Practice · Turn the condition into x+y in terms of xy · Sensim Math

Variant 1 easy

If $\frac{1}{x}+\frac{1}{y}=\frac{3}{2}$ , find the value of the expression below.

$\frac{x(2-y)+2y}{xy-(x+y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(2-y)+2y}{xy-(x+y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{3}{2}\ \Rightarrow\ x+y=\frac{3}{2}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

x+y=(x+y)

.

\frac{x(2-y)+2y}{xy-(x+y)}=\frac{2(x+y)-xy}{xy-(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{3}{2}xy

.

2\times\frac{3}{2}xy-xy=\frac{6}{2}xy-\frac{2}{2}xy=2xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-\frac{3}{2}xy=\frac{2}{2}xy-\frac{3}{2}xy=-\frac{1}{2}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{2xy}{-\frac{1}{2}xy}=-\left(2\div\frac{1}{2}\right)=-4

A negative divided by a positive (or the reverse) is negative.

Answer:

-4

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{4}{3}$ gives $\frac{3}{4}+\frac{3}{4}=\frac{3}{2}$ . Then the numerator is $\frac{4}{3}\times\frac{2}{3}+\frac{8}{3}=\frac{32}{9}$ and the denominator is $\frac{16}{9}-\frac{8}{3}=-\frac{8}{9}$ , and $\frac{32}{9}\div(-\frac{8}{9})=-4$ .

Another way: Set $x+y=3k$ and $xy=2k$ (the ratio $3:2$ ). Then the fraction is $\frac{6k-2k}{2k-3k}=\frac{4k}{-k}=-4$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(2-y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 2 easy

If $\frac{1}{x}+\frac{1}{y}=\frac{5}{2}$ , find the value of the expression below.

$\frac{x(1-2y)+y}{xy-(x+y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(1-2y)+y}{xy-(x+y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{5}{2}\ \Rightarrow\ x+y=\frac{5}{2}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

x+y=(x+y)

.

\frac{x(1-2y)+y}{xy-(x+y)}=\frac{(x+y)-2xy}{xy-(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{5}{2}xy

.

\frac{5}{2}xy-2xy=\frac{5}{2}xy-\frac{4}{2}xy=\frac{1}{2}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-\frac{5}{2}xy=\frac{2}{2}xy-\frac{5}{2}xy=-\frac{3}{2}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{1}{2}xy}{-\frac{3}{2}xy}=-\left(\frac{1}{2}\div\frac{3}{2}\right)=-\frac{1}{3}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{1}{3}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{4}{5}$ gives $\frac{5}{4}+\frac{5}{4}=\frac{5}{2}$ . Then the numerator is $\frac{4}{5}\times(-\frac{3}{5})+\frac{4}{5}=\frac{8}{25}$ and the denominator is $\frac{16}{25}-\frac{8}{5}=-\frac{24}{25}$ , and $\frac{8}{25}\div(-\frac{24}{25})=-\frac{1}{3}$ .

Another way: Set $x+y=5k$ and $xy=2k$ (the ratio $5:2$ ). Then the fraction is $\frac{5k-4k}{2k-5k}=\frac{k}{-3k}=-\frac{1}{3}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(1-2y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 3 easy

If $\frac{1}{x}+\frac{1}{y}=\frac{7}{3}$ , find the value of the expression below.

$\frac{x(1-2y)+y}{xy-(x+y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

$\frac{1}{x}+\frac{1}{y}=\frac{7}{3}$

Unknowns

The value of $\frac{x(1-2y)+y}{xy-(x+y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{7}{3}\ \Rightarrow\ x+y=\frac{7}{3}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

x+y=(x+y)

.

\frac{x(1-2y)+y}{xy-(x+y)}=\frac{(x+y)-2xy}{xy-(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{7}{3}xy

.

