Practice · Express years married two ways, then equate · Sensim Math

Variant 1 easy

A man is now $3$ years older than his wife. $8$ years ago, he had been married for exactly half of the life he had lived so far. $12$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $25$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 3 years older than his wife. One fact about him 8 years ago and one fact about her 12 years from now both describe how long they have been married. From those, find his age when they had been married 25 years.

Givens

The man is $3$ years older than his wife.
$8$ years ago, his years of marriage were half of his age then.
$12$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $25$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 25-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 3

.

\text{man} = x,\quad \text{wife} = x - 3

One unknown is enough, because the two ages differ by a fixed

3

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

8

years ago he was

x - 8

and had been married half of that. 8 more years have passed since, so today the marriage is

8

years longer.

\text{years married now} = \frac{1}{2}(x - 8) + 8

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

12

years she will be

(x - 3) + 12 = x + 9

and married for

\frac{2}{3}

of that. Today is

12

years earlier, so subtract

12

.

\text{years married now} = \frac{2}{3}(x + 9) - 12

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{2}(x-8)+8=\frac{2}{3}(x+9)-12 \;\Rightarrow\; 3x-24+48=4x+36-72 \;\Rightarrow\; 3x+24=4x-36 \;\Rightarrow\; x=60

Clearing denominators turns it into a one-step linear equation.

5Count back to the 25th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(60-8)+8 = 26+8 = 34

years. The 25th anniversary was

34 - 25 = 9

years ago.

60 - 9 = 51

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 51 years old

4 · Reviewdoes it hold up?

$8$ years ago he was $52$ and married $26$ years, exactly half. In $12$ years she will be $57 + 12 = 69$ and married $34 + 12 = 46$ years, and $\frac{2}{3} \times 69 = 46$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 8$ must be even and $x + 9$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 60$ makes them agree (both give $34$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $25$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 2 easy

A man is now $4$ years older than his wife. $6$ years ago, he had been married for exactly half of the life he had lived so far. $13$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $30$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 4 years older than his wife. One fact about him 6 years ago and one fact about her 13 years from now both describe how long they have been married. From those, find his age when they had been married 30 years.

Givens

The man is $4$ years older than his wife.
$6$ years ago, his years of marriage were half of his age then.
$13$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $30$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 30-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 4

.

\text{man} = x,\quad \text{wife} = x - 4

One unknown is enough, because the two ages differ by a fixed

4

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

6

years ago he was

x - 6

and had been married half of that. 6 more years have passed since, so today the marriage is

6

years longer.

\text{years married now} = \frac{1}{2}(x - 6) + 6

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

13

years she will be

(x - 4) + 13 = x + 9

and married for

\frac{2}{3}

of that. Today is

13

years earlier, so subtract

13

.

\text{years married now} = \frac{2}{3}(x + 9) - 13

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{2}(x-6)+6=\frac{2}{3}(x+9)-13 \;\Rightarrow\; 3x-18+36=4x+36-78 \;\Rightarrow\; 3x+18=4x-42 \;\Rightarrow\; x=60

Clearing denominators turns it into a one-step linear equation.

5Count back to the 30th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(60-6)+6 = 27+6 = 33

years. The 30th anniversary was

33 - 30 = 3

years ago.

60 - 3 = 57

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 57 years old

4 · Reviewdoes it hold up?

$6$ years ago he was $54$ and married $27$ years, exactly half. In $13$ years she will be $56 + 13 = 69$ and married $33 + 13 = 46$ years, and $\frac{2}{3} \times 69 = 46$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 6$ must be even and $x + 9$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 60$ makes them agree (both give $33$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $30$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 3 easy

A man is now $4$ years older than his wife. $2$ years ago, he had been married for exactly half of the life he had lived so far. $9$ years from now, his wife will have been married for $\frac{3}{5}$ of her life. How old was the man on the couple's $30$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 4 years older than his wife. One fact about him 2 years ago and one fact about her 9 years from now both describe how long they have been married. From those, find his age when they had been married 30 years.

Givens

The man is $4$ years older than his wife.
$2$ years ago, his years of marriage were half of his age then.
$9$ years from now, her years of marriage will be $\frac{3}{5}$ of her age then.

Unknowns

The man's age at the $30$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 30-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 4

.

