Practice · Bound the smallest variable, then list cases · Sensim Math

Variant 1 easy

Suppose $x\ge y\ge 3$ and $z\ge 3$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{2}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 3$ and $z\ge 3$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{2}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{2}$
$x\ge y\ge 3$ , $z\ge 3$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{2}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{2}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{2}

.

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

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{2}<\frac{2}{y}

, which means

y<4

. With

y\ge 3

,

y=3

.

\frac{1}{2}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<4 \;\Rightarrow\; y=3

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=3

into the inequality.

\frac{1}{x}>\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \;\Rightarrow\; x<6

With

x\ge 3

, the integer

x

is

3

,

4

, or

5

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{3}-\frac{1}{2}=\frac{1}{x}-\frac{1}{6}

. Compute it for each

x

.

x=3: \frac{2}{6}-\frac{1}{6}=\frac{1}{6};\quad x=4: \frac{3}{12}-\frac{2}{12}=\frac{1}{12};\quad x=5: \frac{6}{30}-\frac{5}{30}=\frac{1}{30}

Each case gives a unit fraction, so

z=6

,

12

,

30

, all at least

3

.

Answer:

(3, 3, 6)

,

(4, 3, 12)

,

(5, 3, 30)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{5}+\frac{1}{3}-\frac{1}{30}=\frac{6}{30}+\frac{10}{30}-\frac{1}{30}=\frac{15}{30}=\frac{1}{2}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 3$ and $z\ge 3$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{2}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{4}$ . Trying $y=3$ and then $x=3, \dots, 6$ directly shows $x=6$ already gives $\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$ exactly, which is not more than $\frac{1}{2}$ , so the list stops at $5$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 2 easy

Suppose $x\ge y\ge 5$ and $z\ge 5$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{3}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 5$ and $z\ge 5$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{3}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{3}$
$x\ge y\ge 5$ , $z\ge 5$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{3}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{3}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{3}

.

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

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{3}<\frac{2}{y}

, which means

y<6

. With

y\ge 5

,

y=5

.

\frac{1}{3}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<6 \;\Rightarrow\; y=5

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=5

into the inequality.

\frac{1}{x}>\frac{1}{3}-\frac{1}{5}=\frac{2}{15} \;\Rightarrow\; x<\frac{15}{2}

With

x\ge 5

, the integer

x

is

5

,

6

, or

7

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{5}-\frac{1}{3}=\frac{1}{x}-\frac{2}{15}

. Compute it for each

x

.

x=5: \frac{3}{15}-\frac{2}{15}=\frac{1}{15};\quad x=6: \frac{5}{30}-\frac{4}{30}=\frac{1}{30};\quad x=7: \frac{15}{105}-\frac{14}{105}=\frac{1}{105}

Each case gives a unit fraction, so

z=15

,

30

,

105

, all at least

5

.

Answer:

(5, 5, 15)

,

(6, 5, 30)

,

(7, 5, 105)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{7}+\frac{1}{5}-\frac{1}{105}=\frac{15}{105}+\frac{21}{105}-\frac{1}{105}=\frac{35}{105}=\frac{1}{3}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 5$ and $z\ge 5$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{3}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{6}$ . Trying $y=5$ and then $x=5, \dots, 8$ directly shows $x=8$ already gives $\frac{1}{8}+\frac{1}{5}=\frac{13}{40}<\frac{1}{3}$ , so the list stops at $7$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 3 easy

Suppose $x\ge y\ge 7$ and $z\ge 7$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{4}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 7$ and $z\ge 7$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{4}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{4}$
$x\ge y\ge 7$ , $z\ge 7$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{4}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{4}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{4}

.

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

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{4}<\frac{2}{y}

, which means

y<8

. With

y\ge 7

,

y=7

.

\frac{1}{4}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<8 \;\Rightarrow\; y=7

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=7

into the inequality.

