Practice · Equation fixes the form, inequality bounds it · Sensim Math

Variant 1 easy

A fraction in lowest terms has a whole-number numerator and denominator. Adding $5$ to its denominator gives a fraction equal to $\frac{2}{9}$ , and adding $6$ to its numerator gives a fraction greater than $\frac{1}{2}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $5$ to the denominator makes it equal $\frac{2}{9}$ and adding $6$ to the numerator makes it larger than $\frac{1}{2}$ .

Givens

Numerator over (denominator $+ 5$ ) equals $\frac{2}{9}$ .
(Numerator $+ 6$ ) over denominator is greater than $\frac{1}{2}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 5} = \frac{2}{9}, \quad \text{so} \quad 9y = 2(x + 5)

Cross-multiplying turns the equal fractions into one equation.

2Use that 2 and 9 share no factor

#13 Convert to Algebra 6.NS.B.4

2

divides

9y

but shares no factor with

9

, so

2

divides

y

. Write

y = 2k

; then

x + 5 = 9k

for the same whole number

k

.

y = 2k, \quad x = 9k - 5 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{2}{9}

is

\frac{2k}{9k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

6

to the numerator must give more than

\frac{1}{2}

. The denominator

9k - 5

is positive, so multiplying both sides by

2(9k - 5)

keeps the direction.

\frac{2k + 6}{9k - 5} > \frac{1}{2} \;\Rightarrow\; 4k + 12 > 9k - 5 \;\Rightarrow\; 17 > 5k \;\Rightarrow\; k < 3\frac{2}{5}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{2k}{9k - 5}

.

\frac{2}{4}, \quad \frac{4}{13}, \quad \frac{6}{22}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{4}

and

\frac{6}{22}

can be reduced, because

2

and

4

share the factor

2

, and

6

and

22

share the factor

2

. The numbers

4

and

13

share no factor other than

1

.

\gcd(2, 4) = 2, \quad \gcd(4, 13) = 1, \quad \gcd(6, 22) = 2

Only

\frac{4}{13}

is already in lowest terms.

Answer:

\frac{4}{13}

4 · Reviewdoes it hold up?

Adding $5$ to the denominator: $\frac{4}{18} = \frac{2}{9}$ . Adding $6$ to the numerator: $\frac{10}{13} > \frac{1}{2}$ , since $10 \times 2 = 20 > 13 = 1 \times 13$ . Both conditions hold, and $\frac{4}{13}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{2}{9}$ one by one: $\frac{2}{9}, \frac{4}{18}, \frac{6}{27}, \frac{8}{36}, \ldots$ . Subtract $5$ from each denominator to get $\frac{2}{4}, \frac{4}{13}, \frac{6}{22}, \frac{8}{31}, \ldots$ and test each: adding $6$ to the numerator gives more than $\frac{1}{2}$ only for the first three, and of those only $\frac{4}{13}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $2$ and $9$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $4k + 12 > 9k - 5$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 2 easy

A fraction in lowest terms has a whole-number numerator and denominator. Adding $6$ to its denominator gives a fraction equal to $\frac{4}{11}$ , and adding $4$ to its numerator gives a fraction greater than $\frac{1}{2}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $6$ to the denominator makes it equal $\frac{4}{11}$ and adding $4$ to the numerator makes it larger than $\frac{1}{2}$ .

Givens

Numerator over (denominator $+ 6$ ) equals $\frac{4}{11}$ .
(Numerator $+ 4$ ) over denominator is greater than $\frac{1}{2}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 6} = \frac{4}{11}, \quad \text{so} \quad 11y = 4(x + 6)

Cross-multiplying turns the equal fractions into one equation.

2Use that 4 and 11 share no factor

#13 Convert to Algebra 6.NS.B.4

4

divides

11y

but shares no factor with

11

, so

4

divides

y

. Write

y = 4k

; then

x + 6 = 11k

for the same whole number

k

.

y = 4k, \quad x = 11k - 6 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{4}{11}

is

\frac{4k}{11k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

4

to the numerator must give more than

\frac{1}{2}

. The denominator

11k - 6

is positive, so multiplying both sides by

2(11k - 6)

keeps the direction.

\frac{4k + 4}{11k - 6} > \frac{1}{2} \;\Rightarrow\; 8k + 8 > 11k - 6 \;\Rightarrow\; 14 > 3k \;\Rightarrow\; k < 4\frac{2}{3}

The inequality leaves only

k = 1, 2, 3, 4

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3, 4

into

\frac{4k}{11k - 6}

.