\text{man} = x,\quad \text{wife} = x - 4

One unknown is enough, because the two ages differ by a fixed

4

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

2

years ago he was

x - 2

and had been married half of that. 2 more years have passed since, so today the marriage is

2

years longer.

\text{years married now} = \frac{1}{2}(x - 2) + 2

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

9

years she will be

(x - 4) + 9 = x + 5

and married for

\frac{3}{5}

of that. Today is

9

years earlier, so subtract

9

.

\text{years married now} = \frac{3}{5}(x + 5) - 9

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

10

to clear the fractions.

\frac{1}{2}(x-2)+2=\frac{3}{5}(x+5)-9 \;\Rightarrow\; 5x-10+20=6x+30-90 \;\Rightarrow\; 5x+10=6x-60 \;\Rightarrow\; x=70

Clearing denominators turns it into a one-step linear equation.

5Count back to the 30th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(70-2)+2 = 34+2 = 36

years. The 30th anniversary was

36 - 30 = 6

years ago.

70 - 6 = 64

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 64 years old

4 · Reviewdoes it hold up?

$2$ years ago he was $68$ and married $34$ years, exactly half. In $9$ years she will be $66 + 9 = 75$ and married $36 + 9 = 45$ years, and $\frac{3}{5} \times 75 = 45$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 2$ must be even and $x + 5$ a multiple of $5$ . Testing such ages in both marriage-length expressions, only $x = 70$ makes them agree (both give $36$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $30$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 4 easy

A man is now $5$ years older than his wife. $9$ years ago, he had been married for exactly half of the life he had lived so far. $6$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $30$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 5 years older than his wife. One fact about him 9 years ago and one fact about her 6 years from now both describe how long they have been married. From those, find his age when they had been married 30 years.

Givens

The man is $5$ years older than his wife.
$9$ years ago, his years of marriage were half of his age then.
$6$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $30$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 30-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 5

.

\text{man} = x,\quad \text{wife} = x - 5

One unknown is enough, because the two ages differ by a fixed

5

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

9

years ago he was

x - 9

and had been married half of that. 9 more years have passed since, so today the marriage is

9

years longer.

\text{years married now} = \frac{1}{2}(x - 9) + 9

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

6

years she will be

(x - 5) + 6 = x + 1

and married for

\frac{2}{3}

of that. Today is

6

years earlier, so subtract

6

.

\text{years married now} = \frac{2}{3}(x + 1) - 6

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{2}(x-9)+9=\frac{2}{3}(x+1)-6 \;\Rightarrow\; 3x-27+54=4x+4-36 \;\Rightarrow\; 3x+27=4x-32 \;\Rightarrow\; x=59

Clearing denominators turns it into a one-step linear equation.

5Count back to the 30th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(59-9)+9 = 25+9 = 34

years. The 30th anniversary was

34 - 30 = 4

years ago.

59 - 4 = 55

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 55 years old

4 · Reviewdoes it hold up?

$9$ years ago he was $50$ and married $25$ years, exactly half. In $6$ years she will be $54 + 6 = 60$ and married $34 + 6 = 40$ years, and $\frac{2}{3} \times 60 = 40$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 9$ must be even and $x + 1$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 59$ makes them agree (both give $34$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $30$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 5 medium

A man is now $6$ years older than his wife. $5$ years ago, he had been married for exactly half of the life he had lived so far. $13$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $30$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 6 years older than his wife. One fact about him 5 years ago and one fact about her 13 years from now both describe how long they have been married. From those, find his age when they had been married 30 years.

Givens

The man is $6$ years older than his wife.
$5$ years ago, his years of marriage were half of his age then.
$13$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $30$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 30-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 6

.

\text{man} = x,\quad \text{wife} = x - 6

One unknown is enough, because the two ages differ by a fixed

6

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

5

years ago he was

x - 5

and had been married half of that. 5 more years have passed since, so today the marriage is

5

years longer.

\text{years married now} = \frac{1}{2}(x - 5) + 5

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

13

years she will be

(x - 6) + 13 = x + 7

and married for

\frac{2}{3}

of that. Today is

13

years earlier, so subtract

13

.

\text{years married now} = \frac{2}{3}(x + 7) - 13

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{2}(x-5)+5=\frac{2}{3}(x+7)-13 \;\Rightarrow\; 3x-15+30=4x+28-78 \;\Rightarrow\; 3x+15=4x-50 \;\Rightarrow\; x=65

Clearing denominators turns it into a one-step linear equation.

