frac{22}{3}$$

Answer: \(\dfrac{22}{3}\)

Problem 8. Explain in one sentence why \(\dfrac{5+2}{5+9}\) is not equal to \(\dfrac{2}{9}\).

Solution

Step 1 — Actually simplify \(\dfrac{5+2}{5+9}\): Add inside the numerator and denominator first: \(5+2 = 7\) and \(5+9=14\), so \(\dfrac{5+2}{5+9} = \dfrac{7}{14}\). Since \(\text{GCF}(7,14)=7\), this reduces to \(\dfrac{7}{14} = \dfrac{1}{2}\), not \(\dfrac{2}{9}\).

Step 2 — Identify why cancelling the 5s is invalid: The Equivalent Fractions Property only allows dividing out a factor that multiplies the entire numerator and the entire denominator. Here the \(5\)s are terms being added, not factors being multiplied, so they cannot be cancelled — you cancel factors, not terms.

Answer: \(\dfrac{5+2}{5+9}\) is not \(\dfrac{2}{9}\) because the \(5\)s are added terms rather than common factors of the whole numerator and denominator, so they cannot be cancelled; the fraction actually simplifies to \(\dfrac{1}{2}\).

Problem 9. Rewrite \(\dfrac{4}{-11}\) with the sign in front.

Solution

Step 1 — Locate the negative sign: In \(\dfrac{4}{-11}\), the negative sign sits on the denominator rather than out in front of the fraction.

Step 2 — Move the sign in front: A fraction with exactly one negative part (numerator or denominator) is negative overall, and convention places that sign in front of the fraction bar rather than inside it. $$\frac{4}{-11} = -\frac{4}{11}$$

Answer: \(-\dfrac{4}{11}\)

Problem 10. Multiply: \(\dfrac{3}{8} \cdot \dfrac{2}{9}\).

Solution

Step 1 — Look for common factors to cancel before multiplying: In \(\dfrac{3}{8} \cdot \dfrac{2}{9}\), compare each numerator with the other fraction's denominator. The \(3\) and the \(9\) share a factor of \(3\): the \(3\) becomes \(1\) and the \(9\) becomes \(3\). The \(2\) and the \(8\) share a factor of \(2\): the \(2\) becomes \(1\) and the \(8\) becomes \(4\). Cancelling first keeps the numbers small.

$$\frac{3}{8} \cdot \frac{2}{9} = \frac{1}{4} \cdot \frac{1}{3}$$

Step 2 — Multiply straight across: With common factors already removed, multiply numerator by numerator and denominator by denominator.

$$\frac{1}{4} \cdot \frac{1}{3} = \frac{1 \cdot 1}{4 \cdot 3} = \frac{1}{12}$$

Answer: \(\dfrac{1}{12}\)

Problem 11. Multiply: \(-\dfrac{5}{12} \cdot \dfrac{8}{15}\).

Solution

Step 1 — Determine the sign of the product: One factor, \(-\dfrac{5}{12}\), is negative and the other, \(\dfrac{8}{15}\), is positive. Unlike signs give a negative product, so the answer will be negative once the fraction part is found.

Step 2 — Cancel common factors first: Working with the fraction part \(\dfrac{5}{12} \cdot \dfrac{8}{15}\), compare each numerator to the other denominator. The \(5\) and the \(15\) share a factor of \(5\): the \(5\) becomes \(1\) and the \(15\) becomes \(3\). The \(8\) and the \(12\) share a factor of \(4\): the \(8\) becomes \(2\) and the \(12\) becomes \(3\).

$$\frac{5}{12} \cdot \frac{8}{15} = \frac{1}{3} \cdot \frac{2}{3}$$

Step 3 — Multiply straight across and attach the sign:

$$\frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}, \qquad \text{so} \qquad -\frac{5}{12} \cdot \frac{8}{15} = -\frac{2}{9}$$

Answer: \(-\dfrac{2}{9}\)

Problem 12. Multiply: \(\dfrac{7}{10} \cdot 25\).

Solution

Step 1 — Write the whole number as a fraction: Every whole number \(n\) can be written as \(\dfrac{n}{1}\), so \(25 = \dfrac{25}{1}\). This lets us treat the problem as a fraction times a fraction.

$$\frac{7}{10} \cdot 25 = \frac{7}{10} \cdot \frac{25}{1}$$

Step 2 — Cancel common factors before multiplying: The \(25\) on top and the \(10\) on the bottom share a factor of \(5\): the \(25\) becomes \(5\) and the \(10\) becomes \(2\).

$$\frac{7}{10} \cdot \frac{25}{1} = \frac{7}{2} \cdot \frac{5}{1}$$

Step 3 — Multiply straight across:

$$\frac{7}{2} \cdot \frac{5}{1} = \frac{35}{2}$$

Step 4 — Write it as a mixed number as well. Since \(35 \div 2 = 17\) with remainder \(1\), the improper fraction is \(17\dfrac{1}{2}\).