\frac{4}{5}, \quad \frac{8}{16}, \quad \frac{12}{27}, \quad \frac{16}{38}

Only four fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{8}{16}

,

\frac{12}{27}

, and

\frac{16}{38}

can be reduced, because

8

and

16

share the factor

8

,

12

and

27

share the factor

3

, and

16

and

38

share the factor

2

. The numbers

4

and

5

share no factor other than

1

.

\gcd(4, 5) = 1, \quad \gcd(8, 16) = 8, \quad \gcd(12, 27) = 3, \quad \gcd(16, 38) = 2

Only

\frac{4}{5}

is already in lowest terms.

Answer:

\frac{4}{5}

4 · Reviewdoes it hold up?

Adding $6$ to the denominator: $\frac{4}{11} = \frac{4}{11}$ . Adding $4$ to the numerator: $\frac{8}{5} > \frac{1}{2}$ , since $8 \times 2 = 16 > 5 = 1 \times 5$ . Both conditions hold, and $\frac{4}{5}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{4}{11}$ one by one: $\frac{4}{11}, \frac{8}{22}, \frac{12}{33}, \frac{16}{44}, \frac{20}{55}, \ldots$ . Subtract $6$ from each denominator to get $\frac{4}{5}, \frac{8}{16}, \frac{12}{27}, \frac{16}{38}, \frac{20}{49}, \ldots$ and test each: adding $4$ to the numerator gives more than $\frac{1}{2}$ only for the first four, and of those only $\frac{4}{5}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $4$ and $11$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $8k + 8 > 11k - 6$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 3 easy

A fraction in lowest terms has a whole-number numerator and denominator. Adding $1$ to its denominator gives a fraction equal to $\frac{2}{11}$ , and adding $6$ to its numerator gives a fraction greater than $\frac{1}{3}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $1$ to the denominator makes it equal $\frac{2}{11}$ and adding $6$ to the numerator makes it larger than $\frac{1}{3}$ .

Givens

Numerator over (denominator $+ 1$ ) equals $\frac{2}{11}$ .
(Numerator $+ 6$ ) over denominator is greater than $\frac{1}{3}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 1} = \frac{2}{11}, \quad \text{so} \quad 11y = 2(x + 1)

Cross-multiplying turns the equal fractions into one equation.

2Use that 2 and 11 share no factor

#13 Convert to Algebra 6.NS.B.4

2

divides

11y

but shares no factor with

11

, so

2

divides

y

. Write

y = 2k

; then

x + 1 = 11k

for the same whole number

k

.

y = 2k, \quad x = 11k - 1 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{2}{11}

is

\frac{2k}{11k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

6

to the numerator must give more than

\frac{1}{3}

. The denominator

11k - 1

is positive, so multiplying both sides by

3(11k - 1)

keeps the direction.

\frac{2k + 6}{11k - 1} > \frac{1}{3} \;\Rightarrow\; 6k + 18 > 11k - 1 \;\Rightarrow\; 19 > 5k \;\Rightarrow\; k < 3\frac{4}{5}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{2k}{11k - 1}

.

\frac{2}{10}, \quad \frac{4}{21}, \quad \frac{6}{32}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{10}

and

\frac{6}{32}

can be reduced, because

2

and

10

share the factor

2

, and

6

and

32

share the factor

2

. The numbers

4

and

21

share no factor other than

1

.

\gcd(2, 10) = 2, \quad \gcd(4, 21) = 1, \quad \gcd(6, 32) = 2

Only

\frac{4}{21}

is already in lowest terms.

Answer:

\frac{4}{21}

4 · Reviewdoes it hold up?

Adding $1$ to the denominator: $\frac{4}{22} = \frac{2}{11}$ . Adding $6$ to the numerator: $\frac{10}{21} > \frac{7}{21} = \frac{1}{3}$ . Both conditions hold, and $\frac{4}{21}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{2}{11}$ one by one: $\frac{2}{11}, \frac{4}{22}, \frac{6}{33}, \frac{8}{44}, \ldots$ . Subtract $1$ from each denominator to get $\frac{2}{10}, \frac{4}{21}, \frac{6}{32}, \frac{8}{43}, \ldots$ and test each: adding $6$ to the numerator gives more than $\frac{1}{3}$ only for the first three, and of those only $\frac{4}{21}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $2$ and $11$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $6k + 18 > 11k - 1$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 4 easy

A fraction in lowest terms has a whole-number numerator and denominator. Adding $5$ to its denominator gives a fraction equal to $\frac{2}{11}$ , and adding $6$ to its numerator gives a fraction greater than $\frac{2}{5}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $5$ to the denominator makes it equal $\frac{2}{11}$ and adding $6$ to the numerator makes it larger than $\frac{2}{5}$ .

