Practice · Fix one unknown and count the rest · Sensim Math

Variant 1 easy

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=5$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $5$ . Order matters, so $(1, 2, 2)$ and $(2, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=5$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

5-2=3

.

x \in \{1, 2, 3\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=5-x

that is

4-x

pairs.

y+z=4:\ (1,3),(2,2),(3,1) \;\to\; 3 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

3

: they are

3, 2, 1

.

3 + 2 + 1 = 6

Each case is separate, so the counts simply add.

Answer:

6

4 · Reviewdoes it hold up?

The largest any one unknown can be is $3$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $3, 2, 1$ shows. A total of $6$ fits that list.

Another way: Picture $5$ stars in a row with $4$ gaps between them. Choosing $2$ of the $4$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{4 \times 3}{2} = 6$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 4-x$ to count the pairs $(y, z)$ with $y+z=5-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 2 easy

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=6$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $6$ . Order matters, so $(1, 2, 3)$ and $(3, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=6$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

6-2=4

.

x \in \{1, 2, 3, 4\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=6-x

that is

5-x

pairs.

y+z=5:\ (1,4),(2,3),(3,2),(4,1) \;\to\; 4 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

4

: they are

4, 3, 2, 1

.

4 + 3 + 2 + 1 = 10

Each case is separate, so the counts simply add.

Answer:

10

4 · Reviewdoes it hold up?

The largest any one unknown can be is $4$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $4, 3, 2, 1$ shows. A total of $10$ fits that list.

Another way: Picture $6$ stars in a row with $5$ gaps between them. Choosing $2$ of the $5$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{5 \times 4}{2} = 10$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 5-x$ to count the pairs $(y, z)$ with $y+z=6-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 3 easy

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=7$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $7$ . Order matters, so $(1, 2, 4)$ and $(4, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=7$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

7-2=5

.

x \in \{1, 2, 3, 4, 5\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=7-x

that is

6-x

pairs.

y+z=6:\ (1,5),(2,4),(3,3),(4,2),(5,1) \;\to\; 5 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

5

: they are

5, 4, 3, 2, 1

.

5 + 4 + 3 + 2 + 1 = 15

Each case is separate, so the counts simply add.

Answer:

15

4 · Reviewdoes it hold up?

The largest any one unknown can be is $5$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $5, 4, 3, 2, 1$ shows. A total of $15$ fits that list.

Another way: Picture $7$ stars in a row with $6$ gaps between them. Choosing $2$ of the $6$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{6 \times 5}{2} = 15$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 6-x$ to count the pairs $(y, z)$ with $y+z=7-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 4 easy

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=8$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $8$ . Order matters, so $(1, 2, 5)$ and $(5, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=8$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

8-2=6

.

x \in \{1, 2, 3, 4, 5, 6\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=8-x

that is

7-x

pairs.

y+z=7:\ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \;\to\; 6 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

6

: they are

6, 5, 4, 3, 2, 1

.

6 + 5 + 4 + 3 + 2 + 1 = 21

Each case is separate, so the counts simply add.

Answer:

21

4 · Reviewdoes it hold up?

The largest any one unknown can be is $6$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $6, 5, 4, 3, 2, 1$ shows. A total of $21$ fits that list.

Another way: Picture $8$ stars in a row with $7$ gaps between them. Choosing $2$ of the $7$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{7 \times 6}{2} = 21$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 7-x$ to count the pairs $(y, z)$ with $y+z=8-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 5 medium

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=9$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $9$ . Order matters, so $(1, 2, 6)$ and $(6, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=9$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

9-2=7

.

x \in \{1, 2, \ldots, 7\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=9-x

that is

8-x

pairs.

y+z=8:\ (1,7),(2,6),\ldots,(7,1) \;\to\; 7 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

7

: they are

7, 6, \ldots, 1

.

7 + 6 + 5 + \cdots + 1 = 28

Each case is separate, so the counts simply add.

Answer:

28

4 · Reviewdoes it hold up?

The largest any one unknown can be is $7$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $7, 6, \ldots, 1$ shows. A total of $28$ fits that list.

