Practice · Same numerator: smaller denominator is larger · Sensim Math

Variant 1 easy

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+2}=\frac{1}{5}\qquad\frac{A}{B+8}=\frac{1}{7}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 5 when the bottom is B plus 2; the other equals one over 7 when the bottom is B plus 8. We need both A and B.

Givens

$\frac{A}{B+2} = \frac{1}{5}$ .
$\frac{A}{B+8} = \frac{1}{7}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 5 exactly when that something is 5 times A. The same for the second.

B + 2 = 5A,\qquad B + 8 = 7A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 5A and 7A.

A = 1: 5 and 7, 2: 10 and 14, 3: 15 and 21

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=3

#5 Look for a Pattern 4.OA.A.1

The second bottom is 6 more than the first, and going from 5A to 7A adds 2A. So 2A must be 6.

7A - 5A = 8 - 2 \Rightarrow 2A = 6 \Rightarrow A = 3

A is 3.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 5 \times 3 - 2 = 13,\quad \frac{3}{15} = \frac{1}{5},\quad \frac{3}{21} = \frac{1}{7}

A = 3 and B = 13.

Answer: A = 3, B = 13

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 7 against one over 5 -- and its denominator 21 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 7A - 5A = 6, giving A = 3 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 2 easy

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+4}=\frac{1}{4}\qquad\frac{A}{B+9}=\frac{1}{5}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 4 when the bottom is B plus 4; the other equals one over 5 when the bottom is B plus 9. We need both A and B.

Givens

$\frac{A}{B+4} = \frac{1}{4}$ .
$\frac{A}{B+9} = \frac{1}{5}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 4 exactly when that something is 4 times A. The same for the second.

B + 4 = 4A,\qquad B + 9 = 5A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 4A and 5A.

A = 3: 12 and 15, 4: 16 and 20, 5: 20 and 25

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=5

#5 Look for a Pattern 4.OA.A.1

The second bottom is 5 more than the first, and going from 4A to 5A adds 1A. So 1A must be 5.

5A - 4A = 9 - 4 \Rightarrow 1A = 5 \Rightarrow A = 5

A is 5.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 4 \times 5 - 4 = 16,\quad \frac{5}{20} = \frac{1}{4},\quad \frac{5}{25} = \frac{1}{5}

A = 5 and B = 16.

Answer: A = 5, B = 16

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 5 against one over 4 -- and its denominator 25 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 5A - 4A = 5, giving A = 5 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 1 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 3 easy

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+6}=\frac{1}{7}\qquad\frac{A}{B+10}=\frac{1}{9}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 7 when the bottom is B plus 6; the other equals one over 9 when the bottom is B plus 10. We need both A and B.

Givens

$\frac{A}{B+6} = \frac{1}{7}$ .
$\frac{A}{B+10} = \frac{1}{9}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 7 exactly when that something is 7 times A. The same for the second.

B + 6 = 7A,\qquad B + 10 = 9A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 7A and 9A.

A = 1: 7 and 9, 2: 14 and 18

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=2

#5 Look for a Pattern 4.OA.A.1

The second bottom is 4 more than the first, and going from 7A to 9A adds 2A. So 2A must be 4.

9A - 7A = 10 - 6 \Rightarrow 2A = 4 \Rightarrow A = 2

A is 2.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 7 \times 2 - 6 = 8,\quad \frac{2}{14} = \frac{1}{7},\quad \frac{2}{18} = \frac{1}{9}

A = 2 and B = 8.

Answer: A = 2, B = 8

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 9 against one over 7 -- and its denominator 18 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 9A - 7A = 4, giving A = 2 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 4 easy

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+4}=\frac{1}{6}\qquad\frac{A}{B+14}=\frac{1}{8}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 6 when the bottom is B plus 4; the other equals one over 8 when the bottom is B plus 14. We need both A and B.

Givens

$\frac{A}{B+4} = \frac{1}{6}$ .
$\frac{A}{B+14} = \frac{1}{8}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 6 exactly when that something is 6 times A. The same for the second.