Givens

Numerator over (denominator $+ 5$ ) equals $\frac{2}{11}$ .
(Numerator $+ 6$ ) over denominator is greater than $\frac{2}{5}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 5} = \frac{2}{11}, \quad \text{so} \quad 11y = 2(x + 5)

Cross-multiplying turns the equal fractions into one equation.

2Use that 2 and 11 share no factor

#13 Convert to Algebra 6.NS.B.4

2

divides

11y

but shares no factor with

11

, so

2

divides

y

. Write

y = 2k

; then

x + 5 = 11k

for the same whole number

k

.

y = 2k, \quad x = 11k - 5 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{2}{11}

is

\frac{2k}{11k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

6

to the numerator must give more than

\frac{2}{5}

. The denominator

11k - 5

is positive, so multiplying both sides by

5(11k - 5)

keeps the direction.

\frac{2k + 6}{11k - 5} > \frac{2}{5} \;\Rightarrow\; 10k + 30 > 22k - 10 \;\Rightarrow\; 40 > 12k \;\Rightarrow\; k < 3\frac{1}{3}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{2k}{11k - 5}

.

\frac{2}{6}, \quad \frac{4}{17}, \quad \frac{6}{28}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{6}

and

\frac{6}{28}

can be reduced, because

2

and

6

share the factor

2

, and

6

and

28

share the factor

2

. The numbers

4

and

17

share no factor other than

1

.

\gcd(2, 6) = 2, \quad \gcd(4, 17) = 1, \quad \gcd(6, 28) = 2

Only

\frac{4}{17}

is already in lowest terms.

Answer:

\frac{4}{17}

4 · Reviewdoes it hold up?

Adding $5$ to the denominator: $\frac{4}{22} = \frac{2}{11}$ . Adding $6$ to the numerator: $\frac{10}{17} > \frac{2}{5}$ , since $10 \times 5 = 50 > 34 = 2 \times 17$ . Both conditions hold, and $\frac{4}{17}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{2}{11}$ one by one: $\frac{2}{11}, \frac{4}{22}, \frac{6}{33}, \frac{8}{44}, \ldots$ . Subtract $5$ from each denominator to get $\frac{2}{6}, \frac{4}{17}, \frac{6}{28}, \frac{8}{39}, \ldots$ and test each: adding $6$ to the numerator gives more than $\frac{2}{5}$ only for the first three, and of those only $\frac{4}{17}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $2$ and $11$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $10k + 30 > 22k - 10$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 5 medium

A fraction in lowest terms has a whole-number numerator and denominator. Adding $5$ to its denominator gives a fraction equal to $\frac{4}{15}$ , and adding $5$ to its numerator gives a fraction greater than $\frac{2}{5}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $5$ to the denominator makes it equal $\frac{4}{15}$ and adding $5$ to the numerator makes it larger than $\frac{2}{5}$ .

Givens

Numerator over (denominator $+ 5$ ) equals $\frac{4}{15}$ .
(Numerator $+ 5$ ) over denominator is greater than $\frac{2}{5}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 5} = \frac{4}{15}, \quad \text{so} \quad 15y = 4(x + 5)

Cross-multiplying turns the equal fractions into one equation.

2Use that 4 and 15 share no factor

#13 Convert to Algebra 6.NS.B.4

4

divides

15y

but shares no factor with

15

, so

4

divides

y

. Write

y = 4k

; then

x + 5 = 15k

for the same whole number

k

.

y = 4k, \quad x = 15k - 5 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{4}{15}

is

\frac{4k}{15k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

5

to the numerator must give more than

\frac{2}{5}

. The denominator

15k - 5

is positive, so multiplying both sides by

5(15k - 5)

keeps the direction.

\frac{4k + 5}{15k - 5} > \frac{2}{5} \;\Rightarrow\; 20k + 25 > 30k - 10 \;\Rightarrow\; 35 > 10k \;\Rightarrow\; k < 3\frac{1}{2}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{4k}{15k - 5}

.

\frac{4}{10}, \quad \frac{8}{25}, \quad \frac{12}{40}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{4}{10}

and

\frac{12}{40}

can be reduced, because

4

and

10

share the factor

2

, and

12

and

40

share the factor

4

. The numbers

8

and

25

share no factor other than

1

.

\gcd(4, 10) = 2, \quad \gcd(8, 25) = 1, \quad \gcd(12, 40) = 4

Only

\frac{8}{25}

is already in lowest terms.