5Count back to the 30th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(65-5)+5 = 30+5 = 35

years. The 30th anniversary was

35 - 30 = 5

years ago.

65 - 5 = 60

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 60 years old

4 · Reviewdoes it hold up?

$5$ years ago he was $60$ and married $30$ years, exactly half. In $13$ years she will be $59 + 13 = 72$ and married $35 + 13 = 48$ years, and $\frac{2}{3} \times 72 = 48$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 5$ must be even and $x + 7$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 65$ makes them agree (both give $35$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $30$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 6 medium

A man is now $6$ years older than his wife. $8$ years ago, he had been married for exactly half of the life he had lived so far. $14$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $40$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 6 years older than his wife. One fact about him 8 years ago and one fact about her 14 years from now both describe how long they have been married. From those, find his age when they had been married 40 years.

Givens

The man is $6$ years older than his wife.
$8$ years ago, his years of marriage were half of his age then.
$14$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $40$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 40-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 6

.

\text{man} = x,\quad \text{wife} = x - 6

One unknown is enough, because the two ages differ by a fixed

6

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

8

years ago he was

x - 8

and had been married half of that. 8 more years have passed since, so today the marriage is

8

years longer.

\text{years married now} = \frac{1}{2}(x - 8) + 8

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

14

years she will be

(x - 6) + 14 = x + 8

and married for

\frac{2}{3}

of that. Today is

14

years earlier, so subtract

14

.

\text{years married now} = \frac{2}{3}(x + 8) - 14

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{2}(x-8)+8=\frac{2}{3}(x+8)-14 \;\Rightarrow\; 3x-24+48=4x+32-84 \;\Rightarrow\; 3x+24=4x-52 \;\Rightarrow\; x=76

Clearing denominators turns it into a one-step linear equation.

5Count back to the 40th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{2}(76-8)+8 = 34+8 = 42

years. The 40th anniversary was

42 - 40 = 2

years ago.

76 - 2 = 74

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 74 years old

4 · Reviewdoes it hold up?

$8$ years ago he was $68$ and married $34$ years, exactly half. In $14$ years she will be $70 + 14 = 84$ and married $42 + 14 = 56$ years, and $\frac{2}{3} \times 84 = 56$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 8$ must be even and $x + 8$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 76$ makes them agree (both give $42$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $40$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 7 medium

A man is now $5$ years older than his wife. $9$ years ago, he had been married for exactly $\frac{1}{3}$ of the life he had lived so far. $3$ years from now, his wife will have been married for $\frac{1}{2}$ of her life. How old was the man on the couple's $20$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 5 years older than his wife. One fact about him 9 years ago and one fact about her 3 years from now both describe how long they have been married. From those, find his age when they had been married 20 years.

Givens

The man is $5$ years older than his wife.
$9$ years ago, his years of marriage were $\frac{1}{3}$ of his age then.
$3$ years from now, her years of marriage will be $\frac{1}{2}$ of her age then.

Unknowns

The man's age at the $20$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 20-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 5

.

\text{man} = x,\quad \text{wife} = x - 5

One unknown is enough, because the two ages differ by a fixed

5

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

9

years ago he was

x - 9

and had been married

\frac{1}{3}

of that. 9 more years have passed since, so today the marriage is

9

years longer.

\text{years married now} = \frac{1}{3}(x - 9) + 9

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

3

years she will be

(x - 5) + 3 = x - 2

and married for

\frac{1}{2}

of that. Today is

3

years earlier, so subtract

3

.

\text{years married now} = \frac{1}{2}(x - 2) - 3

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

6

to clear the fractions.

\frac{1}{3}(x-9)+9=\frac{1}{2}(x-2)-3 \;\Rightarrow\; 2x-18+54=3x-6-18 \;\Rightarrow\; 2x+36=3x-24 \;\Rightarrow\; x=60

Clearing denominators turns it into a one-step linear equation.

5Count back to the 20th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{1}{3}(60-9)+9 = 17+9 = 26

years. The 20th anniversary was

26 - 20 = 6

years ago.

60 - 6 = 54

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 54 years old

4 · Reviewdoes it hold up?