\frac{1}{x}>\frac{1}{4}-\frac{1}{7}=\frac{3}{28} \;\Rightarrow\; x<\frac{28}{3}

With

x\ge 7

, the integer

x

is

7

,

8

, or

9

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{7}-\frac{1}{4}=\frac{1}{x}-\frac{3}{28}

. Compute it for each

x

.

x=7: \frac{4}{28}-\frac{3}{28}=\frac{1}{28};\quad x=8: \frac{7}{56}-\frac{6}{56}=\frac{1}{56};\quad x=9: \frac{28}{252}-\frac{27}{252}=\frac{1}{252}

Each case gives a unit fraction, so

z=28

,

56

,

252

, all at least

7

.

Answer:

(7, 7, 28)

,

(8, 7, 56)

,

(9, 7, 252)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{9}+\frac{1}{7}-\frac{1}{252}=\frac{28}{252}+\frac{36}{252}-\frac{1}{252}=\frac{63}{252}=\frac{1}{4}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 7$ and $z\ge 7$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{4}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{8}$ . Trying $y=7$ and then $x=7, \dots, 10$ directly shows $x=10$ already gives $\frac{1}{10}+\frac{1}{7}=\frac{17}{70}<\frac{1}{4}$ , so the list stops at $9$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 4 easy

Suppose $x\ge y\ge 9$ and $z\ge 9$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{5}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 9$ and $z\ge 9$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{5}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{5}$
$x\ge y\ge 9$ , $z\ge 9$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{5}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{5}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{5}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{5}+\frac{1}{z}>\frac{1}{5}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{5}<\frac{2}{y}

, which means

y<10

. With

y\ge 9

,

y=9

.

\frac{1}{5}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<10 \;\Rightarrow\; y=9

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=9

into the inequality.

\frac{1}{x}>\frac{1}{5}-\frac{1}{9}=\frac{4}{45} \;\Rightarrow\; x<\frac{45}{4}

With

x\ge 9

, the integer

x

is

9

,

10

, or

11

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{9}-\frac{1}{5}=\frac{1}{x}-\frac{4}{45}

. Compute it for each

x

.

x=9: \frac{5}{45}-\frac{4}{45}=\frac{1}{45};\quad x=10: \frac{9}{90}-\frac{8}{90}=\frac{1}{90};\quad x=11: \frac{45}{495}-\frac{44}{495}=\frac{1}{495}

Each case gives a unit fraction, so

z=45

,

90

,

495

, all at least

9

.

Answer:

(9, 9, 45)

,

(10, 9, 90)

,

(11, 9, 495)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{11}+\frac{1}{9}-\frac{1}{495}=\frac{45}{495}+\frac{55}{495}-\frac{1}{495}=\frac{99}{495}=\frac{1}{5}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 9$ and $z\ge 9$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{5}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{10}$ . Trying $y=9$ and then $x=9, \dots, 12$ directly shows $x=12$ already gives $\frac{1}{12}+\frac{1}{9}=\frac{7}{36}<\frac{1}{5}$ , so the list stops at $11$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 5 medium

Suppose $x\ge y\ge 11$ and $z\ge 11$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{6}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 11$ and $z\ge 11$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{6}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{6}$
$x\ge y\ge 11$ , $z\ge 11$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{6}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{6}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{6}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{6}+\frac{1}{z}>\frac{1}{6}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{6}<\frac{2}{y}

, which means

y<12

. With

y\ge 11

,

y=11

.

\frac{1}{6}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<12 \;\Rightarrow\; y=11

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=11

into the inequality.

\frac{1}{x}>\frac{1}{6}-\frac{1}{11}=\frac{5}{66} \;\Rightarrow\; x<\frac{66}{5}

With

x\ge 11

, the integer

x

is

11

,

12

, or

13

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{11}-\frac{1}{6}=\frac{1}{x}-\frac{5}{66}

. Compute it for each

x

.

x=11: \frac{6}{66}-\frac{5}{66}=\frac{1}{66};\quad x=12: \frac{11}{132}-\frac{10}{132}=\frac{1}{132};\quad x=13: \frac{66}{858}-\frac{65}{858}=\frac{1}{858}

Each case gives a unit fraction, so

z=66

,

132

,

858

, all at least

11

.