Answer:

\frac{8}{25}

4 · Reviewdoes it hold up?

Adding $5$ to the denominator: $\frac{8}{30} = \frac{4}{15}$ . Adding $5$ to the numerator: $\frac{13}{25} > \frac{10}{25} = \frac{2}{5}$ . Both conditions hold, and $\frac{8}{25}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{4}{15}$ one by one: $\frac{4}{15}, \frac{8}{30}, \frac{12}{45}, \frac{16}{60}, \ldots$ . Subtract $5$ from each denominator to get $\frac{4}{10}, \frac{8}{25}, \frac{12}{40}, \frac{16}{55}, \ldots$ and test each: adding $5$ to the numerator gives more than $\frac{2}{5}$ only for the first three, and of those only $\frac{8}{25}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $4$ and $15$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $20k + 25 > 30k - 10$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 6 medium

A fraction in lowest terms has a whole-number numerator and denominator. Adding $6$ to its denominator gives a fraction equal to $\frac{7}{18}$ , and adding $5$ to its numerator gives a fraction greater than $\frac{1}{2}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $6$ to the denominator makes it equal $\frac{7}{18}$ and adding $5$ to the numerator makes it larger than $\frac{1}{2}$ .

Givens

Numerator over (denominator $+ 6$ ) equals $\frac{7}{18}$ .
(Numerator $+ 5$ ) over denominator is greater than $\frac{1}{2}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 6} = \frac{7}{18}, \quad \text{so} \quad 18y = 7(x + 6)

Cross-multiplying turns the equal fractions into one equation.

2Use that 7 and 18 share no factor

#13 Convert to Algebra 6.NS.B.4

7

divides

18y

but shares no factor with

18

, so

7

divides

y

. Write

y = 7k

; then

x + 6 = 18k

for the same whole number

k

.

y = 7k, \quad x = 18k - 6 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{7}{18}

is

\frac{7k}{18k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

5

to the numerator must give more than

\frac{1}{2}

. The denominator

18k - 6

is positive, so multiplying both sides by

2(18k - 6)

keeps the direction.

\frac{7k + 5}{18k - 6} > \frac{1}{2} \;\Rightarrow\; 14k + 10 > 18k - 6 \;\Rightarrow\; 16 > 4k \;\Rightarrow\; k < 4

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{7k}{18k - 6}

.

\frac{7}{12}, \quad \frac{14}{30}, \quad \frac{21}{48}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{14}{30}

and

\frac{21}{48}

can be reduced, because

14

and

30

share the factor

2

, and

21

and

48

share the factor

3

. The numbers

7

and

12

share no factor other than

1

.

\gcd(7, 12) = 1, \quad \gcd(14, 30) = 2, \quad \gcd(21, 48) = 3

Only

\frac{7}{12}

is already in lowest terms.

Answer:

\frac{7}{12}

4 · Reviewdoes it hold up?

Adding $6$ to the denominator: $\frac{7}{18} = \frac{7}{18}$ . Adding $5$ to the numerator: $\frac{12}{12} > \frac{6}{12} = \frac{1}{2}$ . Both conditions hold, and $\frac{7}{12}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{7}{18}$ one by one: $\frac{7}{18}, \frac{14}{36}, \frac{21}{54}, \frac{28}{72}, \ldots$ . Subtract $6$ from each denominator to get $\frac{7}{12}, \frac{14}{30}, \frac{21}{48}, \frac{28}{66}, \ldots$ and test each: adding $5$ to the numerator gives more than $\frac{1}{2}$ only for the first three, and of those only $\frac{7}{12}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $7$ and $18$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $14k + 10 > 18k - 6$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 7 medium

A fraction in lowest terms has a whole-number numerator and denominator. Adding $4$ to its denominator gives a fraction equal to $\frac{5}{18}$ , and adding $2$ to its numerator gives a fraction greater than $\frac{1}{3}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $4$ to the denominator makes it equal $\frac{5}{18}$ and adding $2$ to the numerator makes it larger than $\frac{1}{3}$ .

Givens

Numerator over (denominator $+ 4$ ) equals $\frac{5}{18}$ .
(Numerator $+ 2$ ) over denominator is greater than $\frac{1}{3}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 4} = \frac{5}{18}, \quad \text{so} \quad 18y = 5(x + 4)

Cross-multiplying turns the equal fractions into one equation.