Another way: Picture $9$ stars in a row with $8$ gaps between them. Choosing $2$ of the $8$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{8 \times 7}{2} = 28$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 8-x$ to count the pairs $(y, z)$ with $y+z=9-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 6 medium

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=10$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $10$ . Order matters, so $(1, 2, 7)$ and $(7, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=10$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

10-2=8

.

x \in \{1, 2, \ldots, 8\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=10-x

that is

9-x

pairs.

y+z=9:\ (1,8),(2,7),\ldots,(8,1) \;\to\; 8 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

8

: they are

8, 7, \ldots, 1

.

8 + 7 + 6 + \cdots + 1 = 36

Each case is separate, so the counts simply add.

Answer:

36

4 · Reviewdoes it hold up?

The largest any one unknown can be is $8$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $8, 7, \ldots, 1$ shows. A total of $36$ fits that list.

Another way: Picture $10$ stars in a row with $9$ gaps between them. Choosing $2$ of the $9$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{9 \times 8}{2} = 36$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 9-x$ to count the pairs $(y, z)$ with $y+z=10-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 7 medium

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=11$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $11$ . Order matters, so $(1, 2, 8)$ and $(8, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=11$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

11-2=9

.

x \in \{1, 2, \ldots, 9\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=11-x

that is

10-x

pairs.

y+z=10:\ (1,9),(2,8),\ldots,(9,1) \;\to\; 9 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

9

: they are

9, 8, \ldots, 1

.

9 + 8 + 7 + \cdots + 1 = 45

Each case is separate, so the counts simply add.

Answer:

45

4 · Reviewdoes it hold up?

The largest any one unknown can be is $9$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $9, 8, \ldots, 1$ shows. A total of $45$ fits that list.

Another way: Picture $11$ stars in a row with $10$ gaps between them. Choosing $2$ of the $10$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{10 \times 9}{2} = 45$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 10-x$ to count the pairs $(y, z)$ with $y+z=11-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 8 medium

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=12$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $12$ . Order matters, so $(1, 2, 9)$ and $(9, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=12$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

12-2=10

.

x \in \{1, 2, \ldots, 10\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=12-x

that is

11-x

pairs.

y+z=11:\ (1,10),(2,9),\ldots,(10,1) \;\to\; 10 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

10

: they are

10, 9, \ldots, 1

.

10 + 9 + 8 + \cdots + 1 = 55

Each case is separate, so the counts simply add.

Answer:

55

4 · Reviewdoes it hold up?

The largest any one unknown can be is $10$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $10, 9, \ldots, 1$ shows. A total of $55$ fits that list.

Another way: Picture $12$ stars in a row with $11$ gaps between them. Choosing $2$ of the $11$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{11 \times 10}{2} = 55$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 11-x$ to count the pairs $(y, z)$ with $y+z=12-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 9 hard

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=13$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $13$ . Order matters, so $(1, 2, 10)$ and $(10, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=13$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

13-2=11

.

x \in \{1, 2, \ldots, 11\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=13-x

that is

12-x

pairs.

y+z=12:\ (1,11),(2,10),\ldots,(11,1) \;\to\; 11 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

11

: they are

11, 10, \ldots, 1

.

11 + 10 + 9 + \cdots + 1 = 66

Each case is separate, so the counts simply add.

Answer:

66

4 · Reviewdoes it hold up?

The largest any one unknown can be is $11$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $11, 10, \ldots, 1$ shows. A total of $66$ fits that list.

Another way: Picture $13$ stars in a row with $12$ gaps between them. Choosing $2$ of the $12$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{12 \times 11}{2} = 66$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 12-x$ to count the pairs $(y, z)$ with $y+z=13-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 10 hard

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=14$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $14$ . Order matters, so $(1, 2, 11)$ and $(11, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=14$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

14-2=12

.

x \in \{1, 2, \ldots, 12\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=14-x

that is

13-x

pairs.

y+z=13:\ (1,12),(2,11),\ldots,(12,1) \;\to\; 12 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

12

: they are

12, 11, \ldots, 1

.

12 + 11 + 10 + \cdots + 1 = 78

Each case is separate, so the counts simply add.

