Practice · Rewrite numerators using the given total · Sensim Math

Variant 1 easy

The constants $a$ , $b$ , $c$ satisfy $2a+b+3c=12$ , with $abc\neq 0$ and $\frac{1}{2a}+\frac{1}{b}+\frac{1}{3c}\neq 0$ . Solve the equation $\frac{x}{2a}+\frac{x}{b}+\frac{x}{3c}-6=\frac{2b+6c}{2a}+\frac{4a+6c}{b}+\frac{4a+2b}{3c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $2a+b+3c=12$ . Each numerator on the right side uses two of the three pieces $2a$ , $b$ , $3c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$2a+b+3c=12$ .
$abc\neq 0$ and $\frac{1}{2a}+\frac{1}{b}+\frac{1}{3c}\neq 0$ .
The equation $\frac{x}{2a}+\frac{x}{b}+\frac{x}{3c}-6=\frac{2b+6c}{2a}+\frac{4a+6c}{b}+\frac{4a+2b}{3c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $2b+6c$ is twice $b+3c$ , which is the total $12$ minus $2a$ . Rewriting all three this way makes the right side look exactly like the left side, with $24$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

b+3c=12-2a

, so double it.

2b+6c=2(12-2a)=24-4a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

2a+3c=12-b

and

2a+b=12-3c

.

4a+6c=24-2b, \quad 4a+2b=24-6c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

24

over its denominator minus

2

.

\frac{24-4a}{2a}+\frac{24-2b}{b}+\frac{24-6c}{3c}=\frac{24}{2a}+\frac{24}{b}+\frac{24}{3c}-6

\frac{4a}{2a}=\frac{2b}{b}=\frac{6c}{3c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{2a}+\frac{1}{b}+\frac{1}{3c}\right)=24\left(\frac{1}{2a}+\frac{1}{b}+\frac{1}{3c}\right)

. The bracket is not

0

, so divide by it.

x=24

Dividing by a nonzero number keeps the equation true.

Answer:

x=24

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=5$ , $c=1$ satisfy $4+5+3=12$ . The left side at $x=24$ is $\frac{24}{4}+\frac{24}{5}+\frac{24}{3}-6=\frac{64}{5}$ . The right side is $\frac{16}{4}+\frac{14}{5}+\frac{18}{3}=\frac{64}{5}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=5$ , $c=1$ . The equation becomes $\frac{47}{60}x-6=\frac{64}{5}$ , so $\frac{47}{60}x=\frac{94}{5}$ and $x=24$ . Since the problem promises a single answer for all such constants, this sample already shows it is $24$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $24$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 2 easy

The constants $a$ , $b$ , $c$ satisfy $3a+b+2c=15$ , with $abc\neq 0$ and $\frac{1}{3a}+\frac{1}{b}+\frac{1}{2c}\neq 0$ . Solve the equation $\frac{x}{3a}+\frac{x}{b}+\frac{x}{2c}-6=\frac{2b+4c}{3a}+\frac{6a+4c}{b}+\frac{6a+2b}{2c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $3a+b+2c=15$ . Each numerator on the right side uses two of the three pieces $3a$ , $b$ , $2c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$3a+b+2c=15$ .
$abc\neq 0$ and $\frac{1}{3a}+\frac{1}{b}+\frac{1}{2c}\neq 0$ .
The equation $\frac{x}{3a}+\frac{x}{b}+\frac{x}{2c}-6=\frac{2b+4c}{3a}+\frac{6a+4c}{b}+\frac{6a+2b}{2c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $2b+4c$ is twice $b+2c$ , which is the total $15$ minus $3a$ . Rewriting all three this way makes the right side look exactly like the left side, with $30$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

b+2c=15-3a

, so double it.

2b+4c=2(15-3a)=30-6a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

3a+2c=15-b

and

3a+b=15-2c

.

6a+4c=30-2b, \quad 6a+2b=30-4c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

30

over its denominator minus

2

.