2Use that 5 and 18 share no factor

#13 Convert to Algebra 6.NS.B.4

5

divides

18y

but shares no factor with

18

, so

5

divides

y

. Write

y = 5k

; then

x + 4 = 18k

for the same whole number

k

.

y = 5k, \quad x = 18k - 4 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{5}{18}

is

\frac{5k}{18k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

2

to the numerator must give more than

\frac{1}{3}

. The denominator

18k - 4

is positive, so multiplying both sides by

3(18k - 4)

keeps the direction.

\frac{5k + 2}{18k - 4} > \frac{1}{3} \;\Rightarrow\; 15k + 6 > 18k - 4 \;\Rightarrow\; 10 > 3k \;\Rightarrow\; k < 3\frac{1}{3}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{5k}{18k - 4}

.

\frac{5}{14}, \quad \frac{10}{32}, \quad \frac{15}{50}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{10}{32}

and

\frac{15}{50}

can be reduced, because

10

and

32

share the factor

2

, and

15

and

50

share the factor

5

. The numbers

5

and

14

share no factor other than

1

.

\gcd(5, 14) = 1, \quad \gcd(10, 32) = 2, \quad \gcd(15, 50) = 5

Only

\frac{5}{14}

is already in lowest terms.

Answer:

\frac{5}{14}

4 · Reviewdoes it hold up?

Adding $4$ to the denominator: $\frac{5}{18} = \frac{5}{18}$ . Adding $2$ to the numerator: $\frac{7}{14} > \frac{1}{3}$ , since $7 \times 3 = 21 > 14 = 1 \times 14$ . Both conditions hold, and $\frac{5}{14}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{5}{18}$ one by one: $\frac{5}{18}, \frac{10}{36}, \frac{15}{54}, \frac{20}{72}, \ldots$ . Subtract $4$ from each denominator to get $\frac{5}{14}, \frac{10}{32}, \frac{15}{50}, \frac{20}{68}, \ldots$ and test each: adding $2$ to the numerator gives more than $\frac{1}{3}$ only for the first three, and of those only $\frac{5}{14}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $5$ and $18$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $15k + 6 > 18k - 4$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 8 medium

A fraction in lowest terms has a whole-number numerator and denominator. Adding $3$ to its denominator gives a fraction equal to $\frac{5}{18}$ , and adding $3$ to its numerator gives a fraction greater than $\frac{1}{3}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $3$ to the denominator makes it equal $\frac{5}{18}$ and adding $3$ to the numerator makes it larger than $\frac{1}{3}$ .

Givens

Numerator over (denominator $+ 3$ ) equals $\frac{5}{18}$ .
(Numerator $+ 3$ ) over denominator is greater than $\frac{1}{3}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 3} = \frac{5}{18}, \quad \text{so} \quad 18y = 5(x + 3)

Cross-multiplying turns the equal fractions into one equation.

2Use that 5 and 18 share no factor

#13 Convert to Algebra 6.NS.B.4

5

divides

18y

but shares no factor with

18

, so

5

divides

y

. Write

y = 5k

; then

x + 3 = 18k

for the same whole number

k

.

y = 5k, \quad x = 18k - 3 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{5}{18}

is

\frac{5k}{18k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

3

to the numerator must give more than

\frac{1}{3}

. The denominator

18k - 3

is positive, so multiplying both sides by

3(18k - 3)

keeps the direction.

\frac{5k + 3}{18k - 3} > \frac{1}{3} \;\Rightarrow\; 15k + 9 > 18k - 3 \;\Rightarrow\; 12 > 3k \;\Rightarrow\; k < 4

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{5k}{18k - 3}

.

\frac{5}{15}, \quad \frac{10}{33}, \quad \frac{15}{51}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{5}{15}

and

\frac{15}{51}

can be reduced, because

5

and

15

share the factor

5

, and

15

and

51

share the factor

3

. The numbers

10

and

33

share no factor other than

1

.

\gcd(5, 15) = 5, \quad \gcd(10, 33) = 1, \quad \gcd(15, 51) = 3

Only

\frac{10}{33}

is already in lowest terms.

Answer:

\frac{10}{33}

4 · Reviewdoes it hold up?

Adding $3$ to the denominator: $\frac{10}{36} = \frac{5}{18}$ . Adding $3$ to the numerator: $\frac{13}{33} > \frac{11}{33} = \frac{1}{3}$ . Both conditions hold, and $\frac{10}{33}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{5}{18}$ one by one: $\frac{5}{18}, \frac{10}{36}, \frac{15}{54}, \frac{20}{72}, \ldots$ . Subtract $3$ from each denominator to get $\frac{5}{15}, \frac{10}{33}, \frac{15}{51}, \frac{20}{69}, \ldots$ and test each: adding $3$ to the numerator gives more than $\frac{1}{3}$ only for the first three, and of those only $\frac{10}{33}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $5$ and $18$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $15k + 9 > 18k - 3$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 9 hard

A fraction in lowest terms has a whole-number numerator and denominator. Adding $3$ to its denominator gives a fraction equal to $\frac{4}{19}$ , and adding $2$ to its numerator gives a fraction greater than $\frac{1}{4}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $3$ to the denominator makes it equal $\frac{4}{19}$ and adding $2$ to the numerator makes it larger than $\frac{1}{4}$ .