\frac{7}{3}xy-2xy=\frac{7}{3}xy-\frac{6}{3}xy=\frac{1}{3}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-\frac{7}{3}xy=\frac{3}{3}xy-\frac{7}{3}xy=-\frac{4}{3}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{1}{3}xy}{-\frac{4}{3}xy}=-\left(\frac{1}{3}\div\frac{4}{3}\right)=-\frac{1}{4}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{1}{4}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{6}{7}$ gives $\frac{7}{6}+\frac{7}{6}=\frac{7}{3}$ . Then the numerator is $\frac{6}{7}\times(-\frac{5}{7})+\frac{6}{7}=\frac{12}{49}$ and the denominator is $\frac{36}{49}-\frac{12}{7}=-\frac{48}{49}$ , and $\frac{12}{49}\div(-\frac{48}{49})=-\frac{1}{4}$ .

Another way: Set $x+y=7k$ and $xy=3k$ (the ratio $7:3$ ). Then the fraction is $\frac{7k-6k}{3k-7k}=\frac{k}{-4k}=-\frac{1}{4}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(1-2y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 4 medium

If $\frac{1}{x}+\frac{1}{y}=\frac{5}{4}$ , find the value of the expression below.

$\frac{x(2-3y)+2y}{xy-(x+y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(2-3y)+2y}{xy-(x+y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{5}{4}\ \Rightarrow\ x+y=\frac{5}{4}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

x+y=(x+y)

.

\frac{x(2-3y)+2y}{xy-(x+y)}=\frac{2(x+y)-3xy}{xy-(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{5}{4}xy

.

2\times\frac{5}{4}xy-3xy=\frac{10}{4}xy-\frac{12}{4}xy=-\frac{1}{2}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-\frac{5}{4}xy=\frac{4}{4}xy-\frac{5}{4}xy=-\frac{1}{4}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{-\frac{1}{2}xy}{-\frac{1}{4}xy}=\frac{1}{2}\div\frac{1}{4}=2

Dividing two negatives gives a positive.

Answer:

2

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{8}{5}$ gives $\frac{5}{8}+\frac{5}{8}=\frac{5}{4}$ . Then the numerator is $\frac{8}{5}\times(-\frac{14}{5})+\frac{16}{5}=-\frac{32}{25}$ and the denominator is $\frac{64}{25}-\frac{16}{5}=-\frac{16}{25}$ , and $-\frac{32}{25}\div(-\frac{16}{25})=2$ .

Another way: Set $x+y=5k$ and $xy=4k$ (the ratio $5:4$ ). Then the fraction is $\frac{10k-12k}{4k-5k}=\frac{-2k}{-k}=2$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(2-3y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 5 easy

If $\frac{1}{x}+\frac{1}{y}=\frac{5}{3}$ , find the value of the expression below.

$\frac{x(2-3y)+2y}{xy-(x+y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(2-3y)+2y}{xy-(x+y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{5}{3}\ \Rightarrow\ x+y=\frac{5}{3}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

x+y=(x+y)

.

\frac{x(2-3y)+2y}{xy-(x+y)}=\frac{2(x+y)-3xy}{xy-(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{5}{3}xy

.

2\times\frac{5}{3}xy-3xy=\frac{10}{3}xy-\frac{9}{3}xy=\frac{1}{3}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-\frac{5}{3}xy=\frac{3}{3}xy-\frac{5}{3}xy=-\frac{2}{3}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{1}{3}xy}{-\frac{2}{3}xy}=-\left(\frac{1}{3}\div\frac{2}{3}\right)=-\frac{1}{2}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{1}{2}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{6}{5}$ gives $\frac{5}{6}+\frac{5}{6}=\frac{5}{3}$ . Then the numerator is $\frac{6}{5}\times(-\frac{8}{5})+\frac{12}{5}=\frac{12}{25}$ and the denominator is $\frac{36}{25}-\frac{12}{5}=-\frac{24}{25}$ , and $\frac{12}{25}\div(-\frac{24}{25})=-\frac{1}{2}$ .

Another way: Set $x+y=5k$ and $xy=3k$ (the ratio $5:3$ ). Then the fraction is $\frac{10k-9k}{3k-5k}=\frac{k}{-2k}=-\frac{1}{2}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(2-3y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 6 medium

If $\frac{1}{x}+\frac{1}{y}=\frac{2}{3}$ , find the value of the expression below.

