Practice · Rename repeated reciprocals to get a linear system · Sensim Math

Variant 1 easy

Solve the equation $\frac{2}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{2}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{2}{x+y}-\frac{1}{x-y}=1, \quad \frac{1}{x-y}-\frac{3}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

2A - B = 1, \quad -3A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+3A

. Substitute into the first equation.

2A - (1+3A) = 1 \;\Rightarrow\; -A - 1 = 1 \;\Rightarrow\; A = -2, \quad B = 1 + 3 \times (-2) = -5

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=-2

then

x+y=-\frac{1}{2}

, and if

\frac{1}{x-y}=-5

then

x-y=-\frac{1}{5}

.

x + y = -\frac{1}{2}, \quad x - y = -\frac{1}{5}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = -\frac{5}{10} - \frac{2}{10} = -\frac{7}{10} \Rightarrow x = -\frac{7}{20}, \quad 2y = -\frac{5}{10} + \frac{2}{10} = -\frac{3}{10} \Rightarrow y = -\frac{3}{20}

A sum and a difference always pin down the two numbers.

Answer:

x=-\frac{7}{20}

,

y=-\frac{3}{20}

4 · Reviewdoes it hold up?

Now $x+y=-\frac{10}{20}=-\frac{1}{2}$ and $x-y=-\frac{4}{20}=-\frac{1}{5}$ . Then $\frac{2}{x+y}-\frac{1}{x-y}=-4+5=1$ and $\frac{1}{x-y}-\frac{3}{x+y}=-5+6=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: adding $-3A+B=1$ to $2A-B=1$ gives $-A=2$ directly, so $A=-2$ and $B=-5$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 2 easy

Solve the equation $\frac{3}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{3}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{3}{x+y}-\frac{2}{x-y}=1, \quad \frac{1}{x-y}-\frac{2}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

3A - 2B = 1, \quad -2A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+2A

. Substitute into the first equation.

3A - 2(1+2A) = 1 \;\Rightarrow\; -A - 2 = 1 \;\Rightarrow\; A = -3, \quad B = 1 + 2 \times (-3) = -5

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=-3

then

x+y=-\frac{1}{3}

, and if

\frac{1}{x-y}=-5

then

x-y=-\frac{1}{5}

.

x + y = -\frac{1}{3}, \quad x - y = -\frac{1}{5}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = -\frac{5}{15} - \frac{3}{15} = -\frac{8}{15} \Rightarrow x = -\frac{4}{15}, \quad 2y = -\frac{5}{15} + \frac{3}{15} = -\frac{2}{15} \Rightarrow y = -\frac{1}{15}

A sum and a difference always pin down the two numbers.

Answer:

x=-\frac{4}{15}

,

y=-\frac{1}{15}

4 · Reviewdoes it hold up?

Now $x+y=-\frac{5}{15}=-\frac{1}{3}$ and $x-y=-\frac{3}{15}=-\frac{1}{5}$ . Then $\frac{3}{x+y}-\frac{2}{x-y}=-9+10=1$ and $\frac{1}{x-y}-\frac{2}{x+y}=-5+6=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: doubling $-2A+B=1$ gives $-4A+2B=2$ , and adding it to $3A-2B=1$ gives $-A=3$ directly, so $A=-3$ and $B=-5$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 3 easy

Solve the equation $\frac{3}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{3}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{3}{x+y}-\frac{1}{x-y}=1, \quad \frac{1}{x-y}-\frac{2}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

3A - B = 1, \quad -2A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+2A

. Substitute into the first equation.

3A - (1+2A) = 1 \;\Rightarrow\; A - 1 = 1 \;\Rightarrow\; A = 2, \quad B = 1 + 2 \times 2 = 5

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=2

then

x+y=\frac{1}{2}

, and if

\frac{1}{x-y}=5

then

x-y=\frac{1}{5}

.

x + y = \frac{1}{2}, \quad x - y = \frac{1}{5}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \Rightarrow x = \frac{7}{20}, \quad 2y = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \Rightarrow y = \frac{3}{20}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{7}{20}

,

y=\frac{3}{20}

4 · Reviewdoes it hold up?