Givens

Numerator over (denominator $+ 3$ ) equals $\frac{4}{19}$ .
(Numerator $+ 2$ ) over denominator is greater than $\frac{1}{4}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 3} = \frac{4}{19}, \quad \text{so} \quad 19y = 4(x + 3)

Cross-multiplying turns the equal fractions into one equation.

2Use that 4 and 19 share no factor

#13 Convert to Algebra 6.NS.B.4

4

divides

19y

but shares no factor with

19

, so

4

divides

y

. Write

y = 4k

; then

x + 3 = 19k

for the same whole number

k

.

y = 4k, \quad x = 19k - 3 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{4}{19}

is

\frac{4k}{19k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

2

to the numerator must give more than

\frac{1}{4}

. The denominator

19k - 3

is positive, so multiplying both sides by

4(19k - 3)

keeps the direction.

\frac{4k + 2}{19k - 3} > \frac{1}{4} \;\Rightarrow\; 16k + 8 > 19k - 3 \;\Rightarrow\; 11 > 3k \;\Rightarrow\; k < 3\frac{2}{3}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{4k}{19k - 3}

.

\frac{4}{16}, \quad \frac{8}{35}, \quad \frac{12}{54}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{4}{16}

and

\frac{12}{54}

can be reduced, because

4

and

16

share the factor

4

, and

12

and

54

share the factor

6

. The numbers

8

and

35

share no factor other than

1

.

\gcd(4, 16) = 4, \quad \gcd(8, 35) = 1, \quad \gcd(12, 54) = 6

Only

\frac{8}{35}

is already in lowest terms.

Answer:

\frac{8}{35}

4 · Reviewdoes it hold up?

Adding $3$ to the denominator: $\frac{8}{38} = \frac{4}{19}$ . Adding $2$ to the numerator: $\frac{10}{35} > \frac{1}{4}$ , since $10 \times 4 = 40 > 35 = 1 \times 35$ . Both conditions hold, and $\frac{8}{35}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{4}{19}$ one by one: $\frac{4}{19}, \frac{8}{38}, \frac{12}{57}, \frac{16}{76}, \ldots$ . Subtract $3$ from each denominator to get $\frac{4}{16}, \frac{8}{35}, \frac{12}{54}, \frac{16}{73}, \ldots$ and test each: adding $2$ to the numerator gives more than $\frac{1}{4}$ only for the first three, and of those only $\frac{8}{35}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $4$ and $19$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $16k + 8 > 19k - 3$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 10 hard

A fraction in lowest terms has a whole-number numerator and denominator. Adding $2$ to its denominator gives a fraction equal to $\frac{5}{19}$ , and adding $4$ to its numerator gives a fraction greater than $\frac{1}{3}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $2$ to the denominator makes it equal $\frac{5}{19}$ and adding $4$ to the numerator makes it larger than $\frac{1}{3}$ .

Givens

Numerator over (denominator $+ 2$ ) equals $\frac{5}{19}$ .
(Numerator $+ 4$ ) over denominator is greater than $\frac{1}{3}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 2} = \frac{5}{19}, \quad \text{so} \quad 19y = 5(x + 2)

Cross-multiplying turns the equal fractions into one equation.

2Use that 5 and 19 share no factor

#13 Convert to Algebra 6.NS.B.4

5

divides

19y

but shares no factor with

19

, so

5

divides

y

. Write

y = 5k

; then

x + 2 = 19k

for the same whole number

k

.

y = 5k, \quad x = 19k - 2 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{5}{19}

is

\frac{5k}{19k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

4

to the numerator must give more than

\frac{1}{3}

. The denominator

19k - 2

is positive, so multiplying both sides by

3(19k - 2)

keeps the direction.

\frac{5k + 4}{19k - 2} > \frac{1}{3} \;\Rightarrow\; 15k + 12 > 19k - 2 \;\Rightarrow\; 14 > 4k \;\Rightarrow\; k < 3\frac{1}{2}

The inequality leaves only

k = 1, 2, 3

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3

into

\frac{5k}{19k - 2}

.

