Practice · Sum of reciprocals is sum over product · Sensim Math

Variant 1 easy

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $1$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $1$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $1$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $1$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=1

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=1

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div 1=\frac{a+b+c}{2}

No single number was ever needed.

Answer:

\frac{a+b+c}{2}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=1$ : the product is $1$ , the pairwise sums are $2$ , $2$ , $2$ , and the reciprocal sum is $1+1+1=3$ . The formula gives $6$ times $\frac{1}{2}$ , which is $3$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 2 easy

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $-2$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $-2$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $-2$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $-2$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=-2

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=-2

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \left(-2\right)=-\frac{a+b+c}{4}

No single number was ever needed.

Answer:

-\frac{a+b+c}{4}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=-2$ : the product is $-2$ , the pairwise sums are $2$ , $-1$ , $-1$ , and the reciprocal sum is $1-\frac{1}{2}-\frac{1}{2}=0$ . The formula gives $0$ times $-\frac{1}{4}$ , which is $0$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 3 easy

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $2$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $2$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $2$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $2$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=2

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=2

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div 2=\frac{a+b+c}{4}

No single number was ever needed.

Answer:

\frac{a+b+c}{4}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=2$ : the product is $2$ , the pairwise sums are $2$ , $3$ , $3$ , and the reciprocal sum is $1+\frac{1}{2}+\frac{1}{2}=2$ . The formula gives $8$ times $\frac{1}{4}$ , which is $2$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 4 easy

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $3$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $3$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $3$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $3$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=3

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=3

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div 3=\frac{a+b+c}{6}

No single number was ever needed.

Answer:

\frac{a+b+c}{6}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=3$ : the product is $3$ , the pairwise sums are $2$ , $4$ , $4$ , and the reciprocal sum is $1+\frac{1}{3}+\frac{1}{3}=\frac{5}{3}$ . The formula gives $10$ times $\frac{1}{6}$ , which is $\frac{5}{3}$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 5 medium

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $-3$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $-3$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $-3$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $-3$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=-3

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=-3

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \left(-3\right)=-\frac{a+b+c}{6}

No single number was ever needed.

Answer:

-\frac{a+b+c}{6}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=-3$ : the product is $-3$ , the pairwise sums are $2$ , $-2$ , $-2$ , and the reciprocal sum is $1-\frac{1}{3}-\frac{1}{3}=\frac{1}{3}$ . The formula gives $-2$ times $-\frac{1}{6}$ , which is $\frac{1}{3}$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 6 medium

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $4$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $4$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $4$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $4$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=4

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=4

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div 4=\frac{a+b+c}{8}

No single number was ever needed.

Answer:

\frac{a+b+c}{8}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=4$ : the product is $4$ , the pairwise sums are $2$ , $5$ , $5$ , and the reciprocal sum is $1+\frac{1}{4}+\frac{1}{4}=\frac{3}{2}$ . The formula gives $12$ times $\frac{1}{8}$ , which is $\frac{3}{2}$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 7 medium

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $\frac{1}{2}$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $\frac{1}{2}$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $\frac{1}{2}$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $\frac{1}{2}$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=\frac{1}{2}

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=\frac{1}{2}

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \frac{1}{2}=a+b+c

No single number was ever needed.

Answer:

a+b+c

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=\frac{1}{2}$ : the product is $\frac{1}{2}$ , the pairwise sums are $2$ , $\frac{3}{2}$ , $\frac{3}{2}$ , and the reciprocal sum is $1+2+2=5$ . The formula gives $5$ times $1$ , which is $5$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 8 medium

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $-\frac{1}{2}$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $-\frac{1}{2}$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $-\frac{1}{2}$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $-\frac{1}{2}$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=-\frac{1}{2}

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=-\frac{1}{2}

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \left(-\frac{1}{2}\right)=-(a+b+c)

No single number was ever needed.

Answer:

-(a+b+c)

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=-\frac{1}{2}$ : the product is $-\frac{1}{2}$ , the pairwise sums are $2$ , $\frac{1}{2}$ , $\frac{1}{2}$ , and the reciprocal sum is $1-2-2=-3$ . The formula gives $3$ times $-1$ , which is $-3$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 9 hard

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $5$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $5$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $5$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $5$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=5

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=5

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div 5=\frac{a+b+c}{10}

No single number was ever needed.

