Practice · A relation fixes a ratio, so the expression is constant · Sensim Math

Variant 1 easy

Suppose $a+4b=2(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{a+4b}{a-b}$ is the same as the solution of the equation $\frac{x+6}{2}-ax=\frac{x-a}{3}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{a+4b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+4b=2(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{a+4b}{a-b}$ solves $\frac{x+6}{2}-ax=\frac{x-a}{3}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=6b$ ; use that to show the fraction always equals $2$ ; then put $x=2$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+4b=2a-2b \;\Rightarrow\; 6b=a

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

6b

; the

b

cancels because

b\neq 0

.

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

Whatever

b

is, the ratio comes out the same.

3Put x = 2 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=2

, so substitute it.

\frac{2+6}{2}-2a=\frac{2-a}{3} \;\Rightarrow\; 4-2a=\frac{2-a}{3}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

3

and collect.

12-6a=2-a \;\Rightarrow\; 10=5a \;\Rightarrow\; a=2

A plain linear equation.

5Evaluate

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

Substitute

a=2

.

-2^2+2-1=-3

Note

-a^2

means

-(a^2)

.

Answer:

-3

4 · Reviewdoes it hold up?

With $a=2$ , $b=\frac{1}{3}$ : $a+4b=\frac{10}{3}$ and $2(a-b)=2\times \frac{5}{3}=\frac{10}{3}$ . In the equation with $a=2$ : $\frac{8}{2}-4=0$ and $\frac{2-2}{3}=0$ , so $x=2$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=6b$ means $a:b=6:1$ , so try $a=6$ , $b=1$ : $\frac{10}{5}=2$ . Any pair in that ratio gives the same $2$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=6b$ and the fraction as the number $2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 2 easy

Suppose $a+5b=4(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{3a+3b}{a-b}$ is the same as the solution of the equation $\frac{x+8}{2}-ax=\frac{x-a}{5}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{3a+3b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+5b=4(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{3a+3b}{a-b}$ solves $\frac{x+8}{2}-ax=\frac{x-a}{5}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=3b$ ; use that to show the fraction always equals $6$ ; then put $x=6$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+5b=4a-4b \;\Rightarrow\; 9b=3a \;\Rightarrow\; a=3b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

3b

; the

b

cancels because

b\neq 0

.

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

Whatever

b

is, the ratio comes out the same.

3Put x = 6 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=6

, so substitute it.

\frac{6+8}{2}-6a=\frac{6-a}{5} \;\Rightarrow\; 7-6a=\frac{6-a}{5}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

5

and collect.

35-30a=6-a \;\Rightarrow\; 29=29a \;\Rightarrow\; a=1

A plain linear equation.

5Evaluate

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

Substitute

a=1

.

-1^2+1-1=-1

Note

-a^2

means

-(a^2)

.

Answer:

-1

4 · Reviewdoes it hold up?

With $a=1$ , $b=\frac{1}{3}$ : $a+5b=\frac{8}{3}$ and $4(a-b)=4\times \frac{2}{3}=\frac{8}{3}$ . In the equation with $a=1$ : $\frac{14}{2}-6=1$ and $\frac{6-1}{5}=1$ , so $x=6$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=3b$ means $a:b=3:1$ , so try $a=3$ , $b=1$ : $\frac{12}{2}=6$ . Any pair in that ratio gives the same $6$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=3b$ and the fraction as the number $6$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 3 easy

Suppose $a+8b=4(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{a+5b}{a-b}$ is the same as the solution of the equation $\frac{x+15}{2}-ax=\frac{x-a}{4}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{a+5b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+8b=4(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{a+5b}{a-b}$ solves $\frac{x+15}{2}-ax=\frac{x-a}{4}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=4b$ ; use that to show the fraction always equals $3$ ; then put $x=3$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+8b=4a-4b \;\Rightarrow\; 12b=3a \;\Rightarrow\; a=4b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

4b

; the

b

cancels because

b\neq 0

.

\frac{a+5b}{a-b}=\frac{4b+5b}{4b-b}=\frac{9b}{3b}=3

Whatever

b

is, the ratio comes out the same.

3Put x = 3 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=3

, so substitute it.

\frac{3+15}{2}-3a=\frac{3-a}{4} \;\Rightarrow\; 9-3a=\frac{3-a}{4}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

4

and collect.

36-12a=3-a \;\Rightarrow\; 33=11a \;\Rightarrow\; a=3

A plain linear equation.