$\frac{x(3-y)+3y}{xy-(2x+2y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(3-y)+3y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{3}\ \Rightarrow\ x+y=\frac{2}{3}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(3-y)+3y}{xy-(2x+2y)}=\frac{3(x+y)-xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{2}{3}xy

.

3\times\frac{2}{3}xy-xy=\frac{6}{3}xy-\frac{3}{3}xy=xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{2}{3}xy=\frac{3}{3}xy-\frac{4}{3}xy=-\frac{1}{3}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{xy}{-\frac{1}{3}xy}=-\left(1\div\frac{1}{3}\right)=-3

A negative divided by a positive (or the reverse) is negative.

Answer:

-3

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=3$ gives $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ . Then the numerator is $3\times0+9=9$ and the denominator is $9-12=-3$ , and $9\div(-3)=-3$ .

Another way: Set $x+y=2k$ and $xy=3k$ (the ratio $2:3$ ). Then the fraction is $\frac{6k-3k}{3k-4k}=\frac{3k}{-k}=-3$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(3-y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 7 medium

If $\frac{1}{x}+\frac{1}{y}=\frac{7}{5}$ , find the value of the expression below.

$\frac{x(1-y)+y}{xy-(2x+2y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

$\frac{1}{x}+\frac{1}{y}=\frac{7}{5}$

Unknowns

The value of $\frac{x(1-y)+y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{7}{5}\ \Rightarrow\ x+y=\frac{7}{5}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(1-y)+y}{xy-(2x+2y)}=\frac{(x+y)-xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{7}{5}xy

.

\frac{7}{5}xy-xy=\frac{7}{5}xy-\frac{5}{5}xy=\frac{2}{5}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{7}{5}xy=\frac{5}{5}xy-\frac{14}{5}xy=-\frac{9}{5}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{2}{5}xy}{-\frac{9}{5}xy}=-\left(\frac{2}{5}\div\frac{9}{5}\right)=-\frac{2}{9}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{2}{9}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{10}{7}$ gives $\frac{7}{10}+\frac{7}{10}=\frac{7}{5}$ . Then the numerator is $\frac{10}{7}\times(-\frac{3}{7})+\frac{10}{7}=\frac{40}{49}$ and the denominator is $\frac{100}{49}-\frac{40}{7}=-\frac{180}{49}$ , and $\frac{40}{49}\div(-\frac{180}{49})=-\frac{2}{9}$ .

Another way: Set $x+y=7k$ and $xy=5k$ (the ratio $7:5$ ). Then the fraction is $\frac{7k-5k}{5k-14k}=\frac{2k}{-9k}=-\frac{2}{9}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(1-y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 8 medium

If $\frac{1}{x}+\frac{1}{y}=\frac{3}{4}$ , find the value of the expression below.

$\frac{x(1-2y)+y}{xy-(2x+2y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(1-2y)+y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{3}{4}\ \Rightarrow\ x+y=\frac{3}{4}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(1-2y)+y}{xy-(2x+2y)}=\frac{(x+y)-2xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{3}{4}xy

.

\frac{3}{4}xy-2xy=\frac{3}{4}xy-\frac{8}{4}xy=-\frac{5}{4}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{3}{4}xy=\frac{4}{4}xy-\frac{6}{4}xy=-\frac{1}{2}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{-\frac{5}{4}xy}{-\frac{1}{2}xy}=\frac{5}{4}\div\frac{1}{2}=2\frac{1}{2}

Dividing two negatives gives a positive.

Answer:

2\frac{1}{2}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{8}{3}$ gives $\frac{3}{8}+\frac{3}{8}=\frac{3}{4}$ . Then the numerator is $\frac{8}{3}\times(-\frac{13}{3})+\frac{8}{3}=-\frac{80}{9}$ and the denominator is $\frac{64}{9}-\frac{32}{3}=-\frac{32}{9}$ , and $-\frac{80}{9}\div(-\frac{32}{9})=2\frac{1}{2}$ .