Answer: \(\dfrac{35}{2}\), which is \(17\dfrac{1}{2}\)

Problem 13. Find the reciprocal of \(-\dfrac{8}{3}\).

Solution

Step 1 — Recall what a reciprocal is: The reciprocal of a number is what you multiply it by to get \(1\). For a fraction, that means swapping the numerator and the denominator.

Step 2 — Flip the fraction, keeping the sign: Flipping does not change whether a number is positive or negative, so the reciprocal of \(-\dfrac{8}{3}\) is \(-\dfrac{3}{8}\).

Step 3 — Check the result: Multiplying the original number by this candidate should give \(1\), and like signs give a positive product.

$$-\frac{8}{3} \cdot \left(-\frac{3}{8}\right) = \frac{8 \cdot 3}{3 \cdot 8} = \frac{24}{24} = 1$$

Answer: \(-\dfrac{3}{8}\)

Problem 14. Explain why 0 has no reciprocal.

Solution

Step 1 — Recall the definition of a reciprocal: The reciprocal of a number \(n\) is the number you multiply \(n\) by to get a product of \(1\).

Step 2 — Test whether such a number exists for \(0\): Multiplying \(0\) by any number always gives \(0\), since \(0 \cdot x = 0\) for every \(x\). There is no value of \(x\) that makes \(0 \cdot x = 1\) true.

Step 3 — Connect this to division: This is the same fact as "you cannot divide by zero" wearing a different hat, which is why every division rule in this section carries a nonzero condition.

Answer: \(0\) has no reciprocal because no number multiplied by \(0\) can ever produce \(1\).

Problem 15. Divide: \(\dfrac{5}{6} \div \dfrac{10}{9}\).

Solution

Step 1 — Rewrite the division as multiplication by the reciprocal: To divide by a fraction, multiply by the reciprocal of the second fraction only. The reciprocal of \(\dfrac{10}{9}\) is \(\dfrac{9}{10}\).

$$\frac{5}{6} \div \frac{10}{9} = \frac{5}{6} \cdot \frac{9}{10}$$

Step 2 — Cancel common factors before multiplying: The \(5\) on top and the \(10\) on the bottom share a factor of \(5\), becoming \(1\) and \(2\). The \(9\) on top and the \(6\) on the bottom share a factor of \(3\), becoming \(3\) and \(2\).

$$\frac{5}{6} \cdot \frac{9}{10} = \frac{1}{2} \cdot \frac{3}{2}$$

Step 3 — Multiply straight across:

$$\frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$$

Answer: \(\dfrac{3}{4}\)

Problem 16. Divide: \(9 \div \dfrac{3}{4}\).

Solution

Step 1 — Write the whole number over 1: Since any whole number \(n\) can be written as \(\dfrac{n}{1}\), the dividend \(9\) becomes \(\dfrac{9}{1}\).

Step 2 — Multiply by the reciprocal of the divisor: The reciprocal of \(\dfrac{3}{4}\) is \(\dfrac{4}{3}\).

$$9 \div \frac{3}{4} = \frac{9}{1} \cdot \frac{4}{3}$$

Step 3 — Cancel before multiplying: The \(9\) on top and the \(3\) on the bottom share a factor of \(3\), becoming \(3\) and \(1\).

$$\frac{3}{1} \cdot \frac{4}{1} = \frac{12}{1} = 12$$

Step 4 — Sanity-check the size. Dividing by a positive number smaller than \(1\) must give an answer bigger than \(9\), and \(12\) is.

Answer: \(12\)

Problem 17. Add: \(\dfrac{4}{15} + \dfrac{7}{15}\).

Solution

Step 1 — Check the denominators: Both fractions, \(\dfrac{4}{15}\) and \(\dfrac{7}{15}\), already have the same denominator, \(15\), so no building up is needed.

Step 2 — Add the numerators and keep the denominator: The denominator names the size of the pieces, so it stays put while the counts combine.

$$\frac{4}{15} + \frac{7}{15} = \frac{4+7}{15} = \frac{11}{15}$$

Step 3 — Check for simplification: \(11\) is prime and does not divide \(15\), so the fraction is already in lowest terms.

Answer: \(\dfrac{11}{15}\)

Problem 18. Find the least common denominator of 8 and 20.

Solution

Step 1 — Find the prime factorization of each number:

$$8 = 2 \cdot 2 \cdot 2 = 2^3 \qquad\qquad 20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$$

Step 2 — Take each prime the greatest number of times it appears in any one factorization: The prime \(2\) appears at most three times (in \(8 = 2^3\)); the prime \(5\) appears at most once (in \(20 = 2^2 \cdot 5\)).