Now $x+y=\frac{10}{20}=\frac{1}{2}$ and $x-y=\frac{4}{20}=\frac{1}{5}$ . Then $\frac{3}{x+y}-\frac{1}{x-y}=6-5=1$ and $\frac{1}{x-y}-\frac{2}{x+y}=5-4=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: adding $-2A+B=1$ to $3A-B=1$ gives $A=2$ directly, so $A=2$ and $B=5$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 4 medium

Solve the equation $\frac{4}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=2$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{4}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=2$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $2$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

2

.

\frac{4}{x+y}-\frac{2}{x-y}=2, \quad \frac{1}{x-y}-\frac{3}{x+y}=2

Pairing each part with the plain number

2

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

4A - 2B = 2, \quad -3A + B = 2

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=2+3A

. Substitute into the first equation.

4A - 2(2+3A) = 2 \;\Rightarrow\; -2A - 4 = 2 \;\Rightarrow\; A = -3, \quad B = 2 + 3 \times (-3) = -7

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=-3

then

x+y=-\frac{1}{3}

, and if

\frac{1}{x-y}=-7

then

x-y=-\frac{1}{7}

.

x + y = -\frac{1}{3}, \quad x - y = -\frac{1}{7}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = -\frac{7}{21} - \frac{3}{21} = -\frac{10}{21} \Rightarrow x = -\frac{5}{21}, \quad 2y = -\frac{7}{21} + \frac{3}{21} = -\frac{4}{21} \Rightarrow y = -\frac{2}{21}

A sum and a difference always pin down the two numbers.

Answer:

x=-\frac{5}{21}

,

y=-\frac{2}{21}

4 · Reviewdoes it hold up?

Now $x+y=-\frac{7}{21}=-\frac{1}{3}$ and $x-y=-\frac{3}{21}=-\frac{1}{7}$ . Then $\frac{4}{x+y}-\frac{2}{x-y}=-12+14=2$ and $\frac{1}{x-y}-\frac{3}{x+y}=-7+9=2$ , so every part of the chain equals $2$ .

Another way: Eliminate instead of substituting: doubling $-3A+B=2$ gives $-6A+2B=4$ , and adding it to $4A-2B=2$ gives $-2A=6$ directly, so $A=-3$ and $B=-7$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 5 easy

Solve the equation $\frac{4}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{4}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{4}{x+y}-\frac{1}{x-y}=1, \quad \frac{1}{x-y}-\frac{3}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

4A - B = 1, \quad -3A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+3A

. Substitute into the first equation.

4A - (1+3A) = 1 \;\Rightarrow\; A - 1 = 1 \;\Rightarrow\; A = 2, \quad B = 1 + 3 \times 2 = 7

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=2

then

x+y=\frac{1}{2}

, and if

\frac{1}{x-y}=7

then

x-y=\frac{1}{7}

.

x + y = \frac{1}{2}, \quad x - y = \frac{1}{7}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{7}{14} + \frac{2}{14} = \frac{9}{14} \Rightarrow x = \frac{9}{28}, \quad 2y = \frac{7}{14} - \frac{2}{14} = \frac{5}{14} \Rightarrow y = \frac{5}{28}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{9}{28}

,

y=\frac{5}{28}

4 · Reviewdoes it hold up?

Now $x+y=\frac{14}{28}=\frac{1}{2}$ and $x-y=\frac{4}{28}=\frac{1}{7}$ . Then $\frac{4}{x+y}-\frac{1}{x-y}=8-7=1$ and $\frac{1}{x-y}-\frac{3}{x+y}=7-6=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: adding $-3A+B=1$ to $4A-B=1$ gives $A=2$ directly, so $A=2$ and $B=7$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 6 medium

Solve the equation $\frac{4}{x+y}-\frac{3}{x-y}=\frac{1}{x-y}-\frac{1}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{4}{x+y}-\frac{3}{x-y}=\frac{1}{x-y}-\frac{1}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{4}{x+y}-\frac{3}{x-y}=1, \quad \frac{1}{x-y}-\frac{1}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

4A - 3B = 1, \quad -A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+A

. Substitute into the first equation.

