How to Find a Common Denominator and Simplify

A researcher writes a fox count as $\frac{84}{120}$ , which is correct but bulky, while a teammate records his own count over a denominator of $45$ . Two jobs show up here. A single fraction should be written in its smallest, cleanest form. Two fractions cut into different-sized pieces need a shared piece size before you can compare them. Both jobs use the same Chapter 3 tools, just pointed in opposite directions.

Problem

The fuel reading is

\frac{84}{120}

. Find

\gcd (84, 120)

, divide top and bottom by it in one move. Write

\frac{84}{120}

in simplest form as

\frac{a}{b}

Show a hint

It all comes down to one number, the $\gcd$ of $84$ and $120$ . One way to find it is to list the prime factors of each and collect every prime they both carry. $84 = 2 \times 2 \times 3 \times 7$ and $120 = 2 \times 2 \times 2 \times 3 \times 5$ . Which primes show up in both, and how many times?
Both numbers carry two $2$ s and one $3$ , so $\gcd (84, 120) = 2 \times 2 \times 3 = 12$ . Now make the single move, divide the top by $12$ and the bottom by $12$ . What is $84 \div 12$ over $120 \div 12$ ?

Show the full solution

84 = 2 \times 2 \times 3 \times 7

and

120 = 2 \times 2 \times 2 \times 3 \times 5

, so the primes they both carry are two

2

s and one

3

, which makes

\gcd (84, 120) = 12

. Divide both by

12

. Since

84 \div 12 = 7

and

120 \div 12 = 10

, the fraction is

7 / 10

Dividing by the gcd finishes in one move, because the biggest shared factor already contains every smaller one.

Problem

Panels generated

168

kWh, and the building used

480

kWh. Divide

\gcd (168, 480)

into both top and bottom once. Write

\frac{168}{480}

in simplest form as

\frac{a}{b}

Show a hint

You only need one division to finish, but it has to be division by the right number. Ask what the largest number is that goes evenly into both $168$ and $480$ . Find that $\gcd$ first, then divide once.
Break each number into prime factors. $168 = 2^{3} \cdot 3 \cdot 7$ and $480 = 2^{5} \cdot 3 \cdot 5$ . The shared part is $2^{3} \cdot 3 = 24$ , so the $\gcd$ is $24$ . Now divide both $168$ and $480$ by $24$ and read off the answer.

Show the full solution

168 = 2^{3} \cdot 3 \cdot 7

and

480 = 2^{5} \cdot 3 \cdot 5

, so the part they share is

2^{3} \cdot 3 = 24

. Divide both by

24

. Since

168 \div 24 = 7

and

480 \div 24 = 20

, the panels supplied

\frac{7}{20}

of the building's electricity. Nothing more cancels, since

7

is prime and does not divide

20

Twelve is

2 \times 2 \times 3

and eighteen is

2 \times 3 \times 3

. They share one

2

and one

3

, and that shared

6

is the gcd. Cancelling it from top and bottom leaves

\frac{2}{3}

Problem

Divide

\frac{9}{20} \div \frac{3}{8} = \frac{9}{20} \times \frac{8}{3}

. Cancel before multiplying. Write the result as

\frac{a}{b}

in simplest form.

Show a hint

Dividing by a fraction is the same as multiplying by its reciprocal, so rewrite $\frac{9}{20} \div \frac{3}{8}$ as $\frac{9}{20} \times \frac{8}{3}$ . Once it is a multiplication, any top number can cancel with any bottom number.
Cancel before multiplying. Replace the $9$ and $3$ with $3$ and $1$ , since both lose a factor of $3$ . Replace the $8$ and $20$ with $2$ and $5$ , since both lose a factor of $4$ . Now multiply the new tops and the new bottoms straight across.

Show the full solution

Flip the divisor, so

\frac{9}{20} \div \frac{3}{8} = \frac{9}{20} \times \frac{8}{3}

. The

9

and the

3

both lose a factor of

3

, leaving

3

and

1

. The

8

and the

20

both lose a factor of

4

, leaving

2

and

5

. That gives

\frac{3}{5} \times \frac{2}{1}

, and multiplying across gives

6 / 5

. Once a division is rewritten as a multiplication, any top can cancel with any bottom, even across the two fractions.