Step 3 — Multiply those prime powers together:

$$\text{LCD} = 2^3 \cdot 5 = 8 \cdot 5 = 40$$

Answer: The least common denominator of \(8\) and \(20\) is \(40\).

Problem 19. Add: \(\dfrac{3}{8} + \dfrac{5}{12}\).

Solution

Step 1 — Break each denominator into primes: \(8 = 2^3\) and \(12 = 2^2 \cdot 3\). Taking each prime the greatest number of times it appears gives an LCD of \(2^3 \cdot 3 = 24\).

Step 2 — Build each fraction up to the LCD using the Equivalent Fractions Property: $$\frac{3}{8} = \frac{3 \cdot 3}{8 \cdot 3} = \frac{9}{24}, \qquad \frac{5}{12} = \frac{5 \cdot 2}{12 \cdot 2} = \frac{10}{24}$$ Multiplying top and bottom by the same number renames the fraction without changing its value.

Step 3 — Add the numerators over the shared denominator: $$\frac{9}{24} + \frac{10}{24} = \frac{19}{24}$$

Step 4 — Simplify. \(19\) is prime and does not divide \(24\), so this is in lowest terms.

Answer: \(\dfrac{19}{24}\)

Problem 20. Subtract: \(\dfrac{5}{9} - \dfrac{5}{6}\).

Solution

Step 1 — Find the LCD by prime factorization: \(9 = 3^2\) and \(6 = 2 \cdot 3\). Taking the highest power of each prime gives \(\text{LCD} = 2 \cdot 3^2 = 18\).

Step 2 — Rewrite each fraction with denominator 18: $$\frac{5}{9} = \frac{5 \cdot 2}{9 \cdot 2} = \frac{10}{18}, \qquad \frac{5}{6} = \frac{5 \cdot 3}{6 \cdot 3} = \frac{15}{18}$$

Step 3 — Subtract the numerators as ordinary signed integers: $$\frac{10}{18} - \frac{15}{18} = \frac{10 - 15}{18} = \frac{-5}{18}$$

Step 4 — Move the sign out front. \(5\) and \(18\) share no common factor, so nothing reduces.

$$\frac{-5}{18} = -\frac{5}{18}$$

Answer: \(-\dfrac{5}{18}\)

Problem 21. Add: \(-\dfrac{7}{10} + \dfrac{2}{5}\).

Solution

Step 1 — Locate the LCD: Since \(10 = 2 \cdot 5\) and \(5\) is already a factor of \(10\), the larger denominator is itself the LCD, so \(\text{LCD} = 10\).

Step 2 — Build the second fraction up to a denominator of 10: $$\frac{2}{5} = \frac{2 \cdot 2}{5 \cdot 2} = \frac{4}{10}$$ The first term, \(-\dfrac{7}{10}\), already has the LCD, so it needs no change.

Step 3 — Combine the numerators as signed integers over the common denominator: $$-\frac{7}{10} + \frac{4}{10} = \frac{-7 + 4}{10} = \frac{-3}{10} = -\frac{3}{10}$$

Answer: \(-\dfrac{3}{10}\)

Problem 22. Explain in one sentence why \(\dfrac{1}{3} + \dfrac{1}{3}\) is not \(\dfrac{2}{6}\).

Solution

Step 1 — Recall what the denominator is doing: the denominator names the size of the pieces being counted, and only the numerators — the counts — combine when you add.

Step 2 — Apply that to \(\dfrac{1}{3} + \dfrac{1}{3}\): both terms already share the denominator \(3\), so the numerators add directly: \(1 + 1 = 2\), giving \(\dfrac{2}{3}\).

Step 3 — See why \(\dfrac{2}{6}\) has to be wrong: \(\dfrac{2}{6}\) comes from also adding the denominators, and it simplifies back to \(\dfrac{1}{3}\) — a sum that equals one of the things being added, which cannot happen.

Answer: Adding fractions combines only the numerators once the denominators match, so \(\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\), not \(\dfrac{2}{6}\), which is just \(\dfrac{1}{3}\) again.

Problem 23. Simplify the complex fraction \(\dfrac{\;\frac{2}{5}\;}{\;\frac{8}{15}\;}\).

Solution

Step 1 — Read the main bar as a division: the long bar separating the two small fractions is a division symbol, so the complex fraction means \(\dfrac{2}{5} \div \dfrac{8}{15}\).