\frac{30-6a}{3a}+\frac{30-2b}{b}+\frac{30-4c}{2c}=\frac{30}{3a}+\frac{30}{b}+\frac{30}{2c}-6

\frac{6a}{3a}=\frac{2b}{b}=\frac{4c}{2c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{3a}+\frac{1}{b}+\frac{1}{2c}\right)=30\left(\frac{1}{3a}+\frac{1}{b}+\frac{1}{2c}\right)

. The bracket is not

0

, so divide by it.

x=30

Dividing by a nonzero number keeps the equation true.

Answer:

x=30

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=7$ , $c=1$ satisfy $6+7+2=15$ . The left side at $x=30$ is $\frac{30}{6}+\frac{30}{7}+\frac{30}{2}-6=\frac{128}{7}$ . The right side is $\frac{18}{6}+\frac{16}{7}+\frac{26}{2}=\frac{128}{7}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=7$ , $c=1$ . The equation becomes $\frac{17}{21}x-6=\frac{128}{7}$ , so $\frac{17}{21}x=\frac{170}{7}$ and $x=30$ . Since the problem promises a single answer for all such constants, this sample already shows it is $30$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $30$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 3 easy

The constants $a$ , $b$ , $c$ satisfy $3a+2b+c=20$ , with $abc\neq 0$ and $\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{3a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{3a}+\frac{6a+2c}{2b}+\frac{6a+4b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $3a+2b+c=20$ . Each numerator on the right side uses two of the three pieces $3a$ , $2b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$3a+2b+c=20$ .
$abc\neq 0$ and $\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{3a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{3a}+\frac{6a+2c}{2b}+\frac{6a+4b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $4b+2c$ is twice $2b+c$ , which is the total $20$ minus $3a$ . Rewriting all three this way makes the right side look exactly like the left side, with $40$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

2b+c=20-3a

, so double it.

4b+2c=2(20-3a)=40-6a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

3a+c=20-2b

and

3a+2b=20-c

.

6a+2c=40-4b, \quad 6a+4b=40-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

40

over its denominator minus

2

.

\frac{40-6a}{3a}+\frac{40-4b}{2b}+\frac{40-2c}{c}=\frac{40}{3a}+\frac{40}{2b}+\frac{40}{c}-6

\frac{6a}{3a}=\frac{4b}{2b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\right)=40\left(\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=40

Dividing by a nonzero number keeps the equation true.

Answer:

x=40

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=8$ satisfy $6+6+8=20$ . The left side at $x=40$ is $\frac{40}{6}+\frac{40}{6}+\frac{40}{8}-6=\frac{37}{3}$ . The right side is $\frac{28}{6}+\frac{28}{6}+\frac{24}{8}=\frac{37}{3}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=8$ . The equation becomes $\frac{11}{24}x-6=\frac{37}{3}$ , so $\frac{11}{24}x=\frac{55}{3}$ and $x=40$ . Since the problem promises a single answer for all such constants, this sample already shows it is $40$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $40$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 4 easy

The constants $a$ , $b$ , $c$ satisfy $4a+3b+c=24$ , with $abc\neq 0$ and $\frac{1}{4a}+\frac{1}{3b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{4a}+\frac{x}{3b}+\frac{x}{c}-6=\frac{6b+2c}{4a}+\frac{8a+2c}{3b}+\frac{8a+6b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $4a+3b+c=24$ . Each numerator on the right side uses two of the three pieces $4a$ , $3b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$4a+3b+c=24$ .
$abc\neq 0$ and $\frac{1}{4a}+\frac{1}{3b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{4a}+\frac{x}{3b}+\frac{x}{c}-6=\frac{6b+2c}{4a}+\frac{8a+2c}{3b}+\frac{8a+6b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $6b+2c$ is twice $3b+c$ , which is the total $24$ minus $4a$ . Rewriting all three this way makes the right side look exactly like the left side, with $48$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

3b+c=24-4a

, so double it.

6b+2c=2(24-4a)=48-8a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

4a+c=24-3b

and

4a+3b=24-c

.

8a+2c=48-6b, \quad 8a+6b=48-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

48

over its denominator minus

2

.