4A - 3(1+A) = 1 \;\Rightarrow\; A - 3 = 1 \;\Rightarrow\; A = 4, \quad B = 1 + 4 = 5

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=4

then

x+y=\frac{1}{4}

, and if

\frac{1}{x-y}=5

then

x-y=\frac{1}{5}

.

x + y = \frac{1}{4}, \quad x - y = \frac{1}{5}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{5}{20} + \frac{4}{20} = \frac{9}{20} \Rightarrow x = \frac{9}{40}, \quad 2y = \frac{5}{20} - \frac{4}{20} = \frac{1}{20} \Rightarrow y = \frac{1}{40}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{9}{40}

,

y=\frac{1}{40}

4 · Reviewdoes it hold up?

Now $x+y=\frac{10}{40}=\frac{1}{4}$ and $x-y=\frac{8}{40}=\frac{1}{5}$ . Then $\frac{4}{x+y}-\frac{3}{x-y}=16-15=1$ and $\frac{1}{x-y}-\frac{1}{x+y}=5-4=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: multiplying $-A+B=1$ by $3$ gives $-3A+3B=3$ , and adding it to $4A-3B=1$ gives $A=4$ directly, so $A=4$ and $B=5$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 7 hard

Solve the equation $\frac{5}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=2$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{5}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=2$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $2$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

2

.

\frac{5}{x+y}-\frac{1}{x-y}=2, \quad \frac{1}{x-y}-\frac{3}{x+y}=2

Pairing each part with the plain number

2

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

5A - B = 2, \quad -3A + B = 2

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=2+3A

. Substitute into the first equation.

5A - (2+3A) = 2 \;\Rightarrow\; 2A - 2 = 2 \;\Rightarrow\; A = 2, \quad B = 2 + 3 \times 2 = 8

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=2

then

x+y=\frac{1}{2}

, and if

\frac{1}{x-y}=8

then

x-y=\frac{1}{8}

.

x + y = \frac{1}{2}, \quad x - y = \frac{1}{8}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} \Rightarrow x = \frac{5}{16}, \quad 2y = \frac{4}{8} - \frac{1}{8} = \frac{3}{8} \Rightarrow y = \frac{3}{16}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{5}{16}

,

y=\frac{3}{16}

4 · Reviewdoes it hold up?

Now $x+y=\frac{8}{16}=\frac{1}{2}$ and $x-y=\frac{2}{16}=\frac{1}{8}$ . Then $\frac{5}{x+y}-\frac{1}{x-y}=10-8=2$ and $\frac{1}{x-y}-\frac{3}{x+y}=8-6=2$ , so every part of the chain equals $2$ .

Another way: Eliminate instead of substituting: adding $-3A+B=2$ to $5A-B=2$ gives $2A=4$ directly, so $A=2$ and $B=8$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 8 medium

Solve the equation $\frac{5}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{5}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{5}{x+y}-\frac{2}{x-y}=1, \quad \frac{1}{x-y}-\frac{3}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

5A - 2B = 1, \quad -3A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+3A

. Substitute into the first equation.

5A - 2(1+3A) = 1 \;\Rightarrow\; -A - 2 = 1 \;\Rightarrow\; A = -3, \quad B = 1 + 3 \times (-3) = -8

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=-3

then

x+y=-\frac{1}{3}

, and if

\frac{1}{x-y}=-8

then

x-y=-\frac{1}{8}

.

x + y = -\frac{1}{3}, \quad x - y = -\frac{1}{8}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = -\frac{8}{24} - \frac{3}{24} = -\frac{11}{24} \Rightarrow x = -\frac{11}{48}, \quad 2y = -\frac{8}{24} + \frac{3}{24} = -\frac{5}{24} \Rightarrow y = -\frac{5}{48}

A sum and a difference always pin down the two numbers.

Answer:

x=-\frac{11}{48}

,

y=-\frac{5}{48}

4 · Reviewdoes it hold up?