Answer:

78

4 · Reviewdoes it hold up?

The largest any one unknown can be is $12$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $12, 11, \ldots, 1$ shows. A total of $78$ fits that list.

Another way: Picture $14$ stars in a row with $13$ gaps between them. Choosing $2$ of the $13$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{13 \times 12}{2} = 78$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 13-x$ to count the pairs $(y, z)$ with $y+z=14-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 11 hard

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=15$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $15$ . Order matters, so $(1, 2, 12)$ and $(12, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=15$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

15-2=13

.

x \in \{1, 2, \ldots, 13\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=15-x

that is

14-x

pairs.

y+z=14:\ (1,13),(2,12),\ldots,(13,1) \;\to\; 13 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

13

: they are

13, 12, \ldots, 1

.

13 + 12 + 11 + \cdots + 1 = 91

Each case is separate, so the counts simply add.

Answer:

91

4 · Reviewdoes it hold up?

The largest any one unknown can be is $13$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $13, 12, \ldots, 1$ shows. A total of $91$ fits that list.

Another way: Picture $15$ stars in a row with $14$ gaps between them. Choosing $2$ of the $14$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{14 \times 13}{2} = 91$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 14-x$ to count the pairs $(y, z)$ with $y+z=15-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Variant 12 hard

If $x$ , $y$ , and $z$ are natural numbers, how many ordered triples $(x, y, z)$ satisfy the linear equation $x+y+z=16$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count every ordered triple of natural numbers $(x, y, z)$ whose sum is $16$ . Order matters, so $(1, 2, 13)$ and $(13, 2, 1)$ are different.

Givens

$x$ , $y$ , $z$ are natural numbers.
$x+y+z=16$ .

Unknowns

The number of ordered triples $(x, y, z)$ .

Constraints

Each unknown is at least $1$ .
Ordered triples: different orders count separately.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems#5 Look for a Pattern

Three unknowns at once is hard to list, so split the problem by the value of $x$ . Each case leaves a two-unknown equation $y+z=n$ , whose count follows a simple pattern. A systematic list of the cases then gives the total.

3 · Execute3 carry out the plan

1Find the possible values of $x$

#7 Identify Subproblems 6.EE.B.5

Since

y

and

z

are each at least

1

,

y+z

is at least

2

, so

x

is at most

16-2=14

.

x \in \{1, 2, \ldots, 14\}

Fixing one unknown turns one hard count into several easy ones.

2Count the solutions of $y+z=n$

#5 Look for a Pattern 7.EE.B.4

For

y+z=n

with natural numbers,

y

can be

1, 2, \ldots, n-1

and then

z

is decided, so there are

n-1

pairs. With

n=16-x

that is

15-x

pairs.

y+z=15:\ (1,14),(2,13),\ldots,(14,1) \;\to\; 14 \text{ pairs}

The pattern

n-1

is visible from one small case.

3Add the cases

#2 Make a Systematic List 6.EE.B.5

List the counts for

x=1

to

14

: they are

14, 13, \ldots, 1

.

14 + 13 + 12 + \cdots + 1 = 105

Each case is separate, so the counts simply add.

Answer:

105

4 · Reviewdoes it hold up?

The largest any one unknown can be is $14$ (the other two at least $1$ each), and the counts shrink by one as $x$ grows, which is what the list $14, 13, \ldots, 1$ shows. A total of $105$ fits that list.

Another way: Picture $16$ stars in a row with $15$ gaps between them. Choosing $2$ of the $15$ gaps as dividers splits the stars into three non-empty groups $x$ , $y$ , $z$ , and there are $\frac{15 \times 14}{2} = 105$ ways to choose the two gaps.

Standardsmin grade 7

6.EE.B.5 Solve equations/inequalities by finding values that make them true — Testing which natural-number values make the equation true.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Fixing $x$ and bounding $y$ by $1\le y\le 15-x$ to count the pairs $(y, z)$ with $y+z=16-x$.

💡Takeaway. Hold one unknown still, count what is left, and add up the cases.

Fix one unknown and count the rest · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?