5Evaluate

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

Substitute

a=3

.

-3^2+3-1=-7

Note

-a^2

means

-(a^2)

.

Answer:

-7

4 · Reviewdoes it hold up?

With $a=3$ , $b=\frac{3}{4}$ : $a+8b=9$ and $4(a-b)=4\times \frac{9}{4}=9$ . In the equation with $a=3$ : $\frac{18}{2}-9=0$ and $\frac{3-3}{4}=0$ , so $x=3$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=4b$ means $a:b=4:1$ , so try $a=4$ , $b=1$ : $\frac{9}{3}=3$ . Any pair in that ratio gives the same $3$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=4b$ and the fraction as the number $3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 4 easy

Suppose $a+8b=4(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{a+2b}{a-b}$ is the same as the solution of the equation $\frac{x+18}{5}-ax=\frac{x-a}{2}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{a+2b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+8b=4(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{a+2b}{a-b}$ solves $\frac{x+18}{5}-ax=\frac{x-a}{2}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=4b$ ; use that to show the fraction always equals $2$ ; then put $x=2$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+8b=4a-4b \;\Rightarrow\; 12b=3a \;\Rightarrow\; a=4b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

4b

; the

b

cancels because

b\neq 0

.

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

Whatever

b

is, the ratio comes out the same.

3Put x = 2 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=2

, so substitute it.

\frac{2+18}{5}-2a=\frac{2-a}{2} \;\Rightarrow\; 4-2a=\frac{2-a}{2}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

2

and collect.

8-4a=2-a \;\Rightarrow\; 6=3a \;\Rightarrow\; a=2

A plain linear equation.

5Evaluate

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

Substitute

a=2

.

-2^2+2-1=-3

Note

-a^2

means

-(a^2)

.

Answer:

-3

4 · Reviewdoes it hold up?

With $a=2$ , $b=\frac{1}{2}$ : $a+8b=6$ and $4(a-b)=4\times \frac{3}{2}=6$ . In the equation with $a=2$ : $\frac{20}{5}-4=0$ and $\frac{2-2}{2}=0$ , so $x=2$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=4b$ means $a:b=4:1$ , so try $a=4$ , $b=1$ : $\frac{6}{3}=2$ . Any pair in that ratio gives the same $2$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=4b$ and the fraction as the number $2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 5 medium

Suppose $a+b=2(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{3a+5b}{a-b}$ is the same as the solution of the equation $\frac{x+23}{3}-ax=\frac{x-a}{2}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{3a+5b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+b=2(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{3a+5b}{a-b}$ solves $\frac{x+23}{3}-ax=\frac{x-a}{2}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=3b$ ; use that to show the fraction always equals $7$ ; then put $x=7$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+b=2a-2b \;\Rightarrow\; 3b=a

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

3b

; the

b

cancels because

b\neq 0

.

\frac{3a+5b}{a-b}=\frac{9b+5b}{3b-b}=\frac{14b}{2b}=7

Whatever

b

is, the ratio comes out the same.

3Put x = 7 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=7

, so substitute it.

\frac{7+23}{3}-7a=\frac{7-a}{2} \;\Rightarrow\; 10-7a=\frac{7-a}{2}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

2

and collect.

20-14a=7-a \;\Rightarrow\; 13=13a \;\Rightarrow\; a=1

A plain linear equation.

5Evaluate

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

Substitute

a=1

.

-1^2+1-1=-1

Note

-a^2

means

-(a^2)

.

Answer:

-1

4 · Reviewdoes it hold up?

With $a=1$ , $b=\frac{1}{3}$ : $a+b=\frac{4}{3}$ and $2(a-b)=2\times \frac{2}{3}=\frac{4}{3}$ . In the equation with $a=1$ : $\frac{30}{3}-7=3$ and $\frac{7-1}{2}=3$ , so $x=7$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=3b$ means $a:b=3:1$ , so try $a=3$ , $b=1$ : $\frac{14}{2}=7$ . Any pair in that ratio gives the same $7$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=3b$ and the fraction as the number $7$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 6 medium

Suppose $a+7b=2(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{4a+4b}{a-b}$ is the same as the solution of the equation $\frac{x+13}{3}-ax=\frac{x-a}{4}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{4a+4b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+7b=2(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{4a+4b}{a-b}$ solves $\frac{x+13}{3}-ax=\frac{x-a}{4}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=9b$ ; use that to show the fraction always equals $5$ ; then put $x=5$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+7b=2a-2b \;\Rightarrow\; 9b=a

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

9b

; the

b

cancels because

b\neq 0

.