Problem

Evaluate

\frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{8}{9}

. Every interior bottom is the next top and cancels. Write the collapsed result as

\frac{a}{b}

in simplest form.

Show a hint

You do not have to multiply this out. Write the whole product as one giant fraction, with all six tops on top and all six bottoms underneath. Now hunt for matching numbers. Does anything on top also appear on the bottom?
Every middle number shows up once on top and once on the bottom, so each of those pairs cancels to $1$ . Cross out the $4$ , $5$ , $6$ , $7$ , and $8$ wherever they appear. The only top with no matching bottom is the first $3$ , and the only bottom with no matching top is the last $9$ . What single fraction is left?

Show the full solution

Stack all the tops over all the bottoms.

\frac{3 \times 4 \times 5 \times 6 \times 7 \times 8}{4 \times 5 \times 6 \times 7 \times 8 \times 9}

Every number from

4

through

8

sits once on top and once on the bottom, so all of those pairs cancel. That leaves

\frac{3}{9}

, which reduces to

\frac{1}{3}

. In a chain like this only the first top and the last bottom ever survive, so there is nothing to multiply out.

Problem

1, 485 = 3^{3} \times 5 \times 11

and

1, 890 = 2 \times 3^{3} \times 5 \times 7

. Cancel shared primes, read what remains. Write

\frac{1, 485}{1, 890}

in simplest form as

\frac{a}{b}

Show a hint

Line the two prime lists up factor by factor. Both the top and the bottom carry three $3$ s and one $5$ . Those are exactly the primes you are allowed to cross off, one from the top for each one you cross off the bottom.
Strike the shared $3 \times 3 \times 3 \times 5$ from both floors. On top you are left with just $11$ . On the bottom you are left with $2 \times 7$ . Multiply each leftover pile back up and write the fraction.

Show the full solution

Line the prime lists up. The top is

1, 485 = 3 \times 3 \times 3 \times 5 \times 11

and the bottom is

1, 890 = 2 \times 3 \times 3 \times 3 \times 5 \times 7

, so the shared primes are three

3

s and one

5

. Cancel each one against its twin.

\frac{1, 485}{1, 890} = \frac{3 \times 3 \times 3 \times 5 \times 11}{2 \times 3 \times 3 \times 3 \times 5 \times 7} = \frac{11}{2 \times 7}

What survives is

\frac{11}{14}

. The cancelled primes multiply to

135

, which is the gcd, so factoring hands it to you without any searching.

Problem

Simplify

\frac{k^{4} \times 10}{k^{2} \times 15}

: cancel two

k

s and the shared factor

5

. Result is

\frac{2 k^{2}}{3}

. Now set

k = 3

. What single whole number does the fraction equal?

Show a hint

You already simplified to $\frac{2 k^{2}}{3}$ . Now $k^{2}$ just means $k \times k$ , so with $k = 3$ that piece is $3 \times 3$ . Build the top first, then divide by $3$ .
With $k = 3$ , $k^{2} = 9$ . The top is $2 \times 9 = 18$ . Then $\frac{18}{3}$ is the final number. Divide it out.

Show the full solution

There are four

k

s on top and two on the bottom, so two pairs cancel and

k^{2}

is left upstairs. The numbers give

\frac{10}{15} = \frac{2}{3}

, so the expression is

\frac{2 k^{2}}{3}

. Now put in

k = 3

. Then

k^{2} = 9

, the top is

2 \times 9 = 18

, and

18 \div 3

gives

6

Cancelling before substituting keeps the numbers small, and it works on letters because a letter is just a factor you have not filled in yet.

Simplest form gives a single fraction its smallest name. But $\frac{5}{6}$ and $\frac{7}{8}$ are both already in simplest form, and you still cannot say which is larger, because sixths and eighths are different-sized pieces and counting one kind against the other tells you nothing. So this time we go the other way. Instead of shrinking the denominators, we grow them until both fractions are cut into pieces of the same size.