Now $x+y=-\frac{16}{48}=-\frac{1}{3}$ and $x-y=-\frac{6}{48}=-\frac{1}{8}$ . Then $\frac{5}{x+y}-\frac{2}{x-y}=-15+16=1$ and $\frac{1}{x-y}-\frac{3}{x+y}=-8+9=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: doubling $-3A+B=1$ gives $-6A+2B=2$ , and adding it to $5A-2B=1$ gives $-A=3$ directly, so $A=-3$ and $B=-8$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 9 hard

Solve the equation $\frac{5}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{4}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{5}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{4}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{5}{x+y}-\frac{1}{x-y}=1, \quad \frac{1}{x-y}-\frac{4}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

5A - B = 1, \quad -4A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+4A

. Substitute into the first equation.

5A - (1+4A) = 1 \;\Rightarrow\; A - 1 = 1 \;\Rightarrow\; A = 2, \quad B = 1 + 4 \times 2 = 9

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=2

then

x+y=\frac{1}{2}

, and if

\frac{1}{x-y}=9

then

x-y=\frac{1}{9}

.

x + y = \frac{1}{2}, \quad x - y = \frac{1}{9}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{9}{18} + \frac{2}{18} = \frac{11}{18} \Rightarrow x = \frac{11}{36}, \quad 2y = \frac{9}{18} - \frac{2}{18} = \frac{7}{18} \Rightarrow y = \frac{7}{36}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{11}{36}

,

y=\frac{7}{36}

4 · Reviewdoes it hold up?

Now $x+y=\frac{18}{36}=\frac{1}{2}$ and $x-y=\frac{4}{36}=\frac{1}{9}$ . Then $\frac{5}{x+y}-\frac{1}{x-y}=10-9=1$ and $\frac{1}{x-y}-\frac{4}{x+y}=9-8=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: adding $-4A+B=1$ to $5A-B=1$ gives $A=2$ directly, so $A=2$ and $B=9$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 10 hard

Solve the equation $\frac{5}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{5}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{5}{x+y}-\frac{2}{x-y}=1, \quad \frac{1}{x-y}-\frac{2}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

5A - 2B = 1, \quad -2A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+2A

. Substitute into the first equation.

5A - 2(1+2A) = 1 \;\Rightarrow\; A - 2 = 1 \;\Rightarrow\; A = 3, \quad B = 1 + 2 \times 3 = 7

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=3

then

x+y=\frac{1}{3}

, and if

\frac{1}{x-y}=7

then

x-y=\frac{1}{7}

.

x + y = \frac{1}{3}, \quad x - y = \frac{1}{7}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{7}{21} + \frac{3}{21} = \frac{10}{21} \Rightarrow x = \frac{5}{21}, \quad 2y = \frac{7}{21} - \frac{3}{21} = \frac{4}{21} \Rightarrow y = \frac{2}{21}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{5}{21}

,

y=\frac{2}{21}

4 · Reviewdoes it hold up?

Now $x+y=\frac{7}{21}=\frac{1}{3}$ and $x-y=\frac{3}{21}=\frac{1}{7}$ . Then $\frac{5}{x+y}-\frac{2}{x-y}=15-14=1$ and $\frac{1}{x-y}-\frac{2}{x+y}=7-6=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: doubling $-2A+B=1$ gives $-4A+2B=2$ , and adding it to $5A-2B=1$ gives $A=3$ directly, so $A=3$ and $B=7$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 11 medium

Solve the equation $\frac{5}{x+y}-\frac{3}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{5}{x+y}-\frac{3}{x-y}=\frac{1}{x-y}-\frac{2}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{5}{x+y}-\frac{3}{x-y}=1, \quad \frac{1}{x-y}-\frac{2}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

5A - 3B = 1, \quad -2A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+2A

. Substitute into the first equation.

5A - 3(1+2A) = 1 \;\Rightarrow\; -A - 3 = 1 \;\Rightarrow\; A = -4, \quad B = 1 + 2 \times (-4) = -7

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=-4

then

x+y=-\frac{1}{4}

, and if

\frac{1}{x-y}=-7

then

x-y=-\frac{1}{7}

.

x + y = -\frac{1}{4}, \quad x - y = -\frac{1}{7}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = -\frac{7}{28} - \frac{4}{28} = -\frac{11}{28} \Rightarrow x = -\frac{11}{56}, \quad 2y = -\frac{7}{28} + \frac{4}{28} = -\frac{3}{28} \Rightarrow y = -\frac{3}{56}

A sum and a difference always pin down the two numbers.