\frac{4a+4b}{a-b}=\frac{36b+4b}{9b-b}=\frac{40b}{8b}=5

Whatever

b

is, the ratio comes out the same.

3Put x = 5 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=5

, so substitute it.

\frac{5+13}{3}-5a=\frac{5-a}{4} \;\Rightarrow\; 6-5a=\frac{5-a}{4}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

4

and collect.

24-20a=5-a \;\Rightarrow\; 19=19a \;\Rightarrow\; a=1

A plain linear equation.

5Evaluate

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

Substitute

a=1

.

-1^2+1-1=-1

Note

-a^2

means

-(a^2)

.

Answer:

-1

4 · Reviewdoes it hold up?

With $a=1$ , $b=\frac{1}{9}$ : $a+7b=\frac{16}{9}$ and $2(a-b)=2\times \frac{8}{9}=\frac{16}{9}$ . In the equation with $a=1$ : $\frac{18}{3}-5=1$ and $\frac{5-1}{4}=1$ , so $x=5$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=9b$ means $a:b=9:1$ , so try $a=9$ , $b=1$ : $\frac{40}{8}=5$ . Any pair in that ratio gives the same $5$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=9b$ and the fraction as the number $5$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 7 medium

Suppose $a+7b=3(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{a+3b}{a-b}$ is the same as the solution of the equation $\frac{x+26}{7}-ax=\frac{x-a}{3}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{a+3b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+7b=3(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{a+3b}{a-b}$ solves $\frac{x+26}{7}-ax=\frac{x-a}{3}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=5b$ ; use that to show the fraction always equals $2$ ; then put $x=2$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+7b=3a-3b \;\Rightarrow\; 10b=2a \;\Rightarrow\; a=5b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

5b

; the

b

cancels because

b\neq 0

.

\frac{a+3b}{a-b}=\frac{5b+3b}{5b-b}=\frac{8b}{4b}=2

Whatever

b

is, the ratio comes out the same.

3Put x = 2 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=2

, so substitute it.

\frac{2+26}{7}-2a=\frac{2-a}{3} \;\Rightarrow\; 4-2a=\frac{2-a}{3}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

3

and collect.

12-6a=2-a \;\Rightarrow\; 10=5a \;\Rightarrow\; a=2

A plain linear equation.

5Evaluate

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

Substitute

a=2

.

-2^2+2-1=-3

Note

-a^2

means

-(a^2)

.

Answer:

-3

4 · Reviewdoes it hold up?

With $a=2$ , $b=\frac{2}{5}$ : $a+7b=\frac{24}{5}$ and $3(a-b)=3\times \frac{8}{5}=\frac{24}{5}$ . In the equation with $a=2$ : $\frac{28}{7}-4=0$ and $\frac{2-2}{3}=0$ , so $x=2$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=5b$ means $a:b=5:1$ , so try $a=5$ , $b=1$ : $\frac{8}{4}=2$ . Any pair in that ratio gives the same $2$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=5b$ and the fraction as the number $2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 8 medium

Suppose $a+4b=2(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{3a+2b}{a-b}$ is the same as the solution of the equation $\frac{x+21}{5}-ax=\frac{x-a}{3}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{3a+2b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+4b=2(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{3a+2b}{a-b}$ solves $\frac{x+21}{5}-ax=\frac{x-a}{3}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=6b$ ; use that to show the fraction always equals $4$ ; then put $x=4$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+4b=2a-2b \;\Rightarrow\; 6b=a

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

6b

; the

b

cancels because

b\neq 0

.

\frac{3a+2b}{a-b}=\frac{18b+2b}{6b-b}=\frac{20b}{5b}=4

Whatever

b

is, the ratio comes out the same.

3Put x = 4 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=4

, so substitute it.

\frac{4+21}{5}-4a=\frac{4-a}{3} \;\Rightarrow\; 5-4a=\frac{4-a}{3}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

3

and collect.

15-12a=4-a \;\Rightarrow\; 11=11a \;\Rightarrow\; a=1

A plain linear equation.

5Evaluate

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

Substitute

a=1

.

-1^2+1-1=-1

Note

-a^2

means

-(a^2)

.

Answer:

-1

4 · Reviewdoes it hold up?