\frac{48-8a}{4a}+\frac{48-6b}{3b}+\frac{48-2c}{c}=\frac{48}{4a}+\frac{48}{3b}+\frac{48}{c}-6

\frac{8a}{4a}=\frac{6b}{3b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{4a}+\frac{1}{3b}+\frac{1}{c}\right)=48\left(\frac{1}{4a}+\frac{1}{3b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=48

Dividing by a nonzero number keeps the equation true.

Answer:

x=48

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=7$ satisfy $8+9+7=24$ . The left side at $x=48$ is $\frac{48}{8}+\frac{48}{9}+\frac{48}{7}-6=\frac{256}{21}$ . The right side is $\frac{32}{8}+\frac{30}{9}+\frac{34}{7}=\frac{256}{21}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=7$ . The equation becomes $\frac{191}{504}x-6=\frac{256}{21}$ , so $\frac{191}{504}x=\frac{382}{21}$ and $x=48$ . Since the problem promises a single answer for all such constants, this sample already shows it is $48$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $48$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 5 medium

The constants $a$ , $b$ , $c$ satisfy $4a+2b+c=25$ , with $abc\neq 0$ and $\frac{1}{4a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{4a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{4a}+\frac{8a+2c}{2b}+\frac{8a+4b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $4a+2b+c=25$ . Each numerator on the right side uses two of the three pieces $4a$ , $2b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$4a+2b+c=25$ .
$abc\neq 0$ and $\frac{1}{4a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{4a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{4a}+\frac{8a+2c}{2b}+\frac{8a+4b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $4b+2c$ is twice $2b+c$ , which is the total $25$ minus $4a$ . Rewriting all three this way makes the right side look exactly like the left side, with $50$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

2b+c=25-4a

, so double it.

4b+2c=2(25-4a)=50-8a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

4a+c=25-2b

and

4a+2b=25-c

.

8a+2c=50-4b, \quad 8a+4b=50-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

50

over its denominator minus

2

.

\frac{50-8a}{4a}+\frac{50-4b}{2b}+\frac{50-2c}{c}=\frac{50}{4a}+\frac{50}{2b}+\frac{50}{c}-6

\frac{8a}{4a}=\frac{4b}{2b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{4a}+\frac{1}{2b}+\frac{1}{c}\right)=50\left(\frac{1}{4a}+\frac{1}{2b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=50

Dividing by a nonzero number keeps the equation true.

Answer:

x=50

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=11$ satisfy $8+6+11=25$ . The left side at $x=50$ is $\frac{50}{8}+\frac{50}{6}+\frac{50}{11}-6=\frac{1733}{132}$ . The right side is $\frac{34}{8}+\frac{38}{6}+\frac{28}{11}=\frac{1733}{132}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=11$ . The equation becomes $\frac{101}{264}x-6=\frac{1733}{132}$ , so $\frac{101}{264}x=\frac{2525}{132}$ and $x=50$ . Since the problem promises a single answer for all such constants, this sample already shows it is $50$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $50$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 6 medium

The constants $a$ , $b$ , $c$ satisfy $3a+4b+c=28$ , with $abc\neq 0$ and $\frac{1}{3a}+\frac{1}{4b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{3a}+\frac{x}{4b}+\frac{x}{c}-9=\frac{12b+3c}{3a}+\frac{9a+3c}{4b}+\frac{9a+12b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $3a+4b+c=28$ . Each numerator on the right side uses two of the three pieces $3a$ , $4b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$3a+4b+c=28$ .
$abc\neq 0$ and $\frac{1}{3a}+\frac{1}{4b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{3a}+\frac{x}{4b}+\frac{x}{c}-9=\frac{12b+3c}{3a}+\frac{9a+3c}{4b}+\frac{9a+12b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $12b+3c$ is $3$ times $4b+c$ , which is the total $28$ minus $3a$ . Rewriting all three this way makes the right side look exactly like the left side, with $84$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

4b+c=28-3a

, so multiply it by

3

.

12b+3c=3(28-3a)=84-9a

Each numerator is

3

times "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

3a+c=28-4b

and

3a+4b=28-c

.

9a+3c=84-12b, \quad 9a+12b=84-3c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

84

over its denominator minus

3

.