Answer:

x=-\frac{11}{56}

,

y=-\frac{3}{56}

4 · Reviewdoes it hold up?

Now $x+y=-\frac{14}{56}=-\frac{1}{4}$ and $x-y=-\frac{8}{56}=-\frac{1}{7}$ . Then $\frac{5}{x+y}-\frac{3}{x-y}=-20+21=1$ and $\frac{1}{x-y}-\frac{2}{x+y}=-7+8=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: multiplying $-2A+B=1$ by $3$ gives $-6A+3B=3$ , and adding it to $5A-3B=1$ gives $-A=4$ directly, so $A=-4$ and $B=-7$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Variant 12 hard

Solve the equation $\frac{6}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{4}{x+y}=1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A chain of three equal expressions hides two equations in $x$ and $y$ . The unknowns sit in denominators, so the equations are not linear as written.

Givens

$\frac{6}{x+y}-\frac{1}{x-y}=\frac{1}{x-y}-\frac{4}{x+y}=1$

Unknowns

The values of $x$ and $y$ .

Constraints

$x+y \neq 0$ and $x-y \neq 0$ , so the fractions are defined.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#7 Identify Subproblems

Reorganize the chain $A=B=C$ into two equations that each equal $1$ . Only two building blocks appear, $\frac{1}{x+y}$ and $\frac{1}{x-y}$ , so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for $x$ and $y$ .

3 · Execute5 carry out the plan

1Split the chain into two equations

#15 Organize Information in More Ways 8.EE.C.8

Each of the first two expressions equals the last one,

1

.

\frac{6}{x+y}-\frac{1}{x-y}=1, \quad \frac{1}{x-y}-\frac{4}{x+y}=1

Pairing each part with the plain number

1

keeps the equations simple.

2Rename the repeated reciprocals

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

Let

A=\frac{1}{x+y}

and

B=\frac{1}{x-y}

. Both equations become linear in

A

and

B

.

6A - B = 1, \quad -4A + B = 1

The fractions were never the real difficulty; they are just two quantities used twice.

3Solve for $A$ and $B$

#13 Convert to Algebra 8.EE.C.8

From the second equation

B=1+4A

. Substitute into the first equation.

6A - (1+4A) = 1 \;\Rightarrow\; 2A - 1 = 1 \;\Rightarrow\; A = 1, \quad B = 1 + 4 \times 1 = 5

This is now a routine two-variable linear system.

4Undo the renaming

#7 Identify Subproblems 8.EE.C.8

If

\frac{1}{x+y}=1

then

x+y=1

, and if

\frac{1}{x-y}=5

then

x-y=\frac{1}{5}

.

x + y = 1, \quad x - y = \frac{1}{5}

Taking reciprocals of both sides is safe because neither side is

0

.

5Add and subtract to get $x$ and $y$

#7 Identify Subproblems 8.EE.C.8

Adding the two equations gives

2x

; subtracting them gives

2y

.

2x = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \Rightarrow x = \frac{3}{5}, \quad 2y = \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \Rightarrow y = \frac{2}{5}

A sum and a difference always pin down the two numbers.

Answer:

x=\frac{3}{5}

,

y=\frac{2}{5}

4 · Reviewdoes it hold up?

Now $x+y=\frac{5}{5}=1$ and $x-y=\frac{1}{5}$ . Then $\frac{6}{x+y}-\frac{1}{x-y}=6-5=1$ and $\frac{1}{x-y}-\frac{4}{x+y}=5-4=1$ , so every part of the chain equals $1$ .

Another way: Eliminate instead of substituting: adding $-4A+B=1$ to $6A-B=1$ gives $2A=2$ directly, so $A=1$ and $B=5$ as before.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two linear systems: first in $A$ and $B$, then in $x+y$ and $x-y$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the equations by renaming the repeated reciprocals $\frac{1}{x+y}$ and $\frac{1}{x-y}$.

💡Takeaway. When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

Rename repeated reciprocals to get a linear system · 12 practice problems

Generated variants — 12

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