With $a=1$ , $b=\frac{1}{6}$ : $a+4b=\frac{5}{3}$ and $2(a-b)=2\times \frac{5}{6}=\frac{5}{3}$ . In the equation with $a=1$ : $\frac{25}{5}-4=1$ and $\frac{4-1}{3}=1$ , so $x=4$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=6b$ means $a:b=6:1$ , so try $a=6$ , $b=1$ : $\frac{20}{5}=4$ . Any pair in that ratio gives the same $4$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=6b$ and the fraction as the number $4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 9 hard

Suppose $a+7b=3(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{2a+2b}{a-b}$ is the same as the solution of the equation $\frac{x+24}{3}-ax=\frac{x-a}{2}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{2a+2b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+7b=3(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{2a+2b}{a-b}$ solves $\frac{x+24}{3}-ax=\frac{x-a}{2}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=5b$ ; use that to show the fraction always equals $3$ ; then put $x=3$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+7b=3a-3b \;\Rightarrow\; 10b=2a \;\Rightarrow\; a=5b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

5b

; the

b

cancels because

b\neq 0

.

\frac{2a+2b}{a-b}=\frac{10b+2b}{5b-b}=\frac{12b}{4b}=3

Whatever

b

is, the ratio comes out the same.

3Put x = 3 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=3

, so substitute it.

\frac{3+24}{3}-3a=\frac{3-a}{2} \;\Rightarrow\; 9-3a=\frac{3-a}{2}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

2

and collect.

18-6a=3-a \;\Rightarrow\; 15=5a \;\Rightarrow\; a=3

A plain linear equation.

5Evaluate

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

Substitute

a=3

.

-3^2+3-1=-7

Note

-a^2

means

-(a^2)

.

Answer:

-7

4 · Reviewdoes it hold up?

With $a=3$ , $b=\frac{3}{5}$ : $a+7b=\frac{36}{5}$ and $3(a-b)=3\times \frac{12}{5}=\frac{36}{5}$ . In the equation with $a=3$ : $\frac{27}{3}-9=0$ and $\frac{3-3}{2}=0$ , so $x=3$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=5b$ means $a:b=5:1$ , so try $a=5$ , $b=1$ : $\frac{12}{4}=3$ . Any pair in that ratio gives the same $3$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=5b$ and the fraction as the number $3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 10 hard

Suppose $a+3b=3(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{2a+2b}{a-b}$ is the same as the solution of the equation $\frac{x+26}{6}-ax=\frac{x-a}{3}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{2a+2b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+3b=3(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{2a+2b}{a-b}$ solves $\frac{x+26}{6}-ax=\frac{x-a}{3}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=3b$ ; use that to show the fraction always equals $4$ ; then put $x=4$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+3b=3a-3b \;\Rightarrow\; 6b=2a \;\Rightarrow\; a=3b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

3b

; the

b

cancels because

b\neq 0

.

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

Whatever

b

is, the ratio comes out the same.

3Put x = 4 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=4

, so substitute it.

\frac{4+26}{6}-4a=\frac{4-a}{3} \;\Rightarrow\; 5-4a=\frac{4-a}{3}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

3

and collect.

15-12a=4-a \;\Rightarrow\; 11=11a \;\Rightarrow\; a=1

A plain linear equation.

5Evaluate

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

Substitute

a=1

.

-1^2+1-1=-1

Note

-a^2

means

-(a^2)

.

Answer:

-1

4 · Reviewdoes it hold up?

With $a=1$ , $b=\frac{1}{3}$ : $a+3b=2$ and $3(a-b)=3\times \frac{2}{3}=2$ . In the equation with $a=1$ : $\frac{30}{6}-4=1$ and $\frac{4-1}{3}=1$ , so $x=4$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=3b$ means $a:b=3:1$ , so try $a=3$ , $b=1$ : $\frac{8}{2}=4$ . Any pair in that ratio gives the same $4$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=3b$ and the fraction as the number $4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 11 hard

Suppose $a+8b=4(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{2a+4b}{a-b}$ is the same as the solution of the equation $\frac{x+28}{2}-ax=\frac{x-a}{3}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{2a+4b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+8b=4(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{2a+4b}{a-b}$ solves $\frac{x+28}{2}-ax=\frac{x-a}{3}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=4b$ ; use that to show the fraction always equals $4$ ; then put $x=4$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+8b=4a-4b \;\Rightarrow\; 12b=3a \;\Rightarrow\; a=4b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

4b

; the

b

cancels because

b\neq 0

.