\frac{84-9a}{3a}+\frac{84-12b}{4b}+\frac{84-3c}{c}=\frac{84}{3a}+\frac{84}{4b}+\frac{84}{c}-9

\frac{9a}{3a}=\frac{12b}{4b}=\frac{3c}{c}=3

, and three

3

s make

9

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

9

to both sides and factor:

x\left(\frac{1}{3a}+\frac{1}{4b}+\frac{1}{c}\right)=84\left(\frac{1}{3a}+\frac{1}{4b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=84

Dividing by a nonzero number keeps the equation true.

Answer:

x=84

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=10$ satisfy $6+12+10=28$ . The left side at $x=84$ is $\frac{84}{6}+\frac{84}{12}+\frac{84}{10}-9=\frac{102}{5}$ . The right side is $\frac{66}{6}+\frac{48}{12}+\frac{54}{10}=\frac{102}{5}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=10$ . The equation becomes $\frac{7}{20}x-9=\frac{102}{5}$ , so $\frac{7}{20}x=\frac{147}{5}$ and $x=84$ . Since the problem promises a single answer for all such constants, this sample already shows it is $84$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $84$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 7 medium

The constants $a$ , $b$ , $c$ satisfy $5a+2b+c=30$ , with $abc\neq 0$ and $\frac{1}{5a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{5a}+\frac{x}{2b}+\frac{x}{c}-9=\frac{6b+3c}{5a}+\frac{15a+3c}{2b}+\frac{15a+6b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $5a+2b+c=30$ . Each numerator on the right side uses two of the three pieces $5a$ , $2b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$5a+2b+c=30$ .
$abc\neq 0$ and $\frac{1}{5a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{5a}+\frac{x}{2b}+\frac{x}{c}-9=\frac{6b+3c}{5a}+\frac{15a+3c}{2b}+\frac{15a+6b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $6b+3c$ is $3$ times $2b+c$ , which is the total $30$ minus $5a$ . Rewriting all three this way makes the right side look exactly like the left side, with $90$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

2b+c=30-5a

, so multiply it by

3

.

6b+3c=3(30-5a)=90-15a

Each numerator is

3

times "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

5a+c=30-2b

and

5a+2b=30-c

.

15a+3c=90-6b, \quad 15a+6b=90-3c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

90

over its denominator minus

3

.

\frac{90-15a}{5a}+\frac{90-6b}{2b}+\frac{90-3c}{c}=\frac{90}{5a}+\frac{90}{2b}+\frac{90}{c}-9

\frac{15a}{5a}=\frac{6b}{2b}=\frac{3c}{c}=3

, and three

3

s make

9

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

9

to both sides and factor:

x\left(\frac{1}{5a}+\frac{1}{2b}+\frac{1}{c}\right)=90\left(\frac{1}{5a}+\frac{1}{2b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=90

Dividing by a nonzero number keeps the equation true.

Answer:

x=90

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=14$ satisfy $10+6+14=30$ . The left side at $x=90$ is $\frac{90}{10}+\frac{90}{6}+\frac{90}{14}-9=\frac{150}{7}$ . The right side is $\frac{60}{10}+\frac{72}{6}+\frac{48}{14}=\frac{150}{7}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=14$ . The equation becomes $\frac{71}{210}x-9=\frac{150}{7}$ , so $\frac{71}{210}x=\frac{213}{7}$ and $x=90$ . Since the problem promises a single answer for all such constants, this sample already shows it is $90$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $90$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 8 hard

The constants $a$ , $b$ , $c$ satisfy $6a+2b+c=30$ , with $abc\neq 0$ and $\frac{1}{6a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{6a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{6a}+\frac{12a+2c}{2b}+\frac{12a+4b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $6a+2b+c=30$ . Each numerator on the right side uses two of the three pieces $6a$ , $2b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$6a+2b+c=30$ .
$abc\neq 0$ and $\frac{1}{6a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{6a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{6a}+\frac{12a+2c}{2b}+\frac{12a+4b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $4b+2c$ is twice $2b+c$ , which is the total $30$ minus $6a$ . Rewriting all three this way makes the right side look exactly like the left side, with $60$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

2b+c=30-6a

, so double it.