B + 4 = 6A,\qquad B + 14 = 8A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 6A and 8A.

A = 3: 18 and 24, 4: 24 and 32, 5: 30 and 40

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=5

#5 Look for a Pattern 4.OA.A.1

The second bottom is 10 more than the first, and going from 6A to 8A adds 2A. So 2A must be 10.

8A - 6A = 14 - 4 \Rightarrow 2A = 10 \Rightarrow A = 5

A is 5.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 6 \times 5 - 4 = 26,\quad \frac{5}{30} = \frac{1}{6},\quad \frac{5}{40} = \frac{1}{8}

A = 5 and B = 26.

Answer: A = 5, B = 26

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 8 against one over 6 -- and its denominator 40 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 8A - 6A = 10, giving A = 5 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 5 medium

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+5}=\frac{1}{8}\qquad\frac{A}{B+14}=\frac{1}{11}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 8 when the bottom is B plus 5; the other equals one over 11 when the bottom is B plus 14. We need both A and B.

Givens

$\frac{A}{B+5} = \frac{1}{8}$ .
$\frac{A}{B+14} = \frac{1}{11}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 8 exactly when that something is 8 times A. The same for the second.

B + 5 = 8A,\qquad B + 14 = 11A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 8A and 11A.

A = 1: 8 and 11, 2: 16 and 22, 3: 24 and 33

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=3

#5 Look for a Pattern 4.OA.A.1

The second bottom is 9 more than the first, and going from 8A to 11A adds 3A. So 3A must be 9.

11A - 8A = 14 - 5 \Rightarrow 3A = 9 \Rightarrow A = 3

A is 3.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 8 \times 3 - 5 = 19,\quad \frac{3}{24} = \frac{1}{8},\quad \frac{3}{33} = \frac{1}{11}

A = 3 and B = 19.

Answer: A = 3, B = 19

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 11 against one over 8 -- and its denominator 33 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 11A - 8A = 9, giving A = 3 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 3 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 6 medium

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+3}=\frac{1}{4}\qquad\frac{A}{B+15}=\frac{1}{6}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 4 when the bottom is B plus 3; the other equals one over 6 when the bottom is B plus 15. We need both A and B.

Givens

$\frac{A}{B+3} = \frac{1}{4}$ .
$\frac{A}{B+15} = \frac{1}{6}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 4 exactly when that something is 4 times A. The same for the second.

B + 3 = 4A,\qquad B + 15 = 6A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 4A and 6A.

A = 4: 16 and 24, 5: 20 and 30, 6: 24 and 36

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=6

#5 Look for a Pattern 4.OA.A.1

The second bottom is 12 more than the first, and going from 4A to 6A adds 2A. So 2A must be 12.

6A - 4A = 15 - 3 \Rightarrow 2A = 12 \Rightarrow A = 6

A is 6.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 4 \times 6 - 3 = 21,\quad \frac{6}{24} = \frac{1}{4},\quad \frac{6}{36} = \frac{1}{6}

A = 6 and B = 21.

Answer: A = 6, B = 21

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 6 against one over 4 -- and its denominator 36 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 6A - 4A = 12, giving A = 6 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 7 medium

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+5}=\frac{1}{3}\qquad\frac{A}{B+17}=\frac{1}{6}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 3 when the bottom is B plus 5; the other equals one over 6 when the bottom is B plus 17. We need both A and B.

Givens

$\frac{A}{B+5} = \frac{1}{3}$ .
$\frac{A}{B+17} = \frac{1}{6}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 3 exactly when that something is 3 times A. The same for the second.

B + 5 = 3A,\qquad B + 17 = 6A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 3A and 6A.

A = 2: 6 and 12, 3: 9 and 18, 4: 12 and 24

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=4

#5 Look for a Pattern 4.OA.A.1

The second bottom is 12 more than the first, and going from 3A to 6A adds 3A. So 3A must be 12.