Answer:

(11, 11, 66)

,

(12, 11, 132)

,

(13, 11, 858)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{13}+\frac{1}{11}-\frac{1}{858}=\frac{66}{858}+\frac{78}{858}-\frac{1}{858}=\frac{143}{858}=\frac{1}{6}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 11$ and $z\ge 11$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{6}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{12}$ . Trying $y=11$ and then $x=11, \dots, 14$ directly shows $x=14$ already gives $\frac{1}{14}+\frac{1}{11}=\frac{25}{154}<\frac{1}{6}$ , so the list stops at $13$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 6 medium

Suppose $x\ge y\ge 13$ and $z\ge 13$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{7}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 13$ and $z\ge 13$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{7}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{7}$
$x\ge y\ge 13$ , $z\ge 13$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{7}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{7}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{7}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{7}+\frac{1}{z}>\frac{1}{7}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{7}<\frac{2}{y}

, which means

y<14

. With

y\ge 13

,

y=13

.

\frac{1}{7}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<14 \;\Rightarrow\; y=13

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=13

into the inequality.

\frac{1}{x}>\frac{1}{7}-\frac{1}{13}=\frac{6}{91} \;\Rightarrow\; x<\frac{91}{6}

With

x\ge 13

, the integer

x

is

13

,

14

, or

15

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{13}-\frac{1}{7}=\frac{1}{x}-\frac{6}{91}

. Compute it for each

x

.

x=13: \frac{7}{91}-\frac{6}{91}=\frac{1}{91};\quad x=14: \frac{13}{182}-\frac{12}{182}=\frac{1}{182};\quad x=15: \frac{91}{1365}-\frac{90}{1365}=\frac{1}{1365}

Each case gives a unit fraction, so

z=91

,

182

,

1365

, all at least

13

.

Answer:

(13, 13, 91)

,

(14, 13, 182)

,

(15, 13, 1365)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{15}+\frac{1}{13}-\frac{1}{1365}=\frac{91}{1365}+\frac{105}{1365}-\frac{1}{1365}=\frac{195}{1365}=\frac{1}{7}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 13$ and $z\ge 13$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{7}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{14}$ . Trying $y=13$ and then $x=13, \dots, 16$ directly shows $x=16$ already gives $\frac{1}{16}+\frac{1}{13}=\frac{29}{208}<\frac{1}{7}$ , so the list stops at $15$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 7 medium

Suppose $x\ge y\ge 15$ and $z\ge 15$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{8}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 15$ and $z\ge 15$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{8}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{8}$
$x\ge y\ge 15$ , $z\ge 15$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{8}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{8}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{8}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{8}+\frac{1}{z}>\frac{1}{8}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{8}<\frac{2}{y}

, which means

y<16

. With

y\ge 15

,

y=15

.

\frac{1}{8}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<16 \;\Rightarrow\; y=15

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=15

into the inequality.

\frac{1}{x}>\frac{1}{8}-\frac{1}{15}=\frac{7}{120} \;\Rightarrow\; x<\frac{120}{7}

With

x\ge 15

, the integer

x

is

15

,

16

, or

17

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{15}-\frac{1}{8}=\frac{1}{x}-\frac{7}{120}

. Compute it for each

x

.

x=15: \frac{8}{120}-\frac{7}{120}=\frac{1}{120};\quad x=16: \frac{15}{240}-\frac{14}{240}=\frac{1}{240};\quad x=17: \frac{120}{2040}-\frac{119}{2040}=\frac{1}{2040}

Each case gives a unit fraction, so

z=120

,

240

,

2040

, all at least

15

.