4b+2c=2(30-6a)=60-12a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

6a+c=30-2b

and

6a+2b=30-c

.

12a+2c=60-4b, \quad 12a+4b=60-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

60

over its denominator minus

2

.

\frac{60-12a}{6a}+\frac{60-4b}{2b}+\frac{60-2c}{c}=\frac{60}{6a}+\frac{60}{2b}+\frac{60}{c}-6

\frac{12a}{6a}=\frac{4b}{2b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{6a}+\frac{1}{2b}+\frac{1}{c}\right)=60\left(\frac{1}{6a}+\frac{1}{2b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=60

Dividing by a nonzero number keeps the equation true.

Answer:

x=60

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=12$ satisfy $12+6+12=30$ . The left side at $x=60$ is $\frac{60}{12}+\frac{60}{6}+\frac{60}{12}-6=14$ . The right side is $\frac{36}{12}+\frac{48}{6}+\frac{36}{12}=14$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=12$ . The equation becomes $\frac{1}{3}x-6=14$ , so $\frac{1}{3}x=20$ and $x=60$ . Since the problem promises a single answer for all such constants, this sample already shows it is $60$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $60$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 9 medium

The constants $a$ , $b$ , $c$ satisfy $3a+2b+c=30$ , with $abc\neq 0$ and $\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{3a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{3a}+\frac{6a+2c}{2b}+\frac{6a+4b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $3a+2b+c=30$ . Each numerator on the right side uses two of the three pieces $3a$ , $2b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$3a+2b+c=30$ .
$abc\neq 0$ and $\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{3a}+\frac{x}{2b}+\frac{x}{c}-6=\frac{4b+2c}{3a}+\frac{6a+2c}{2b}+\frac{6a+4b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $4b+2c$ is twice $2b+c$ , which is the total $30$ minus $3a$ . Rewriting all three this way makes the right side look exactly like the left side, with $60$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

2b+c=30-3a

, so double it.

4b+2c=2(30-3a)=60-6a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

3a+c=30-2b

and

3a+2b=30-c

.

6a+2c=60-4b, \quad 6a+4b=60-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

60

over its denominator minus

2

.

\frac{60-6a}{3a}+\frac{60-4b}{2b}+\frac{60-2c}{c}=\frac{60}{3a}+\frac{60}{2b}+\frac{60}{c}-6

\frac{6a}{3a}=\frac{4b}{2b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\right)=60\left(\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=60

Dividing by a nonzero number keeps the equation true.

Answer:

x=60

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=18$ satisfy $6+6+18=30$ . The left side at $x=60$ is $\frac{60}{6}+\frac{60}{6}+\frac{60}{18}-6=\frac{52}{3}$ . The right side is $\frac{48}{6}+\frac{48}{6}+\frac{24}{18}=\frac{52}{3}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=18$ . The equation becomes $\frac{7}{18}x-6=\frac{52}{3}$ , so $\frac{7}{18}x=\frac{70}{3}$ and $x=60$ . Since the problem promises a single answer for all such constants, this sample already shows it is $60$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $60$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 10 hard

The constants $a$ , $b$ , $c$ satisfy $5a+4b+c=50$ , with $abc\neq 0$ and $\frac{1}{5a}+\frac{1}{4b}+\frac{1}{c}\neq 0$ . Solve the equation $\frac{x}{5a}+\frac{x}{4b}+\frac{x}{c}-6=\frac{8b+2c}{5a}+\frac{10a+2c}{4b}+\frac{10a+8b}{c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $5a+4b+c=50$ . Each numerator on the right side uses two of the three pieces $5a$ , $4b$ , $c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$5a+4b+c=50$ .
$abc\neq 0$ and $\frac{1}{5a}+\frac{1}{4b}+\frac{1}{c}\neq 0$ .
The equation $\frac{x}{5a}+\frac{x}{4b}+\frac{x}{c}-6=\frac{8b+2c}{5a}+\frac{10a+2c}{4b}+\frac{10a+8b}{c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $8b+2c$ is twice $4b+c$ , which is the total $50$ minus $5a$ . Rewriting all three this way makes the right side look exactly like the left side, with $100$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

4b+c=50-5a

, so double it.