Problem

The lcd of

6

and

8

24

. Eighths

\times 3

\frac{7}{8} = \frac{21}{24}

. Do the same for

\frac{5}{6}

. What does

\frac{5}{6}

become over

24

Show a hint

You want to rebuild $\frac{5}{6}$ so the bottom reads $24$ instead of $6$ . Ask yourself what you multiply $6$ by to land on $24$ , then remember the equivalent fraction rule says the top has to get the exact same treatment.
Since $6 \times 4 = 24$ , the multiplier is $4$ . Multiply the top by $4$ as well, so $5 \times 4 = 20$ , and the rebuilt fraction is $\frac{20}{24}$ .

Show the full solution

Since

6 \times 4 = 24

, the bottom was multiplied by

4

, so the top gets the same

4

. That gives

5 \times 4 = 20

on top.

\frac{5}{6} = 20 / 24

Same ribbon, smaller pieces. Each sixth was cut into

4

parts, which is why the count went from

5

20

Problem

To add

\frac{1}{6} + \frac{4}{9}

find a shared denominator.

6 \times 9 = 54

works but overshoots, and

6

and

9

3

. What is

lcm (6, 9)

Show a hint

You are hunting for the smallest number that both $6$ and $9$ divide into evenly. Try walking up the multiples of the bigger one, $9, 18, 27, \dots$ , and stop at the first that $6$ also divides.
List the multiples of $9$ : $9, 18, 27$ . Is $9$ a multiple of $6$ ? No. Is $18$ a multiple of $6$ ? Yes, $18 = 6 \times 3$ . That first match is your answer.

Show the full solution

Break each denominator into primes,

6 = 2 \times 3

and

9 = 3 \times 3

. Covering both takes one

2

and two

3

s, so

lcm (6, 9) = 2 \times 3 \times 3 = 18

. The product

6 \times 9 = 54

also works as a common denominator, but it counts the shared

3

twice, so it comes out

3

times bigger than it needs to be.

Seven tenths and four fifteenths have different-size pieces. Re-slicing both to the LCD

30

makes the pieces match:

\frac{7}{10} = \frac{21}{30}

and

\frac{4}{15} = \frac{8}{30}

Problem

lcm (4, 6, 9)

: primes are

2^{2}

and

3^{2}

. Multiply the highest powers. What is the lcd of

\frac{1}{4}

\frac{5}{6}

\frac{7}{9}

Show a hint

Line up the prime factorizations, $4 = 2^{2}$ , $6 = 2 \times 3$ , $9 = 3^{2}$ . The lcd has to be divisible by every one of them, so it needs enough $2$ s for the $4$ and enough $3$ s for the $9$ at the same time.
Take the highest power of each prime seen anywhere. The biggest stack of $2$ s is $2^{2}$ and the biggest stack of $3$ s is $3^{2}$ . The lcd is $2^{2} \times 3^{2} = 4 \times 9$ . Just multiply.

Show the full solution

Write each denominator in primes,

4 = 2^{2}

6 = 2 \times 3

, and

9 = 3^{2}

. Take the highest power of each prime,

2^{2}

from the

4

and

3^{2}

from the

9

, then multiply,

2^{2} \times 3^{2} = 4 \times 9 = 36

. The lcd is

36

. It checks out, since

36 \div 4 = 9

36 \div 6 = 6

, and

36 \div 9 = 4

, and dropping any factor would leave one denominator unable to divide in.

Problem

lcd is

36

\frac{1}{4} \to \frac{9}{36}

\frac{7}{9} \to \frac{28}{36}

. What does

\frac{5}{6}

become over

36

Show a hint

Ask the bottom first. What times $6$ lands on $36$ ? Whatever that number is, the top has to ride along with the exact same multiplier.
The bottom $6$ needs $\times 6$ to become $36$ , so multiply the top by $6$ too. That is $5 \times 6$ over $6 \times 6$ .