Answer:

(15, 15, 120)

,

(16, 15, 240)

,

(17, 15, 2040)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{17}+\frac{1}{15}-\frac{1}{2040}=\frac{120}{2040}+\frac{136}{2040}-\frac{1}{2040}=\frac{255}{2040}=\frac{1}{8}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 15$ and $z\ge 15$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{8}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{16}$ . Trying $y=15$ and then $x=15, \dots, 18$ directly shows $x=18$ already gives $\frac{1}{18}+\frac{1}{15}=\frac{11}{90}<\frac{1}{8}$ , so the list stops at $17$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 8 medium

Suppose $x\ge y\ge 17$ and $z\ge 17$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{9}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 17$ and $z\ge 17$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{9}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{9}$
$x\ge y\ge 17$ , $z\ge 17$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{9}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{9}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{9}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{9}+\frac{1}{z}>\frac{1}{9}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{9}<\frac{2}{y}

, which means

y<18

. With

y\ge 17

,

y=17

.

\frac{1}{9}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<18 \;\Rightarrow\; y=17

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=17

into the inequality.

\frac{1}{x}>\frac{1}{9}-\frac{1}{17}=\frac{8}{153} \;\Rightarrow\; x<\frac{153}{8}

With

x\ge 17

, the integer

x

is

17

,

18

, or

19

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{17}-\frac{1}{9}=\frac{1}{x}-\frac{8}{153}

. Compute it for each

x

.

x=17: \frac{9}{153}-\frac{8}{153}=\frac{1}{153};\quad x=18: \frac{17}{306}-\frac{16}{306}=\frac{1}{306};\quad x=19: \frac{153}{2907}-\frac{152}{2907}=\frac{1}{2907}

Each case gives a unit fraction, so

z=153

,

306

,

2907

, all at least

17

.

Answer:

(17, 17, 153)

,

(18, 17, 306)

,

(19, 17, 2907)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{19}+\frac{1}{17}-\frac{1}{2907}=\frac{153}{2907}+\frac{171}{2907}-\frac{1}{2907}=\frac{323}{2907}=\frac{1}{9}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 17$ and $z\ge 17$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{9}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{18}$ . Trying $y=17$ and then $x=17, \dots, 20$ directly shows $x=20$ already gives $\frac{1}{20}+\frac{1}{17}=\frac{37}{340}<\frac{1}{9}$ , so the list stops at $19$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 9 hard

Suppose $x\ge y\ge 19$ and $z\ge 19$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{10}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 19$ and $z\ge 19$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{10}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{10}$
$x\ge y\ge 19$ , $z\ge 19$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{10}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{10}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{10}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{10}+\frac{1}{z}>\frac{1}{10}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{10}<\frac{2}{y}

, which means

y<20

. With

y\ge 19

,

y=19

.

\frac{1}{10}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<20 \;\Rightarrow\; y=19

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=19

into the inequality.

\frac{1}{x}>\frac{1}{10}-\frac{1}{19}=\frac{9}{190} \;\Rightarrow\; x<\frac{190}{9}

With

x\ge 19

, the integer

x

is

19

,

20

, or

21

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{19}-\frac{1}{10}=\frac{1}{x}-\frac{9}{190}

. Compute it for each

x

.

x=19: \frac{10}{190}-\frac{9}{190}=\frac{1}{190};\quad x=20: \frac{19}{380}-\frac{18}{380}=\frac{1}{380};\quad x=21: \frac{190}{3990}-\frac{189}{3990}=\frac{1}{3990}

Each case gives a unit fraction, so

z=190

,

380

,

3990

, all at least

19

.