8b+2c=2(50-5a)=100-10a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

5a+c=50-4b

and

5a+4b=50-c

.

10a+2c=100-8b, \quad 10a+8b=100-2c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

100

over its denominator minus

2

.

\frac{100-10a}{5a}+\frac{100-8b}{4b}+\frac{100-2c}{c}=\frac{100}{5a}+\frac{100}{4b}+\frac{100}{c}-6

\frac{10a}{5a}=\frac{8b}{4b}=\frac{2c}{c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{5a}+\frac{1}{4b}+\frac{1}{c}\right)=100\left(\frac{1}{5a}+\frac{1}{4b}+\frac{1}{c}\right)

. The bracket is not

0

, so divide by it.

x=100

Dividing by a nonzero number keeps the equation true.

Answer:

x=100

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=3$ , $c=28$ satisfy $10+12+28=50$ . The left side at $x=100$ is $\frac{100}{10}+\frac{100}{12}+\frac{100}{28}-6=\frac{334}{21}$ . The right side is $\frac{80}{10}+\frac{76}{12}+\frac{44}{28}=\frac{334}{21}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=3$ , $c=28$ . The equation becomes $\frac{23}{105}x-6=\frac{334}{21}$ , so $\frac{23}{105}x=\frac{460}{21}$ and $x=100$ . Since the problem promises a single answer for all such constants, this sample already shows it is $100$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $100$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 11 hard

The constants $a$ , $b$ , $c$ satisfy $4a+3b+2c=36$ , with $abc\neq 0$ and $\frac{1}{4a}+\frac{1}{3b}+\frac{1}{2c}\neq 0$ . Solve the equation $\frac{x}{4a}+\frac{x}{3b}+\frac{x}{2c}-9=\frac{9b+6c}{4a}+\frac{12a+6c}{3b}+\frac{12a+9b}{2c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $4a+3b+2c=36$ . Each numerator on the right side uses two of the three pieces $4a$ , $3b$ , $2c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$4a+3b+2c=36$ .
$abc\neq 0$ and $\frac{1}{4a}+\frac{1}{3b}+\frac{1}{2c}\neq 0$ .
The equation $\frac{x}{4a}+\frac{x}{3b}+\frac{x}{2c}-9=\frac{9b+6c}{4a}+\frac{12a+6c}{3b}+\frac{12a+9b}{2c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $9b+6c$ is $3$ times $3b+2c$ , which is the total $36$ minus $4a$ . Rewriting all three this way makes the right side look exactly like the left side, with $108$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

3b+2c=36-4a

, so multiply it by

3

.

9b+6c=3(36-4a)=108-12a

Each numerator is

3

times "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

4a+2c=36-3b

and

4a+3b=36-2c

.

12a+6c=108-9b, \quad 12a+9b=108-6c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

108

over its denominator minus

3

.

\frac{108-12a}{4a}+\frac{108-9b}{3b}+\frac{108-6c}{2c}=\frac{108}{4a}+\frac{108}{3b}+\frac{108}{2c}-9

\frac{12a}{4a}=\frac{9b}{3b}=\frac{6c}{2c}=3

, and three

3

s make

9

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

9

to both sides and factor:

x\left(\frac{1}{4a}+\frac{1}{3b}+\frac{1}{2c}\right)=108\left(\frac{1}{4a}+\frac{1}{3b}+\frac{1}{2c}\right)

. The bracket is not

0

, so divide by it.

x=108

Dividing by a nonzero number keeps the equation true.

