← Same numerator: smaller denominator is larger · Find Two Unknowns from Sum and Difference

Same numerator: smaller denominator is larger · 12 practice problems

3.NF.A.33.OA.A.44.OA.A.1

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: A = 3, B = 13

Find the values of AA and BB that satisfy both of the following equations.

AB+2=15AB+8=17\frac{A}{B+2}=\frac{1}{5}\qquad\frac{A}{B+8}=\frac{1}{7}

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
  • AB+2=15\frac{A}{B+2} = \frac{1}{5}.
  • AB+8=17\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,B+8=7AB + 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=1:5and7,2:10and14,3:15and21A = 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⇒2A=6⇒A=37A - 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=5×3−2=13,315=15,321=17B = 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 answer: A = 5, B = 16

Find the values of AA and BB that satisfy both of the following equations.

AB+4=14AB+9=15\frac{A}{B+4}=\frac{1}{4}\qquad\frac{A}{B+9}=\frac{1}{5}

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
  • AB+4=14\frac{A}{B+4} = \frac{1}{4}.
  • AB+9=15\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,B+9=5AB + 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=3:12and15,4:16and20,5:20and25A = 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⇒1A=5⇒A=55A - 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=4×5−4=16,520=14,525=15B = 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 answer: A = 2, B = 8

Find the values of AA and BB that satisfy both of the following equations.

AB+6=17AB+10=19\frac{A}{B+6}=\frac{1}{7}\qquad\frac{A}{B+10}=\frac{1}{9}

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
  • AB+6=17\frac{A}{B+6} = \frac{1}{7}.
  • AB+10=19\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,B+10=9AB + 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=1:7and9,2:14and18A = 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⇒2A=4⇒A=29A - 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=7×2−6=8,214=17,218=19B = 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 answer: A = 5, B = 26

Find the values of AA and BB that satisfy both of the following equations.

AB+4=16AB+14=18\frac{A}{B+4}=\frac{1}{6}\qquad\frac{A}{B+14}=\frac{1}{8}

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
  • AB+4=16\frac{A}{B+4} = \frac{1}{6}.
  • AB+14=18\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,B+14=8AB + 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=3:18and24,4:24and32,5:30and40A = 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⇒2A=10⇒A=58A - 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=6×5−4=26,530=16,540=18B = 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 answer: A = 3, B = 19

Find the values of AA and BB that satisfy both of the following equations.

AB+5=18AB+14=111\frac{A}{B+5}=\frac{1}{8}\qquad\frac{A}{B+14}=\frac{1}{11}

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
  • AB+5=18\frac{A}{B+5} = \frac{1}{8}.
  • AB+14=111\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,B+14=11AB + 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=1:8and11,2:16and22,3:24and33A = 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⇒3A=9⇒A=311A - 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=8×3−5=19,324=18,333=111B = 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 answer: A = 6, B = 21

Find the values of AA and BB that satisfy both of the following equations.

AB+3=14AB+15=16\frac{A}{B+3}=\frac{1}{4}\qquad\frac{A}{B+15}=\frac{1}{6}

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
  • AB+3=14\frac{A}{B+3} = \frac{1}{4}.
  • AB+15=16\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,B+15=6AB + 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=4:16and24,5:20and30,6:24and36A = 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⇒2A=12⇒A=66A - 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=4×6−3=21,624=14,636=16B = 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 answer: A = 4, B = 7

Find the values of AA and BB that satisfy both of the following equations.

AB+5=13AB+17=16\frac{A}{B+5}=\frac{1}{3}\qquad\frac{A}{B+17}=\frac{1}{6}

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
  • AB+5=13\frac{A}{B+5} = \frac{1}{3}.
  • AB+17=16\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,B+17=6AB + 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=2:6and12,3:9and18,4:12and24A = 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⇒3A=12⇒A=46A - 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=3×4−5=7,412=13,424=16B = 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 answer: A = 9, B = 10

Find the values of AA and BB that satisfy both of the following equations.

AB+8=12AB+17=13\frac{A}{B+8}=\frac{1}{2}\qquad\frac{A}{B+17}=\frac{1}{3}

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
  • AB+8=12\frac{A}{B+8} = \frac{1}{2}.
  • AB+17=13\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,B+17=3AB + 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=7:14and21,8:16and24,9:18and27A = 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⇒1A=9⇒A=93A - 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=2×9−8=10,918=12,927=13B = 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 answer: A = 4, B = 14

Find the values of AA and BB that satisfy both of the following equations.

AB+6=15AB+22=19\frac{A}{B+6}=\frac{1}{5}\qquad\frac{A}{B+22}=\frac{1}{9}

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
  • AB+6=15\frac{A}{B+6} = \frac{1}{5}.
  • AB+22=19\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,B+22=9AB + 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=2:10and18,3:15and27,4:20and36A = 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⇒4A=16⇒A=49A - 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=5×4−6=14,420=15,436=19B = 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 answer: A = 7, B = 12

Find the values of AA and BB that satisfy both of the following equations.

AB+9=13AB+23=15\frac{A}{B+9}=\frac{1}{3}\qquad\frac{A}{B+23}=\frac{1}{5}

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
  • AB+9=13\frac{A}{B+9} = \frac{1}{3}.
  • AB+23=15\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,B+23=5AB + 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=5:15and25,6:18and30,7:21and35A = 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⇒2A=14⇒A=75A - 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=3×7−9=12,721=13,735=15B = 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 answer: A = 8, B = 9

Find the values of AA and BB that satisfy both of the following equations.

AB+7=12AB+23=14\frac{A}{B+7}=\frac{1}{2}\qquad\frac{A}{B+23}=\frac{1}{4}

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
  • AB+7=12\frac{A}{B+7} = \frac{1}{2}.
  • AB+23=14\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,B+23=4AB + 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=6:12and24,7:14and28,8:16and32A = 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⇒2A=16⇒A=84A - 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=2×8−7=9,816=12,832=14B = 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 answer: A = 6, B = 18

Find the values of AA and BB that satisfy both of the following equations.

AB+12=15AB+24=17\frac{A}{B+12}=\frac{1}{5}\qquad\frac{A}{B+24}=\frac{1}{7}

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
  • AB+12=15\frac{A}{B+12} = \frac{1}{5}.
  • AB+24=17\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,B+24=7AB + 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=4:20and28,5:25and35,6:30and42A = 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⇒2A=12⇒A=67A - 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=5×6−12=18,630=15,642=17B = 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.