Answer:

(19, 19, 190)

,

(20, 19, 380)

,

(21, 19, 3990)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{21}+\frac{1}{19}-\frac{1}{3990}=\frac{190}{3990}+\frac{210}{3990}-\frac{1}{3990}=\frac{399}{3990}=\frac{1}{10}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 19$ and $z\ge 19$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{10}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{20}$ . Trying $y=19$ and then $x=19, \dots, 22$ directly shows $x=22$ already gives $\frac{1}{22}+\frac{1}{19}=\frac{41}{418}<\frac{1}{10}$ , so the list stops at $21$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 10 hard

Suppose $x\ge y\ge 21$ and $z\ge 21$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{11}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 21$ and $z\ge 21$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{11}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{11}$
$x\ge y\ge 21$ , $z\ge 21$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{11}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{11}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{11}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{11}+\frac{1}{z}>\frac{1}{11}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{11}<\frac{2}{y}

, which means

y<22

. With

y\ge 21

,

y=21

.

\frac{1}{11}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<22 \;\Rightarrow\; y=21

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=21

into the inequality.

\frac{1}{x}>\frac{1}{11}-\frac{1}{21}=\frac{10}{231} \;\Rightarrow\; x<\frac{231}{10}

With

x\ge 21

, the integer

x

is

21

,

22

, or

23

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{21}-\frac{1}{11}=\frac{1}{x}-\frac{10}{231}

. Compute it for each

x

.

x=21: \frac{11}{231}-\frac{10}{231}=\frac{1}{231};\quad x=22: \frac{21}{462}-\frac{20}{462}=\frac{1}{462};\quad x=23: \frac{231}{5313}-\frac{230}{5313}=\frac{1}{5313}

Each case gives a unit fraction, so

z=231

,

462

,

5313

, all at least

21

.

Answer:

(21, 21, 231)

,

(22, 21, 462)

,

(23, 21, 5313)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{23}+\frac{1}{21}-\frac{1}{5313}=\frac{231}{5313}+\frac{253}{5313}-\frac{1}{5313}=\frac{483}{5313}=\frac{1}{11}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 21$ and $z\ge 21$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{11}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{22}$ . Trying $y=21$ and then $x=21, \dots, 24$ directly shows $x=24$ already gives $\frac{1}{24}+\frac{1}{21}=\frac{5}{56}<\frac{1}{11}$ , so the list stops at $23$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 11 hard

Suppose $x\ge y\ge 23$ and $z\ge 23$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{12}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 23$ and $z\ge 23$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{12}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{12}$
$x\ge y\ge 23$ , $z\ge 23$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{12}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{12}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{12}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{12}+\frac{1}{z}>\frac{1}{12}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{12}<\frac{2}{y}

, which means

y<24

. With

y\ge 23

,

y=23

.

\frac{1}{12}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<24 \;\Rightarrow\; y=23

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=23

into the inequality.

\frac{1}{x}>\frac{1}{12}-\frac{1}{23}=\frac{11}{276} \;\Rightarrow\; x<\frac{276}{11}

With

x\ge 23

, the integer

x

is

23

,

24

, or

25

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{23}-\frac{1}{12}=\frac{1}{x}-\frac{11}{276}

. Compute it for each

x

.

x=23: \frac{12}{276}-\frac{11}{276}=\frac{1}{276};\quad x=24: \frac{23}{552}-\frac{22}{552}=\frac{1}{552};\quad x=25: \frac{276}{6900}-\frac{275}{6900}=\frac{1}{6900}

Each case gives a unit fraction, so

z=276

,

552

,

6900

, all at least

23

.

Answer:

(23, 23, 276)

,

(24, 23, 552)

,

(25, 23, 6900)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{25}+\frac{1}{23}-\frac{1}{6900}=\frac{276}{6900}+\frac{300}{6900}-\frac{1}{6900}=\frac{575}{6900}=\frac{1}{12}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 23$ and $z\ge 23$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{12}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{24}$ . Trying $y=23$ and then $x=23, \dots, 26$ directly shows $x=26$ already gives $\frac{1}{26}+\frac{1}{23}=\frac{49}{598}<\frac{1}{12}$ , so the list stops at $25$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Variant 12 hard

Suppose $x\ge y\ge 25$ and $z\ge 25$ . Find all ordered triples of integers $(x, y, z)$ that satisfy $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{13}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every integer triple with $x\ge y\ge 25$ and $z\ge 25$ for which $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{13}$ . Use inequalities to squeeze $y$ and then $x$ into a few values, and compute $z$ in each case.