Answer:

x=108

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=4$ , $c=8$ satisfy $8+12+16=36$ . The left side at $x=108$ is $\frac{108}{8}+\frac{108}{12}+\frac{108}{16}-9=\frac{81}{4}$ . The right side is $\frac{84}{8}+\frac{72}{12}+\frac{60}{16}=\frac{81}{4}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=4$ , $c=8$ . The equation becomes $\frac{13}{48}x-9=\frac{81}{4}$ , so $\frac{13}{48}x=\frac{117}{4}$ and $x=108$ . Since the problem promises a single answer for all such constants, this sample already shows it is $108$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $108$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Variant 12 hard

The constants $a$ , $b$ , $c$ satisfy $5a+3b+2c=40$ , with $abc\neq 0$ and $\frac{1}{5a}+\frac{1}{3b}+\frac{1}{2c}\neq 0$ . Solve the equation $\frac{x}{5a}+\frac{x}{3b}+\frac{x}{2c}-6=\frac{6b+4c}{5a}+\frac{10a+4c}{3b}+\frac{10a+6b}{2c}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The constants are tied together by $5a+3b+2c=40$ . Each numerator on the right side uses two of the three pieces $5a$ , $3b$ , $2c$ , so it can be rewritten using the total. Solve for $x$ .

Givens

$5a+3b+2c=40$ .
$abc\neq 0$ and $\frac{1}{5a}+\frac{1}{3b}+\frac{1}{2c}\neq 0$ .
The equation $\frac{x}{5a}+\frac{x}{3b}+\frac{x}{2c}-6=\frac{6b+4c}{5a}+\frac{10a+4c}{3b}+\frac{10a+6b}{2c}$ .

Unknowns

The solution $x$ .

Constraints

$a$ , $b$ , $c$ are not given individually, so the answer must not depend on them.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

Instead of clearing denominators, look at each numerator: $6b+4c$ is twice $3b+2c$ , which is the total $40$ minus $5a$ . Rewriting all three this way makes the right side look exactly like the left side, with $80$ in place of $x$ .

3 · Execute4 carry out the plan

1Rewrite the first numerator

#16 Count the Complement 7.EE.A.2

3b+2c=40-5a

, so double it.

6b+4c=2(40-5a)=80-10a

Each numerator is twice "the total minus the denominator".

2Rewrite the other two

#7 Identify Subproblems 7.EE.A.2

In the same way,

5a+2c=40-3b

and

5a+3b=40-2c

.

10a+4c=80-6b, \quad 10a+6b=80-4c

Same move, three times.

3Split each fraction

#16 Count the Complement 7.EE.A.2

Each fraction becomes

80

over its denominator minus

2

.

\frac{80-10a}{5a}+\frac{80-6b}{3b}+\frac{80-4c}{2c}=\frac{80}{5a}+\frac{80}{3b}+\frac{80}{2c}-6

\frac{10a}{5a}=\frac{6b}{3b}=\frac{4c}{2c}=2

, and three

2

s make

6

.

4Compare the two sides

#16 Count the Complement 7.EE.B.4

Add

6

to both sides and factor:

x\left(\frac{1}{5a}+\frac{1}{3b}+\frac{1}{2c}\right)=80\left(\frac{1}{5a}+\frac{1}{3b}+\frac{1}{2c}\right)

. The bracket is not

0

, so divide by it.

x=80

Dividing by a nonzero number keeps the equation true.

Answer:

x=80

4 · Reviewdoes it hold up?

Try particular constants: $a=2$ , $b=4$ , $c=9$ satisfy $10+12+18=40$ . The left side at $x=80$ is $\frac{80}{10}+\frac{80}{12}+\frac{80}{18}-6=\frac{118}{9}$ . The right side is $\frac{60}{10}+\frac{56}{12}+\frac{44}{18}=\frac{118}{9}$ . Equal.

Another way: Solve an easier related problem: pick any allowed constants, such as $a=2$ , $b=4$ , $c=9$ . The equation becomes $\frac{43}{180}x-6=\frac{118}{9}$ , so $\frac{43}{180}x=\frac{172}{9}$ and $x=80$ . Since the problem promises a single answer for all such constants, this sample already shows it is $80$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting each numerator as $80$ minus a multiple of its denominator to reveal the structure.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the simplified equation for $x$.

💡Takeaway. When a total is given, look for pieces that are "the total minus something" and rewrite them.

Rewrite numerators using the given total · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?