6A - 3A = 17 - 5 \Rightarrow 3A = 12 \Rightarrow A = 4

A is 4.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 3 \times 4 - 5 = 7,\quad \frac{4}{12} = \frac{1}{3},\quad \frac{4}{24} = \frac{1}{6}

A = 4 and B = 7.

Answer: A = 4, B = 7

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 6 against one over 3 -- and its denominator 24 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 6A - 3A = 12, giving A = 4 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 3 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 8 medium

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+8}=\frac{1}{2}\qquad\frac{A}{B+17}=\frac{1}{3}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 2 when the bottom is B plus 8; the other equals one over 3 when the bottom is B plus 17. We need both A and B.

Givens

$\frac{A}{B+8} = \frac{1}{2}$ .
$\frac{A}{B+17} = \frac{1}{3}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 2 exactly when that something is 2 times A. The same for the second.

B + 8 = 2A,\qquad B + 17 = 3A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 2A and 3A.

A = 7: 14 and 21, 8: 16 and 24, 9: 18 and 27

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=9

#5 Look for a Pattern 4.OA.A.1

The second bottom is 9 more than the first, and going from 2A to 3A adds 1A. So 1A must be 9.

3A - 2A = 17 - 8 \Rightarrow 1A = 9 \Rightarrow A = 9

A is 9.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 2 \times 9 - 8 = 10,\quad \frac{9}{18} = \frac{1}{2},\quad \frac{9}{27} = \frac{1}{3}

A = 9 and B = 10.

Answer: A = 9, B = 10

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 3 against one over 2 -- and its denominator 27 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 3A - 2A = 9, giving A = 9 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 1 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 9 hard

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+6}=\frac{1}{5}\qquad\frac{A}{B+22}=\frac{1}{9}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 5 when the bottom is B plus 6; the other equals one over 9 when the bottom is B plus 22. We need both A and B.

Givens

$\frac{A}{B+6} = \frac{1}{5}$ .
$\frac{A}{B+22} = \frac{1}{9}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 5 exactly when that something is 5 times A. The same for the second.

B + 6 = 5A,\qquad B + 22 = 9A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 5A and 9A.

A = 2: 10 and 18, 3: 15 and 27, 4: 20 and 36

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=4

#5 Look for a Pattern 4.OA.A.1

The second bottom is 16 more than the first, and going from 5A to 9A adds 4A. So 4A must be 16.

9A - 5A = 22 - 6 \Rightarrow 4A = 16 \Rightarrow A = 4

A is 4.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 5 \times 4 - 6 = 14,\quad \frac{4}{20} = \frac{1}{5},\quad \frac{4}{36} = \frac{1}{9}

A = 4 and B = 14.

Answer: A = 4, B = 14

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 9 against one over 5 -- and its denominator 36 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 9A - 5A = 16, giving A = 4 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 4 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 10 hard

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+9}=\frac{1}{3}\qquad\frac{A}{B+23}=\frac{1}{5}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 3 when the bottom is B plus 9; the other equals one over 5 when the bottom is B plus 23. We need both A and B.

Givens

$\frac{A}{B+9} = \frac{1}{3}$ .
$\frac{A}{B+23} = \frac{1}{5}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 3 exactly when that something is 3 times A. The same for the second.

B + 9 = 3A,\qquad B + 23 = 5A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 3A and 5A.

A = 5: 15 and 25, 6: 18 and 30, 7: 21 and 35

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=7

#5 Look for a Pattern 4.OA.A.1

The second bottom is 14 more than the first, and going from 3A to 5A adds 2A. So 2A must be 14.

5A - 3A = 23 - 9 \Rightarrow 2A = 14 \Rightarrow A = 7

A is 7.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 3 \times 7 - 9 = 12,\quad \frac{7}{21} = \frac{1}{3},\quad \frac{7}{35} = \frac{1}{5}

A = 7 and B = 12.