\frac{5}{17}, \quad \frac{10}{36}, \quad \frac{15}{55}

Only three fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{10}{36}

and

\frac{15}{55}

can be reduced, because

10

and

36

share the factor

2

, and

15

and

55

share the factor

5

. The numbers

5

and

17

share no factor other than

1

.

\gcd(5, 17) = 1, \quad \gcd(10, 36) = 2, \quad \gcd(15, 55) = 5

Only

\frac{5}{17}

is already in lowest terms.

Answer:

\frac{5}{17}

4 · Reviewdoes it hold up?

Adding $2$ to the denominator: $\frac{5}{19} = \frac{5}{19}$ . Adding $4$ to the numerator: $\frac{9}{17} > \frac{1}{3}$ , since $9 \times 3 = 27 > 17 = 1 \times 17$ . Both conditions hold, and $\frac{5}{17}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{5}{19}$ one by one: $\frac{5}{19}, \frac{10}{38}, \frac{15}{57}, \frac{20}{76}, \ldots$ . Subtract $2$ from each denominator to get $\frac{5}{17}, \frac{10}{36}, \frac{15}{55}, \frac{20}{74}, \ldots$ and test each: adding $4$ to the numerator gives more than $\frac{1}{3}$ only for the first three, and of those only $\frac{5}{17}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $5$ and $19$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $15k + 12 > 19k - 2$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 11 hard

A fraction in lowest terms has a whole-number numerator and denominator. Adding $6$ to its denominator gives a fraction equal to $\frac{5}{22}$ , and adding $1$ to its numerator gives a fraction greater than $\frac{1}{4}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $6$ to the denominator makes it equal $\frac{5}{22}$ and adding $1$ to the numerator makes it larger than $\frac{1}{4}$ .

Givens

Numerator over (denominator $+ 6$ ) equals $\frac{5}{22}$ .
(Numerator $+ 1$ ) over denominator is greater than $\frac{1}{4}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 6} = \frac{5}{22}, \quad \text{so} \quad 22y = 5(x + 6)

Cross-multiplying turns the equal fractions into one equation.

2Use that 5 and 22 share no factor

#13 Convert to Algebra 6.NS.B.4

5

divides

22y

but shares no factor with

22

, so

5

divides

y

. Write

y = 5k

; then

x + 6 = 22k

for the same whole number

k

.

y = 5k, \quad x = 22k - 6 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{5}{22}

is

\frac{5k}{22k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

1

to the numerator must give more than

\frac{1}{4}

. The denominator

22k - 6

is positive, so multiplying both sides by

4(22k - 6)

keeps the direction.

\frac{5k + 1}{22k - 6} > \frac{1}{4} \;\Rightarrow\; 20k + 4 > 22k - 6 \;\Rightarrow\; 10 > 2k \;\Rightarrow\; k < 5

The inequality leaves only

k = 1, 2, 3, 4

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3, 4

into

\frac{5k}{22k - 6}

.

\frac{5}{16}, \quad \frac{10}{38}, \quad \frac{15}{60}, \quad \frac{20}{82}

Only four fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{10}{38}

,

\frac{15}{60}

, and

\frac{20}{82}

can be reduced, because

10

and

38

share the factor

2

,

15

and

60

share the factor

15

, and

20

and

82

share the factor

2

. The numbers

5

and

16

share no factor other than

1

.

\gcd(5, 16) = 1, \quad \gcd(10, 38) = 2, \quad \gcd(15, 60) = 15, \quad \gcd(20, 82) = 2

Only

\frac{5}{16}

is already in lowest terms.

Answer:

\frac{5}{16}

4 · Reviewdoes it hold up?

Adding $6$ to the denominator: $\frac{5}{22} = \frac{5}{22}$ . Adding $1$ to the numerator: $\frac{6}{16} > \frac{4}{16} = \frac{1}{4}$ . Both conditions hold, and $\frac{5}{16}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{5}{22}$ one by one: $\frac{5}{22}, \frac{10}{44}, \frac{15}{66}, \frac{20}{88}, \frac{25}{110}, \ldots$ . Subtract $6$ from each denominator to get $\frac{5}{16}, \frac{10}{38}, \frac{15}{60}, \frac{20}{82}, \frac{25}{104}, \ldots$ and test each: adding $1$ to the numerator gives more than $\frac{1}{4}$ only for the first four, and of those only $\frac{5}{16}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $5$ and $22$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $20k + 4 > 22k - 6$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Variant 12 hard

A fraction in lowest terms has a whole-number numerator and denominator. Adding $6$ to its denominator gives a fraction equal to $\frac{7}{24}$ , and adding $3$ to its numerator gives a fraction greater than $\frac{1}{3}$ . Find the fraction.