Show the full solution

The bottom needs

6 \times 6 = 36

, so the multiplier is

6

and the top takes the same

6

. That gives

5 \times 6 = 30

, so

\frac{5}{6} = \frac{5 \times 6}{6 \times 6} = 30 / 36

. Scaling both parts by the same number renames a fraction without changing its value, and now all three read over

36

, ready to compare or add.

The single point three quarters of the way to

1

answers to

\frac{3}{4}, \frac{6}{8}, \frac{9}{12},

and

\frac{75}{100}

alike. Multiplying top and bottom by the same number renames it without moving it.

Problem

Simplify

\frac{a^{4} b^{2}}{a^{2} b^{5}}

: cancel

a

s and

b

s in pairs. Then set

a = 3

and

b = 2

. What single fraction in simplest form does the whole expression equal?

Show a hint

The two letters never mix, so deal with them separately. There are four $a$ s on top and two on the bottom, so two $a$ s upstairs survive. There are two $b$ s on top and five on the bottom, so three $b$ s downstairs survive. Write what is left as a tidy fraction before you put numbers in.
After cancelling, you are left with $\frac{a \times a}{b \times b \times b}$ , which is $\frac{a^{2}}{b^{3}}$ . Now $a = 3$ gives $a^{2} = 3 \times 3 = 9$ , and $b = 2$ gives $b^{3} = 2 \times 2 \times 2 = 8$ . Put the top over the bottom.

Show the full solution

Four

a

s on top against two on the bottom leaves

a^{2}

upstairs. Two

b

s on top against five on the bottom leaves

b^{3}

downstairs, so the expression is

\frac{a^{2}}{b^{3}}

. With

a = 3

the top is

3 \times 3 = 9

, and with

b = 2

the bottom is

2 \times 2 \times 2 = 8

9 / 8

Each letter cancels only against itself, so count the

a

s and the

b

s as two separate piles.

Problem

\frac{5}{9}

\frac{7}{12}

. Which is larger? Write it as

\frac{a}{b}

Show a hint

The two fractions are hard to compare because they count different-sized pieces. Rewrite each one over $36$ . What do you multiply the top and bottom of $\frac{5}{9}$ by to land on $36$ ? What about $\frac{7}{12}$ ?
Since $9 \times 4 = 36$ , $\frac{5}{9} = \frac{20}{36}$ . Since $12 \times 3 = 36$ , $\frac{7}{12} = \frac{21}{36}$ . Now both count thirty-sixths, so the one with more of them wins. Compare $20$ and $21$ .

Show the full solution

The least common denominator of

9

and

12

36

. Grow both,

\frac{5}{9} = \frac{5 \times 4}{9 \times 4} = \frac{20}{36}

and

\frac{7}{12} = \frac{7 \times 3}{12 \times 3} = \frac{21}{36}

. Both count thirty-sixths now, and

21 > 20

, so the larger fraction is

\frac{7}{12}

. Numerators only race fairly once the denominators match.

Practice these ideas

Practice

A contact sheet has

24

thumbnails, and

18

are keepers. Reduce

\frac{18}{24}

by dividing by

\gcd (18, 24)

. Write the answer as

\frac{a}{b}

Show the solution

The divisors of

18

are

1, 2, 3, 6, 9, 18

and the divisors of

24

are

1, 2, 3, 4, 6, 8, 12, 24

, so

\gcd (18, 24) = 6

. Divide top and bottom by

6

\frac{18}{24} = \frac{18 \div 6}{24 \div 6} = 3 / 4

Three of every four pictures made the cut. Going straight to the gcd does in one step what dividing by

2

and then by

3

does in two.

Practice

A big batch uses

360

g of one ingredient per

504

g of another. Simplify

\frac{360}{504}

to simplest form and write as

\frac{a}{b}

Show the solution

Chase the difference to get the gcd. Since

504 - 360 = 144

\gcd (360, 504) = \gcd (360, 144)

, and

360 - 2 \cdot 144 = 72

, so that equals

\gcd (144, 72) = 72

. Divide both by

72

, where

360 \div 72 = 5

and

504 \div 72 = 7

, so

\frac{360}{504} = 5 / 7

. Subtracting the smaller number from the larger never changes the gcd, which is what makes that loop safe.