Step 2 — Multiply by the reciprocal of the divisor: $$\frac{2}{5} \div \frac{8}{15} = \frac{2}{5} \cdot \frac{15}{8}$$

Step 3 — Cancel before multiplying: the \(15\) and the \(5\) share a factor of \(5\), becoming \(3\) and \(1\); the \(2\) and the \(8\) share a factor of \(2\), becoming \(1\) and \(4\).

$$\frac{1}{1} \cdot \frac{3}{4} = \frac{3}{4}$$

Answer: \(\dfrac{3}{4}\)

Problem 24. Simplify \(\dfrac{7 - 3(4)}{2^2 + 1}\).

Solution

Step 1 — Treat the fraction bar as a grouping symbol: the entire numerator and the entire denominator each get simplified on their own, as if wrapped in invisible parentheses, before any dividing happens.

Step 2 — Simplify the numerator, multiplication before subtraction: $$7 - 3(4) = 7 - 12 = -5$$

Step 3 — Simplify the denominator, exponent before addition: $$2^2 + 1 = 4 + 1 = 5$$

Step 4 — Now divide. Unlike signs give a negative quotient. $$\frac{-5}{5} = -1$$

Answer: \(-1\)

Problem 25. Simplify \(\;\dfrac{2}{3} + \dfrac{1}{4}\left(\dfrac{2}{3}\right)^2\).

Solution

Step 1 — Exponents first, applied to numerator and denominator both: $$\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$$

Step 2 — Then the multiplication, cancelling the 4 against the 4 first: $$\frac{1}{4} \cdot \frac{4}{9} = \frac{1}{9}$$

Step 3 — Addition last, so find the LCD. Since \(9 = 3^2\) and \(3\) divides \(9\), the LCD is \(9\); build the first fraction up with the Equivalent Fractions Property: \(\dfrac{2}{3} = \dfrac{6}{9}\).

Step 4 — Add the numerators over the common denominator: $$\frac{6}{9} + \frac{1}{9} = \frac{7}{9}$$

Answer: \(\dfrac{7}{9}\)

Problem 26. Evaluate \(\;\dfrac{3x}{x+2}\;\) when \(x = -\dfrac{1}{3}\).

Solution

Step 1 — Substitute \(x = -\dfrac{1}{3}\), wrapped in parentheses, everywhere \(x\) appears: $$\frac{3x}{x+2} = \frac{3\left(-\frac{1}{3}\right)}{\left(-\frac{1}{3}\right)+2}$$ The parentheses keep the negative sign attached to the value as it is carried through.

Step 2 — Simplify the numerator. Unlike signs give a negative product. $$3\left(-\frac{1}{3}\right) = -1$$

Step 3 — Simplify the denominator by writing 2 over the same denominator: $$-\frac{1}{3} + \frac{6}{3} = \frac{5}{3}$$

Step 4 — The bar is a division, so multiply by the reciprocal: $$-1 \cdot \frac{3}{5} = -\frac{3}{5}$$

Answer: \(-\dfrac{3}{5}\)

Problem 27. Evaluate \(\;\dfrac{1}{2}n + \dfrac{3}{4}\;\) when \(n = -\dfrac{1}{2}\).

Solution

Step 1 — Substitute \(n = -\dfrac{1}{2}\), wrapped in parentheses: $$\frac{1}{2}n + \frac{3}{4} = \frac{1}{2}\left(-\frac{1}{2}\right) + \frac{3}{4}$$

Step 2 — Multiplication before addition. Unlike signs give a negative product. $$\frac{1}{2}\left(-\frac{1}{2}\right) = -\frac{1}{4}$$

Step 3 — Add. The two fractions already share the denominator \(4\), so combine the numerators as signed integers. $$-\frac{1}{4} + \frac{3}{4} = \frac{-1+3}{4} = \frac{2}{4}$$

Step 4 — Simplify. Both \(2\) and \(4\) are divisible by \(2\). $$\frac{2}{4} = \frac{1}{2}$$

Answer: \(\dfrac{1}{2}\)

Key Terms

fraction — a number written \(\dfrac{a}{b}\) with \(b \neq 0\), naming \(a\) parts of a whole that was divided into \(b\) equal parts.

numerator — the top number of a fraction, counting how many parts are being taken.

denominator — the bottom number of a fraction, naming how many equal parts the whole was divided into.

equivalent fractions — two fractions that name the same amount, such as \(\dfrac{3}{8}\) and \(\dfrac{6}{16}\).

lowest terms — the form of a fraction whose numerator and denominator share no common factor but 1.

proper fraction — a fraction whose numerator is smaller than its denominator, so its value is less than 1.

improper fraction — a fraction whose numerator is greater than or equal to its denominator, so its value is 1 or more.

mixed number — a whole number written beside a proper fraction, meaning their sum.

reciprocal — the number you multiply a given nonzero number by to get 1, found by swapping numerator and denominator.

least common denominator — the smallest positive number that every denominator in a problem divides into evenly.

complex fraction — a fraction whose numerator, denominator, or both are themselves fractions.