Givens

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{13}$
$x\ge y\ge 25$ , $z\ge 25$ , and $x$ , $y$ , $z$ are integers.

Unknowns

All ordered triples $(x, y, z)$ .

Constraints

All three variables are positive, so all three unit fractions are positive.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#2 Make a Systematic List

Move $\frac{1}{z}$ across so that the left side is a sum of two unit fractions that must exceed $\frac{1}{13}$ . Since $x\ge y$ , the sum is at most $\frac{2}{y}$ , which bounds $y$ ; then the same inequality bounds $x$ , and a short list of cases remains.

3 · Execute4 carry out the plan

1Rearrange and compare with $\frac{1}{13}$

#15 Organize Information in More Ways 7.EE.B.4

Add

\frac{1}{z}

to both sides. Because

\frac{1}{z}>0

, the sum

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

is bigger than

\frac{1}{13}

.

\frac{1}{x}+\frac{1}{y}=\frac{1}{13}+\frac{1}{z}>\frac{1}{13}

Seen this way, the equation becomes an inequality with only

x

and

y

.

2Bound $y$

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

Since

x\ge y>0

,

\frac{1}{x}\le\frac{1}{y}

, so the sum is at most

\frac{2}{y}

. Hence

\frac{1}{13}<\frac{2}{y}

, which means

y<26

. With

y\ge 25

,

y=25

.

\frac{1}{13}<\frac{1}{x}+\frac{1}{y}\le\frac{2}{y} \;\Rightarrow\; y<26 \;\Rightarrow\; y=25

The larger of two unit fractions must carry at least half of the sum.

3Bound $x$

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

Put

y=25

into the inequality.

\frac{1}{x}>\frac{1}{13}-\frac{1}{25}=\frac{12}{325} \;\Rightarrow\; x<\frac{325}{12}

With

x\ge 25

, the integer

x

is

25

,

26

, or

27

.

4Find $z$ in each case

#2 Make a Systematic List 7.NS.A.1

From the equation,

\frac{1}{z}=\frac{1}{x}+\frac{1}{25}-\frac{1}{13}=\frac{1}{x}-\frac{12}{325}

. Compute it for each

x

.

x=25: \frac{13}{325}-\frac{12}{325}=\frac{1}{325};\quad x=26: \frac{25}{650}-\frac{24}{650}=\frac{1}{650};\quad x=27: \frac{325}{8775}-\frac{324}{8775}=\frac{1}{8775}

Each case gives a unit fraction, so

z=325

,

650

,

8775

, all at least

25

.

Answer:

(25, 25, 325)

,

(26, 25, 650)

,

(27, 25, 8775)

4 · Reviewdoes it hold up?

Check one: $\frac{1}{27}+\frac{1}{25}-\frac{1}{8775}=\frac{325}{8775}+\frac{351}{8775}-\frac{1}{8775}=\frac{675}{8775}=\frac{1}{13}$ . The other two check the same way, and every triple satisfies $x\ge y\ge 25$ and $z\ge 25$ .

Another way: Think of it as a sharing problem: $\frac{1}{x}+\frac{1}{y}$ must be a little more than $\frac{1}{13}$ , and the smaller denominator $y$ has to supply at least half, so $\frac{1}{y}>\frac{1}{26}$ . Trying $y=25$ and then $x=25, \dots, 28$ directly shows $x=28$ already gives $\frac{1}{28}+\frac{1}{25}=\frac{53}{700}<\frac{1}{13}$ , so the list stops at $27$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Building and solving the inequalities that bound $y$ and $x$.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding and subtracting fractions to compute $\frac{1}{z}$ in each case.

💡Takeaway. An equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.

Bound the smallest variable, then list cases · 12 practice problems

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