Practice

Compute

\frac{8}{21} \times \frac{7}{12}

. Cancel shared factors across tops and bottoms before multiplying. Write the product as

\frac{a}{b}

in simplest form.

Show the solution

Write it as one fraction,

\frac{8 \times 7}{21 \times 12}

, and cancel before multiplying. The

7

and the

21

share a

7

, leaving

1

and

3

. The

8

and the

12

share a

4

, leaving

2

and

3

. What is left is

\frac{2 \times 1}{3 \times 3} = 2 / 9

. Cancelling diagonally is allowed because every top is about to pile onto one top anyway.

Practice

Divide

\frac{15}{16} \div \frac{25}{8}

. Flip the divisor, cancel, then multiply. Write the quotient as

\frac{a}{b}

in simplest form.

Show the solution

Flip the divisor, so

\frac{15}{16} \div \frac{25}{8} = \frac{15}{16} \times \frac{8}{25}

. The

8

and the

16

share an

8

, leaving

1

and

2

. The

15

and the

25

share a

5

, leaving

3

and

5

. So the product is

\frac{3}{2} \times \frac{1}{5} = \frac{3}{10}

. Cancelling first means you never have to write

120

over

400

and reduce it after.

Practice

\frac{2310}{2730}

: top is

2 \times 3 \times 5 \times 7 \times 11

, bottom is

2 \times 3 \times 5 \times 7 \times 13

. Cancel shared primes. Write the reduced fraction as

\frac{a}{b}

Show the solution

Both lists carry

2 \times 3 \times 5 \times 7

, so those four primes cancel and only

11

is left on top with

13

on the bottom.

\frac{2310}{2730} = \frac{2 \times 3 \times 5 \times 7 \times 11}{2 \times 3 \times 5 \times 7 \times 13} = \frac{11}{13}

That shared chunk is

210

, the gcd. Two different primes share no factor, so this is already in lowest terms.

Practice

\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7}

. Cancel matching pairs. Write the result as

\frac{a}{b}

in simplest form.

Show the solution

Multiply across with the factors lined up.

\frac{2 \times 3 \times 4 \times 5 \times 6}{3 \times 4 \times 5 \times 6 \times 7}

The

3, 4, 5, 6

appear on both rows and cancel to

1

, leaving

2 / 7

. In a chain like this only the first top and the last bottom ever go unpartnered, and

2

and

7

share nothing.

Practice

Simplify

\frac{m^{5}}{m^{2} \times 4}

: cancel

m

s, leaving

\frac{m^{3}}{4}

. Set

m = 2

. What single number does the expression equal?

Show the solution

There are five

m

s on top and two on the bottom, so two pairs cancel and

m^{3}

is left over the

4

. Set

m = 2

. The top becomes

2 \times 2 \times 2 = 8

, so the fraction is

\frac{8}{4} = 2

. Letters cancel exactly like numbers, so simplify before you substitute and the arithmetic stays small.

Practice

A measuring strip must mark both

\frac{3}{8}

in and

\frac{1}{10}

in tick marks. Find the smallest shared bottom:

lcm (8, 10)

. What is that least common denominator?

Show the solution

Break each bottom into primes,

8 = 2 \times 2 \times 2

and

10 = 2 \times 5

. The lcm takes the most of each prime that shows up in either one, so

lcm (8, 10) = 2 \times 2 \times 2 \times 5 = 40

. A strip ruled into

40

ths carries both marks exactly, since

\frac{3}{8} = \frac{15}{40}

and

\frac{1}{10} = \frac{4}{40}

Practice

Three rhythm loops land every

4

10

, and

15

beats. The next beat where all three align is

lcm (4, 10, 15)

. What is that number?

Show the solution

In primes,

4 = 2^{2}

10 = 2 \cdot 5

, and

15 = 3 \cdot 5

. Keep the highest power of each prime, two

2

s from the

4

, one

3

from the

15

, and one

5

, so the lcm is

2^{2} \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60

. Check the fits,

60 \div 4 = 15

60 \div 10 = 6

, and

60 \div 15 = 4

, all whole.