Another way: Set $x+y=3k$ and $xy=4k$ (the ratio $3:4$ ). Then the fraction is $\frac{3k-8k}{4k-6k}=\frac{-5k}{-2k}=2\frac{1}{2}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(1-2y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 9 hard

If $\frac{1}{x}+\frac{1}{y}=\frac{2}{5}$ , find the value of the expression below.

$\frac{x(4-y)+4y}{xy-(3x+3y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(4-y)+4y}{xy-(3x+3y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{5}\ \Rightarrow\ x+y=\frac{2}{5}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

3x+3y=3(x+y)

.

\frac{x(4-y)+4y}{xy-(3x+3y)}=\frac{4(x+y)-xy}{xy-3(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{2}{5}xy

.

4\times\frac{2}{5}xy-xy=\frac{8}{5}xy-\frac{5}{5}xy=\frac{3}{5}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-3\times\frac{2}{5}xy=\frac{5}{5}xy-\frac{6}{5}xy=-\frac{1}{5}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{3}{5}xy}{-\frac{1}{5}xy}=-\left(\frac{3}{5}\div\frac{1}{5}\right)=-3

A negative divided by a positive (or the reverse) is negative.

Answer:

-3

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=5$ gives $\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$ . Then the numerator is $5\times(-1)+20=15$ and the denominator is $25-30=-5$ , and $15\div(-5)=-3$ .

Another way: Set $x+y=2k$ and $xy=5k$ (the ratio $2:5$ ). Then the fraction is $\frac{8k-5k}{5k-6k}=\frac{3k}{-k}=-3$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(4-y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 10 hard

If $\frac{1}{x}+\frac{1}{y}=\frac{4}{3}$ , find the value of the expression below.

$\frac{x(2-3y)+2y}{xy-(2x+2y)}$

Your answer

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(2-3y)+2y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{4}{3}\ \Rightarrow\ x+y=\frac{4}{3}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(2-3y)+2y}{xy-(2x+2y)}=\frac{2(x+y)-3xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{4}{3}xy

.

2\times\frac{4}{3}xy-3xy=\frac{8}{3}xy-\frac{9}{3}xy=-\frac{1}{3}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{4}{3}xy=\frac{3}{3}xy-\frac{8}{3}xy=-\frac{5}{3}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{-\frac{1}{3}xy}{-\frac{5}{3}xy}=\frac{1}{3}\div\frac{5}{3}=\frac{1}{5}

Dividing two negatives gives a positive.

Answer:

\frac{1}{5}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{3}{2}$ gives $\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$ . Then the numerator is $\frac{3}{2}\times(-\frac{5}{2})+3=-\frac{3}{4}$ and the denominator is $\frac{9}{4}-6=-\frac{15}{4}$ , and $-\frac{3}{4}\div(-\frac{15}{4})=\frac{1}{5}$ .

Another way: Set $x+y=4k$ and $xy=3k$ (the ratio $4:3$ ). Then the fraction is $\frac{8k-9k}{3k-8k}=\frac{-k}{-5k}=\frac{1}{5}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(2-3y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 11 hard

If $\frac{1}{x}+\frac{1}{y}=\frac{4}{5}$ , find the value of the expression below.

$\frac{x(3-2y)+3y}{xy-(2x+2y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

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

Unknowns

The value of $\frac{x(3-2y)+3y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{4}{5}\ \Rightarrow\ x+y=\frac{4}{5}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(3-2y)+3y}{xy-(2x+2y)}=\frac{3(x+y)-2xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{4}{5}xy

.

3\times\frac{4}{5}xy-2xy=\frac{12}{5}xy-\frac{10}{5}xy=\frac{2}{5}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{4}{5}xy=\frac{5}{5}xy-\frac{8}{5}xy=-\frac{3}{5}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{2}{5}xy}{-\frac{3}{5}xy}=-\left(\frac{2}{5}\div\frac{3}{5}\right)=-\frac{2}{3}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{2}{3}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{5}{2}$ gives $\frac{2}{5}+\frac{2}{5}=\frac{4}{5}$ . Then the numerator is $\frac{5}{2}\times(-2)+\frac{15}{2}=\frac{5}{2}$ and the denominator is $\frac{25}{4}-10=-\frac{15}{4}$ , and $\frac{5}{2}\div(-\frac{15}{4})=-\frac{2}{3}$ .