Answer: A = 7, B = 12

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 5 against one over 3 -- and its denominator 35 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 5A - 3A = 14, giving A = 7 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 11 hard

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+7}=\frac{1}{2}\qquad\frac{A}{B+23}=\frac{1}{4}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 2 when the bottom is B plus 7; the other equals one over 4 when the bottom is B plus 23. We need both A and B.

Givens

$\frac{A}{B+7} = \frac{1}{2}$ .
$\frac{A}{B+23} = \frac{1}{4}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 2 exactly when that something is 2 times A. The same for the second.

B + 7 = 2A,\qquad B + 23 = 4A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 2A and 4A.

A = 6: 12 and 24, 7: 14 and 28, 8: 16 and 32

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=8

#5 Look for a Pattern 4.OA.A.1

The second bottom is 16 more than the first, and going from 2A to 4A adds 2A. So 2A must be 16.

4A - 2A = 23 - 7 \Rightarrow 2A = 16 \Rightarrow A = 8

A is 8.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 2 \times 8 - 7 = 9,\quad \frac{8}{16} = \frac{1}{2},\quad \frac{8}{32} = \frac{1}{4}

A = 8 and B = 9.

Answer: A = 8, B = 9

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 4 against one over 2 -- and its denominator 32 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 4A - 2A = 16, giving A = 8 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Variant 12 hard

Find the values of $A$ and $B$ that satisfy both of the following equations.

$\frac{A}{B+12}=\frac{1}{5}\qquad\frac{A}{B+24}=\frac{1}{7}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two fractions share the same top number A. One equals one over 5 when the bottom is B plus 12; the other equals one over 7 when the bottom is B plus 24. We need both A and B.

Givens

$\frac{A}{B+12} = \frac{1}{5}$ .
$\frac{A}{B+24} = \frac{1}{7}$ .
A and B are the same numbers in both equations.

Unknowns

The values of A and B.

Constraints

Both are whole numbers, and each fraction must reduce exactly.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

Each equation says the bottom is a fixed multiple of A. Since B is shared, the two bottoms sit a known distance apart, so trying A values quickly shows which one fits that distance.

3 · Execute4 carry out the plan

1Turn each fraction equation into a denominator rule

#7 Identify Subproblems 3.NF.A.3

A over something equals one over 5 exactly when that something is 5 times A. The same for the second.

B + 12 = 5A,\qquad B + 24 = 7A

Two statements about B, both in terms of A.

2Guess values of A and build both denominators

#6 Guess and Check 3.OA.A.4

For each A, the two denominators would be 5A and 7A.

A = 4: 20 and 28, 5: 25 and 35, 6: 30 and 42

Only one of these fits the gap the equations demand.

3See the jump pattern and confirm A=6

#5 Look for a Pattern 4.OA.A.1

The second bottom is 12 more than the first, and going from 5A to 7A adds 2A. So 2A must be 12.

7A - 5A = 24 - 12 \Rightarrow 2A = 12 \Rightarrow A = 6

A is 6.

4Find B and double-check both equations

#6 Guess and Check 3.NF.A.3

With A known, the first rule gives B directly.

B = 5 \times 6 - 12 = 18,\quad \frac{6}{30} = \frac{1}{5},\quad \frac{6}{42} = \frac{1}{7}

A = 6 and B = 18.

Answer: A = 6, B = 18

4 · Reviewdoes it hold up?

The second fraction is the smaller one -- one over 7 against one over 5 -- and its denominator 42 is indeed the bigger, which is how same-numerator fractions behave.

Another way: Subtracting the two denominator rules removes B in one line: 7A - 5A = 12, giving A = 6 without any guessing at all.

Standardsmin grade 4

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Reading each fraction equation as a statement about its denominator.
3.OA.A.4 Determine unknown whole number in multiplication or division equation — Trying candidate values of A against both rules.
4.OA.A.1 Interpret a multiplication equation as a comparison — Reading the gap between the denominators as 2 times A.

💡Takeaway. When the same unknown sits in two equations, subtracting them cancels whatever they share and leaves one thing to find.

Same numerator: smaller denominator is larger · 12 practice problems

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