Practice

Rewrite

\frac{7}{12}

as an equivalent fraction over

60

. Find what

12

is multiplied by to reach

60

, then scale the top. What is the new numerator?

Show the solution

Start with the bottoms. Since

60 \div 12 = 5

, the denominator was multiplied by

5

, so the numerator takes the same

5

. That gives

7 \times 5 = 35

, so

\frac{7}{12} = \frac{35}{60}

and the new numerator is

35

. Scaling top and bottom by the same number renames the fraction without changing its value.

Practice

A track is

\frac{4}{9}

done. Rewrite over the common denominator

45

. Find what

9

is multiplied by to reach

45

, then scale the top. What is the new numerator over

45

Show the solution

The denominator went from

9

45

, and

9 \times 5 = 45

, so the multiplier is

5

. The top grows by the same

5

, giving

4 \times 5 = 20

, so

\frac{4}{9} = \frac{20}{45}

and the new numerator is

20

. Dividing

\frac{20}{45}

back by

5

returns

\frac{4}{9}

, so the value never moved.

Practice

Maya finishes

\frac{4}{7}

of a loop, and Theo finishes

\frac{5}{9}

. Grow both to the lcd of

7

and

9

, compare the tops. Which is larger? Write the answer as

\frac{a}{b}

Show the solution

Since

7

and

9

share no common factor, the least common denominator is

7 \times 9 = 63

. Grow both,

\frac{4}{7} = \frac{4 \times 9}{7 \times 9} = \frac{36}{63}

and

\frac{5}{9} = \frac{5 \times 7}{9 \times 7} = \frac{35}{63}

. Since

36 > 35

, Maya covered more, so the larger fraction is

\frac{4}{7}

. When two denominators share no factor, their lcm is simply the product.

Practice

A turbine produced

280

MWh out of

630

regional MWh. Break both into primes, cancel shared factors, and write

\frac{280}{630}

in simplest form.

Show the solution

Break both into primes,

280 = 2 \times 2 \times 2 \times 5 \times 7

and

630 = 2 \times 3 \times 3 \times 5 \times 7

. They both carry one

2

, one

5

, and one

7

, so

\gcd (280, 630) = 2 \times 5 \times 7 = 70

. Divide top and bottom by

70

, where

280 \div 70 = 4

and

630 \div 70 = 9

, giving

4 / 9

. Only shared primes cancel, and

4 = 2 \times 2

and

9 = 3 \times 3

have none left in common.

Practice

Rope 1 reaches

\frac{11}{15}

of a wall, and Rope 2 reaches

\frac{7}{9}

. The lcd is

45

. Rename both over

45

and compare. Which fraction is larger? Write it as

\frac{a}{b}

Show the solution

The least common denominator of

15

and

9

45

, since

45 = 15 \times 3 = 9 \times 5

. Grow both,

\frac{11}{15} = \frac{11 \times 3}{15 \times 3} = \frac{33}{45}

and

\frac{7}{9} = \frac{7 \times 5}{9 \times 5} = \frac{35}{45}

. Since

35 > 33

, Rope 2 reaches farther, so the larger fraction is

7 / 9

. Once the pieces are the same size, the comparison is just the numerators.

Practice

Evaluate

\frac{12}{25} \times \frac{35}{18} \div \frac{7}{15}

. Flip the last fraction, then cancel every shared factor across all tops and bottoms. What single number unlocks the box?

Show the solution

Flip the last fraction, so the expression becomes

\frac{12}{25} \times \frac{35}{18} \times \frac{15}{7} .

The

35

and the

7

share a

7

, leaving

5

on top, and that

5

cuts the

25

down to

5

. Then the

15

over that

5

leaves

3

on top. The

12

and the

18

share a

6

, becoming

2

and

3

, and that bottom

3

cancels the

3

above. Only

\frac{2}{1}

is left, so the box opens with

2

. Cancelling across all three fractions at once keeps a messy product down to single digits.

Common Denominators and Simplification

Practice these ideas