Another way: Set $x+y=4k$ and $xy=5k$ (the ratio $4:5$ ). Then the fraction is $\frac{12k-10k}{5k-8k}=\frac{2k}{-3k}=-\frac{2}{3}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(3-2y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Variant 12 hard

If $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$ , find the value of the expression below.

$\frac{x(3-2y)+3y}{xy-(2x+2y)}$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know the sum of the reciprocals of $x$ and $y$ . We must find the value of a fraction built from $x$ and $y$ , even though $x$ and $y$ themselves are not determined.

Givens

$\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$

Unknowns

The value of $\frac{x(3-2y)+3y}{xy-(2x+2y)}$ .

Constraints

$x\neq0$ and $y\neq0$ , so $xy\neq0$ .
Many pairs $(x,y)$ satisfy the condition, so the answer must not depend on which pair.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#9 Solve an Easier Related Problem

The condition only links $x+y$ and $xy$ . So rewrite the condition as $x+y$ in terms of $xy$ , rewrite the big fraction using only $x+y$ and $xy$ , and substitute. Everything should become a multiple of $xy$ , which cancels.

3 · Execute5 carry out the plan

1Rewrite the condition

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

Add the two fractions over the common denominator

xy

, then multiply both sides by

xy

.

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{5}{6}\ \Rightarrow\ x+y=\frac{5}{6}xy

The condition is really a statement about

x+y

compared with

xy

.

2Regroup the fraction around x+y

#7 Identify Subproblems 6.EE.A.3

Expand the numerator and collect the

x

and

y

terms; the denominator already has

2x+2y=2(x+y)

.

\frac{x(3-2y)+3y}{xy-(2x+2y)}=\frac{3(x+y)-2xy}{xy-2(x+y)}

Group the pieces so the known combination

x+y

shows up.

3Substitute in the numerator

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

Replace

x+y

with

\frac{5}{6}xy

.

3\times\frac{5}{6}xy-2xy=\frac{15}{6}xy-\frac{12}{6}xy=\frac{1}{2}xy

Now the numerator is just a number times

xy

.

4Substitute in the denominator

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

Do the same below the bar.

xy-2\times\frac{5}{6}xy=\frac{6}{6}xy-\frac{10}{6}xy=-\frac{2}{3}xy

The denominator is also a number times

xy

.

5Cancel xy

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

Since

xy\neq0

, it cancels, leaving a ratio of two numbers.

\frac{\frac{1}{2}xy}{-\frac{2}{3}xy}=-\left(\frac{1}{2}\div\frac{2}{3}\right)=-\frac{3}{4}

A negative divided by a positive (or the reverse) is negative.

Answer:

-\frac{3}{4}

4 · Reviewdoes it hold up?

Pick a pair that fits: $x=y=\frac{12}{5}$ gives $\frac{5}{12}+\frac{5}{12}=\frac{5}{6}$ . Then the numerator is $\frac{12}{5}\times(-\frac{9}{5})+\frac{36}{5}=\frac{72}{25}$ and the denominator is $\frac{144}{25}-\frac{48}{5}=-\frac{96}{25}$ , and $\frac{72}{25}\div(-\frac{96}{25})=-\frac{3}{4}$ .

Another way: Set $x+y=5k$ and $xy=6k$ (the ratio $5:6$ ). Then the fraction is $\frac{15k-12k}{6k-10k}=\frac{3k}{-4k}=-\frac{3}{4}$ .

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the condition and the fraction in terms of $x+y$ and $xy$.
6.EE.A.3 Apply properties to generate equivalent expressions — Expanding $x(3-2y)$ and collecting like terms.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying and dividing the rational coefficients.

💡Takeaway. When you can't find

x

and

y

, find the combination the question really needs.

Turn the condition into x+y in terms of xy · 12 practice problems

Generated variants — 12

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Where next?