$9$ years ago he was $51$ and married $17$ years, exactly $\frac{1}{3}$ . In $3$ years she will be $55 + 3 = 58$ and married $26 + 3 = 29$ years, and $\frac{1}{2} \times 58 = 29$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 9$ must be a multiple of $3$ and $x - 2$ even. Testing such ages in both marriage-length expressions, only $x = 60$ makes them agree (both give $26$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $20$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 8 hard

A man is now $4$ years older than his wife. $6$ years ago, he had been married for exactly $\frac{2}{5}$ of the life he had lived so far. $2$ years from now, his wife will have been married for $\frac{1}{2}$ of her life. How old was the man on the couple's $25$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 4 years older than his wife. One fact about him 6 years ago and one fact about her 2 years from now both describe how long they have been married. From those, find his age when they had been married 25 years.

Givens

The man is $4$ years older than his wife.
$6$ years ago, his years of marriage were $\frac{2}{5}$ of his age then.
$2$ years from now, her years of marriage will be $\frac{1}{2}$ of her age then.

Unknowns

The man's age at the $25$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 25-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 4

.

\text{man} = x,\quad \text{wife} = x - 4

One unknown is enough, because the two ages differ by a fixed

4

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

6

years ago he was

x - 6

and had been married

\frac{2}{5}

of that. 6 more years have passed since, so today the marriage is

6

years longer.

\text{years married now} = \frac{2}{5}(x - 6) + 6

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

2

years she will be

(x - 4) + 2 = x - 2

and married for

\frac{1}{2}

of that. Today is

2

years earlier, so subtract

2

.

\text{years married now} = \frac{1}{2}(x - 2) - 2

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

10

to clear the fractions.

\frac{2}{5}(x-6)+6=\frac{1}{2}(x-2)-2 \;\Rightarrow\; 4x-24+60=5x-10-20 \;\Rightarrow\; 4x+36=5x-30 \;\Rightarrow\; x=66

Clearing denominators turns it into a one-step linear equation.

5Count back to the 25th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{2}{5}(66-6)+6 = 24+6 = 30

years. The 25th anniversary was

30 - 25 = 5

years ago.

66 - 5 = 61

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 61 years old

4 · Reviewdoes it hold up?

$6$ years ago he was $60$ and married $24$ years, exactly $\frac{2}{5}$ . In $2$ years she will be $62 + 2 = 64$ and married $30 + 2 = 32$ years, and $\frac{1}{2} \times 64 = 32$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 6$ must be a multiple of $5$ and $x - 2$ even. Testing such ages in both marriage-length expressions, only $x = 66$ makes them agree (both give $30$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $25$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 9 medium

A man is now $6$ years older than his wife. $8$ years ago, he had been married for exactly $\frac{2}{5}$ of the life he had lived so far. $12$ years from now, his wife will have been married for $\frac{5}{8}$ of her life. How old was the man on the couple's $25$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 6 years older than his wife. One fact about him 8 years ago and one fact about her 12 years from now both describe how long they have been married. From those, find his age when they had been married 25 years.

Givens

The man is $6$ years older than his wife.
$8$ years ago, his years of marriage were $\frac{2}{5}$ of his age then.
$12$ years from now, her years of marriage will be $\frac{5}{8}$ of her age then.

Unknowns

The man's age at the $25$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 25-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 6

.

\text{man} = x,\quad \text{wife} = x - 6

One unknown is enough, because the two ages differ by a fixed

6

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

8

years ago he was

x - 8

and had been married

\frac{2}{5}

of that. 8 more years have passed since, so today the marriage is

8

years longer.

\text{years married now} = \frac{2}{5}(x - 8) + 8

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

12

years she will be

(x - 6) + 12 = x + 6

and married for

\frac{5}{8}

of that. Today is

12

years earlier, so subtract

12

.

\text{years married now} = \frac{5}{8}(x + 6) - 12

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

40

to clear the fractions.

\frac{2}{5}(x-8)+8=\frac{5}{8}(x+6)-12 \;\Rightarrow\; 16x-128+320=25x+150-480 \;\Rightarrow\; 16x+192=25x-330 \;\Rightarrow\; x=58

Clearing denominators turns it into a one-step linear equation.

5Count back to the 25th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{2}{5}(58-8)+8 = 20+8 = 28

years. The 25th anniversary was

28 - 25 = 3

years ago.