Your answer

Show solution

1 · Understandwhat's really being asked

Find a fraction in lowest terms, with whole-number numerator and denominator, such that adding $6$ to the denominator makes it equal $\frac{7}{24}$ and adding $3$ to the numerator makes it larger than $\frac{1}{3}$ .

Givens

Numerator over (denominator $+ 6$ ) equals $\frac{7}{24}$ .
(Numerator $+ 3$ ) over denominator is greater than $\frac{1}{3}$ .
The fraction is in lowest terms.

Unknowns

The fraction.

Constraints

Numerator and denominator are whole numbers with no common factor other than $1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

The equation pins down the shape of the fraction up to one whole number $k$ . The inequality then limits $k$ to a few values, and the lowest-terms condition picks out the one that works.

3 · Execute5 carry out the plan

1Name the fraction

#13 Convert to Algebra 7.EE.B.4

Call the fraction

\frac{y}{x}

, where

x

and

y

are whole numbers with no common factor. The first condition says:

\frac{y}{x + 6} = \frac{7}{24}, \quad \text{so} \quad 24y = 7(x + 6)

Cross-multiplying turns the equal fractions into one equation.

2Use that 7 and 24 share no factor

#13 Convert to Algebra 6.NS.B.4

7

divides

24y

but shares no factor with

24

, so

7

divides

y

. Write

y = 7k

; then

x + 6 = 24k

for the same whole number

k

.

y = 7k, \quad x = 24k - 6 \quad (k = 1, 2, 3, \ldots)

Every fraction equal to

\frac{7}{24}

is

\frac{7k}{24k}

for some whole number

k

.

3Turn the second condition into an inequality in k

#13 Convert to Algebra 7.EE.B.4

Adding

3

to the numerator must give more than

\frac{1}{3}

. The denominator

24k - 6

is positive, so multiplying both sides by

3(24k - 6)

keeps the direction.

\frac{7k + 3}{24k - 6} > \frac{1}{3} \;\Rightarrow\; 21k + 9 > 24k - 6 \;\Rightarrow\; 15 > 3k \;\Rightarrow\; k < 5

The inequality leaves only

k = 1, 2, 3, 4

.

4List the candidates

#2 Make a Systematic List 7.EE.B.4

Put

k = 1, 2, 3, 4

into

\frac{7k}{24k - 6}

.

\frac{7}{18}, \quad \frac{14}{42}, \quad \frac{21}{66}, \quad \frac{28}{90}

Only four fractions satisfy both conditions.

5Keep the one in lowest terms

#3 Eliminate Possibilities 6.NS.B.4

\frac{14}{42}

,

\frac{21}{66}

, and

\frac{28}{90}

can be reduced, because

14

and

42

share the factor

14

,

21

and

66

share the factor

3

, and

28

and

90

share the factor

2

. The numbers

7

and

18

share no factor other than

1

.

\gcd(7, 18) = 1, \quad \gcd(14, 42) = 14, \quad \gcd(21, 66) = 3, \quad \gcd(28, 90) = 2

Only

\frac{7}{18}

is already in lowest terms.

Answer:

\frac{7}{18}

4 · Reviewdoes it hold up?

Adding $6$ to the denominator: $\frac{7}{24} = \frac{7}{24}$ . Adding $3$ to the numerator: $\frac{10}{18} > \frac{6}{18} = \frac{1}{3}$ . Both conditions hold, and $\frac{7}{18}$ cannot be reduced.

Another way: Write the fractions equal to $\frac{7}{24}$ one by one: $\frac{7}{24}, \frac{14}{48}, \frac{21}{72}, \frac{28}{96}, \frac{35}{120}, \ldots$ . Subtract $6$ from each denominator to get $\frac{7}{18}, \frac{14}{42}, \frac{21}{66}, \frac{28}{90}, \frac{35}{114}, \ldots$ and test each: adding $3$ to the numerator gives more than $\frac{1}{3}$ only for the first four, and of those only $\frac{7}{18}$ is in lowest terms.

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using that $7$ and $24$ share no factor, and checking which candidate fractions are in lowest terms.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the two conditions as an equation and an inequality in the numerator and denominator, and solving $21k + 9 > 24k - 6$.

💡Takeaway. An equation tells you the shape of the answer, an inequality tells you how far to look, and lowest terms tells you which one to keep.

Equation fixes the form, inequality bounds it · 12 practice problems

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