\frac{2a+4b}{a-b}=\frac{8b+4b}{4b-b}=\frac{12b}{3b}=4

Whatever

b

is, the ratio comes out the same.

3Put x = 4 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=4

, so substitute it.

\frac{4+28}{2}-4a=\frac{4-a}{3} \;\Rightarrow\; 16-4a=\frac{4-a}{3}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

3

and collect.

48-12a=4-a \;\Rightarrow\; 44=11a \;\Rightarrow\; a=4

A plain linear equation.

5Evaluate

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

Substitute

a=4

.

-4^2+4-1=-13

Note

-a^2

means

-(a^2)

.

Answer:

-13

4 · Reviewdoes it hold up?

With $a=4$ , $b=1$ : $a+8b=12$ and $4(a-b)=4\times 3=12$ . In the equation with $a=4$ : $\frac{32}{2}-16=0$ and $\frac{4-4}{3}=0$ , so $x=4$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=4b$ means $a:b=4:1$ , so try $a=4$ , $b=1$ : $\frac{12}{3}=4$ . Any pair in that ratio gives the same $4$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=4b$ and the fraction as the number $4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

Variant 12 hard

Suppose $a+5b=3(a-b)$ , where $ab\neq 0$ and $a\neq b$ . The value of $\frac{2a+4b}{a-b}$ is the same as the solution of the equation $\frac{x+28}{2}-ax=\frac{x-a}{5}$ in $x$ . Find the value of $-a^2+a-1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The relation between $a$ and $b$ makes the fraction $\frac{2a+4b}{a-b}$ a fixed number. That number is the solution of a second equation that contains $a$ . Use it to find $a$ , then evaluate $-a^2+a-1$ .

Givens

$a+5b=3(a-b)$ , with $ab\neq 0$ and $a\neq b$ .
The value of $\frac{2a+4b}{a-b}$ solves $\frac{x+28}{2}-ax=\frac{x-a}{5}$ .

Unknowns

The value of $-a^2+a-1$ .

Constraints

$b\neq 0$ , so $b$ may be cancelled from a fraction.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#16 Count the Complement

Split it into three small problems: turn the relation into $a=4b$ ; use that to show the fraction always equals $4$ ; then put $x=4$ into the second equation to get $a$ .

3 · Execute5 carry out the plan

1Simplify the relation

#13 Convert to Algebra 7.EE.A.2

Expand and collect terms.

a+5b=3a-3b \;\Rightarrow\; 8b=2a \;\Rightarrow\; a=4b

One letter is now a multiple of the other.

2The fraction is a fixed number

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

Replace

a

with

4b

; the

b

cancels because

b\neq 0

.

\frac{2a+4b}{a-b}=\frac{8b+4b}{4b-b}=\frac{12b}{3b}=4

Whatever

b

is, the ratio comes out the same.

3Put x = 4 into the equation

#7 Identify Subproblems 7.EE.B.4

The solution is

x=4

, so substitute it.

\frac{4+28}{2}-4a=\frac{4-a}{5} \;\Rightarrow\; 16-4a=\frac{4-a}{5}

Only

a

is unknown now.

4Solve for a

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

Multiply both sides by

5

and collect.

80-20a=4-a \;\Rightarrow\; 76=19a \;\Rightarrow\; a=4

A plain linear equation.

5Evaluate

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

Substitute

a=4

.

-4^2+4-1=-13

Note

-a^2

means

-(a^2)

.

Answer:

-13

4 · Reviewdoes it hold up?

With $a=4$ , $b=1$ : $a+5b=9$ and $3(a-b)=3\times 3=9$ . In the equation with $a=4$ : $\frac{32}{2}-16=0$ and $\frac{4-4}{5}=0$ , so $x=4$ is indeed the solution. Also $ab\neq 0$ and $a\neq b$ .

Another way: Treat the relation as a ratio: $a=4b$ means $a:b=4:1$ , so try $a=4$ , $b=1$ : $\frac{12}{3}=4$ . Any pair in that ratio gives the same $4$ , which is why the fraction does not depend on $b$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the relation as $a=4b$ and the fraction as the number $4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the equation for $a$ and evaluating the final expression.

💡Takeaway. When a relation ties two letters together, a fraction made of them can turn into a plain number.

A relation fixes a ratio, so the expression is constant · 12 practice problems

Generated variants — 12

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