58 - 3 = 55

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 55 years old

4 · Reviewdoes it hold up?

$8$ years ago he was $50$ and married $20$ years, exactly $\frac{2}{5}$ . In $12$ years she will be $52 + 12 = 64$ and married $28 + 12 = 40$ years, and $\frac{5}{8} \times 64 = 40$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 8$ must be a multiple of $5$ and $x + 6$ a multiple of $8$ . Testing such ages in both marriage-length expressions, only $x = 58$ makes them agree (both give $28$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $25$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 10 hard

A man is now $4$ years older than his wife. $9$ years ago, he had been married for exactly $\frac{2}{5}$ of the life he had lived so far. $15$ years from now, his wife will have been married for $\frac{3}{5}$ of her life. How old was the man on the couple's $30$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 4 years older than his wife. One fact about him 9 years ago and one fact about her 15 years from now both describe how long they have been married. From those, find his age when they had been married 30 years.

Givens

The man is $4$ years older than his wife.
$9$ years ago, his years of marriage were $\frac{2}{5}$ of his age then.
$15$ years from now, her years of marriage will be $\frac{3}{5}$ of her age then.

Unknowns

The man's age at the $30$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 30-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 4

.

\text{man} = x,\quad \text{wife} = x - 4

One unknown is enough, because the two ages differ by a fixed

4

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

9

years ago he was

x - 9

and had been married

\frac{2}{5}

of that. 9 more years have passed since, so today the marriage is

9

years longer.

\text{years married now} = \frac{2}{5}(x - 9) + 9

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

15

years she will be

(x - 4) + 15 = x + 11

and married for

\frac{3}{5}

of that. Today is

15

years earlier, so subtract

15

.

\text{years married now} = \frac{3}{5}(x + 11) - 15

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

5

to clear the fractions.

\frac{2}{5}(x-9)+9=\frac{3}{5}(x+11)-15 \;\Rightarrow\; 2x-18+45=3x+33-75 \;\Rightarrow\; 2x+27=3x-42 \;\Rightarrow\; x=69

Clearing denominators turns it into a one-step linear equation.

5Count back to the 30th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{2}{5}(69-9)+9 = 24+9 = 33

years. The 30th anniversary was

33 - 30 = 3

years ago.

69 - 3 = 66

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 66 years old

4 · Reviewdoes it hold up?

$9$ years ago he was $60$ and married $24$ years, exactly $\frac{2}{5}$ . In $15$ years she will be $65 + 15 = 80$ and married $33 + 15 = 48$ years, and $\frac{3}{5} \times 80 = 48$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 9$ must be a multiple of $5$ and $x + 11$ a multiple of $5$ . Testing such ages in both marriage-length expressions, only $x = 69$ makes them agree (both give $33$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $30$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 11 hard

A man is now $4$ years older than his wife. $2$ years ago, he had been married for exactly $\frac{3}{5}$ of the life he had lived so far. $3$ years from now, his wife will have been married for $\frac{2}{3}$ of her life. How old was the man on the couple's $40$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 4 years older than his wife. One fact about him 2 years ago and one fact about her 3 years from now both describe how long they have been married. From those, find his age when they had been married 40 years.

Givens

The man is $4$ years older than his wife.
$2$ years ago, his years of marriage were $\frac{3}{5}$ of his age then.
$3$ years from now, her years of marriage will be $\frac{2}{3}$ of her age then.

Unknowns

The man's age at the $40$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 40-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 4

.

\text{man} = x,\quad \text{wife} = x - 4

One unknown is enough, because the two ages differ by a fixed

4

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

2

years ago he was

x - 2

and had been married

\frac{3}{5}

of that. 2 more years have passed since, so today the marriage is

2

years longer.

\text{years married now} = \frac{3}{5}(x - 2) + 2

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

3

years she will be

(x - 4) + 3 = x - 1

and married for

\frac{2}{3}

of that. Today is

3

years earlier, so subtract

3

.

\text{years married now} = \frac{2}{3}(x - 1) - 3

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

15

to clear the fractions.

\frac{3}{5}(x-2)+2=\frac{2}{3}(x-1)-3 \;\Rightarrow\; 9x-18+30=10x-10-45 \;\Rightarrow\; 9x+12=10x-55 \;\Rightarrow\; x=67

Clearing denominators turns it into a one-step linear equation.