Answer:

\frac{a+b+c}{10}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=5$ : the product is $5$ , the pairwise sums are $2$ , $6$ , $6$ , and the reciprocal sum is $1+\frac{1}{5}+\frac{1}{5}=\frac{7}{5}$ . The formula gives $14$ times $\frac{1}{10}$ , which is $\frac{7}{5}$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 10 hard

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $\frac{1}{3}$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $\frac{1}{3}$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $\frac{1}{3}$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $\frac{1}{3}$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=\frac{1}{3}

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=\frac{1}{3}

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \frac{1}{3}=\frac{3(a+b+c)}{2}

No single number was ever needed.

Answer:

\frac{3(a+b+c)}{2}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=\frac{1}{3}$ : the product is $\frac{1}{3}$ , the pairwise sums are $2$ , $\frac{4}{3}$ , $\frac{4}{3}$ , and the reciprocal sum is $1+3+3=7$ . The formula gives $\frac{14}{3}$ times $\frac{3}{2}$ , which is $7$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 11 hard

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $\frac{3}{2}$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $\frac{3}{2}$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $\frac{3}{2}$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $\frac{3}{2}$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=\frac{3}{2}

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=\frac{3}{2}

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \frac{3}{2}=\frac{a+b+c}{3}

No single number was ever needed.

Answer:

\frac{a+b+c}{3}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=\frac{3}{2}$ : the product is $\frac{3}{2}$ , the pairwise sums are $2$ , $\frac{5}{2}$ , $\frac{5}{2}$ , and the reciprocal sum is $1+\frac{2}{3}+\frac{2}{3}=\frac{7}{3}$ . The formula gives $7$ times $\frac{1}{3}$ , which is $\frac{7}{3}$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Variant 12 hard

There are three numbers. Adding them two at a time gives $a$ , $b$ , and $c$ , and the product of all three numbers is $\frac{2}{3}$ . Now multiply the three numbers two at a time and add up the reciprocals of those three products. Write this sum in terms of $a$ , $b$ , and $c$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three unknown numbers have pairwise sums $a$ , $b$ , $c$ and product $\frac{2}{3}$ . Express $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$ in terms of $a$ , $b$ , $c$ .

Givens

The three pairwise sums are $a$ , $b$ , and $c$ .
The product of the three numbers is $\frac{2}{3}$ .

Unknowns

The sum of the reciprocals of the three pairwise products, in terms of $a$ , $b$ , $c$ .

Constraints

None of the numbers is $0$ , because their product is $\frac{2}{3}$ .

2 · Planchoose the strategy

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

Name the three numbers. Instead of finding each one, look at what the target needs: over a common denominator the reciprocal sum is just (sum of the numbers) divided by (their product), and both of those come straight from the givens.

3 · Execute4 carry out the plan

1Name the numbers

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

Call the numbers

x

,

y

,

z

and write the givens as equations.

x+y=a,\ y+z=b,\ z+x=c,\ xyz=\frac{2}{3}

Four facts about three letters.

2Add the three sums

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

Adding the three pairwise sums counts every number twice.

2(x+y+z)=a+b+c

So the total of the three numbers is half of

a+b+c

.

3Common denominator

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

Give the three reciprocals the common denominator

xyz

: each missing letter moves to the top.

\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}

The top becomes the sum, the bottom the product.

4Substitute

#7 Identify Subproblems 6.EE.A.3

Put in

x+y+z=\frac{a+b+c}{2}

and

xyz=\frac{2}{3}

.

\frac{x+y+z}{xyz}=\frac{a+b+c}{2} \div \frac{2}{3}=\frac{3(a+b+c)}{4}

No single number was ever needed.

Answer:

\frac{3(a+b+c)}{4}

4 · Reviewdoes it hold up?

Try $x=1$ , $y=1$ , $z=\frac{2}{3}$ : the product is $\frac{2}{3}$ , the pairwise sums are $2$ , $\frac{5}{3}$ , $\frac{5}{3}$ , and the reciprocal sum is $1+\frac{3}{2}+\frac{3}{2}=4$ . The formula gives $\frac{16}{3}$ times $\frac{3}{4}$ , which is $4$ too.

Another way: Solving for $x$ , $y$ , $z$ one by one ( $x=\frac{a-b+c}{2}$ and so on) also works but is much messier; the common-denominator view skips it entirely.

Standardsmin grade 7

7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the reciprocal sum as sum over product, and the three sums as twice the total.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Naming the three numbers and writing the conditions as equations.
6.EE.A.3 Apply properties to generate equivalent expressions — Substituting equal expressions to get the final form.

💡Takeaway. Before solving for each unknown, check whether the question only needs their sum and product.

Sum of reciprocals is sum over product · 12 practice problems

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