5Count back to the 40th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{3}{5}(67-2)+2 = 39+2 = 41

years. The 40th anniversary was

41 - 40 = 1

year ago.

67 - 1 = 66

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 66 years old

4 · Reviewdoes it hold up?

$2$ years ago he was $65$ and married $39$ years, exactly $\frac{3}{5}$ . In $3$ years she will be $63 + 3 = 66$ and married $41 + 3 = 44$ years, and $\frac{2}{3} \times 66 = 44$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 2$ must be a multiple of $5$ and $x - 1$ a multiple of $3$ . Testing such ages in both marriage-length expressions, only $x = 67$ makes them agree (both give $41$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $40$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Variant 12 hard

A man is now $3$ years older than his wife. $11$ years ago, he had been married for exactly $\frac{3}{5}$ of the life he had lived so far. $16$ years from now, his wife will have been married for $\frac{3}{4}$ of her life. How old was the man on the couple's $45$ th wedding anniversary?

Your answer

Show solution

1 · Understandwhat's really being asked

A husband is 3 years older than his wife. One fact about him 11 years ago and one fact about her 16 years from now both describe how long they have been married. From those, find his age when they had been married 45 years.

Givens

The man is $3$ years older than his wife.
$11$ years ago, his years of marriage were $\frac{3}{5}$ of his age then.
$16$ years from now, her years of marriage will be $\frac{3}{4}$ of her age then.

Unknowns

The man's age at the $45$ th wedding anniversary.

Constraints

They married on the same day, so at any moment both have been married the same number of years.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

Call the man's present age $x$ . Each clue, read on a timeline, gives the number of years they have been married up to today. Both expressions count the same marriage, so set them equal and solve. Then walk back along the timeline to the 45-year mark.

3 · Execute5 carry out the plan

1Name the ages

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

Let the man be

x

years old now; then his wife is

x - 3

.

\text{man} = x,\quad \text{wife} = x - 3

One unknown is enough, because the two ages differ by a fixed

3

.

2Read the husband's clue

#1 Draw a Diagram 7.EE.B.4

11

years ago he was

x - 11

and had been married

\frac{3}{5}

of that. 11 more years have passed since, so today the marriage is

11

years longer.

\text{years married now} = \frac{3}{5}(x - 11) + 11

On a timeline, the marriage length grows by exactly the years that pass.

3Read the wife's clue

#1 Draw a Diagram 7.EE.B.4

In

16

years she will be

(x - 3) + 16 = x + 13

and married for

\frac{3}{4}

of that. Today is

16

years earlier, so subtract

16

.

\text{years married now} = \frac{3}{4}(x + 13) - 16

The same timeline, read from the other end.

4Equate and solve

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

Both expressions count the same marriage. Multiply both sides by

20

to clear the fractions.

\frac{3}{5}(x-11)+11=\frac{3}{4}(x+13)-16 \;\Rightarrow\; 12x-132+220=15x+195-320 \;\Rightarrow\; 12x+88=15x-125 \;\Rightarrow\; x=71

Clearing denominators turns it into a one-step linear equation.

5Count back to the 45th anniversary

#7 Identify Subproblems 7.EE.B.3

Today they have been married

\frac{3}{5}(71-11)+11 = 36+11 = 47

years. The 45th anniversary was

47 - 45 = 2

years ago.

71 - 2 = 69

The question asks for a different moment than the one we solved for, so one more subproblem is needed.

Answer: 69 years old

4 · Reviewdoes it hold up?

$11$ years ago he was $60$ and married $36$ years, exactly $\frac{3}{5}$ . In $16$ years she will be $68 + 16 = 84$ and married $47 + 16 = 63$ years, and $\frac{3}{4} \times 84 = 63$ . Both clues hold.

Another way: Guess and check with the divisibility the clues force: $x - 11$ must be a multiple of $5$ and $x + 13$ a multiple of $4$ . Testing such ages in both marriage-length expressions, only $x = 71$ makes them agree (both give $47$ years).

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the marriage length two ways from the two clues.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Clearing the fractions to simplify the equation.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Turning today's marriage length into the age at the $45$th anniversary.

💡Takeaway. When two clues describe the same quantity, write it both ways and set them equal.

Express years married two ways, then equate · 12 practice problems

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