Practice · Write the solution in a, then filter by wholeness · Sensim Math

Variant 1 easy

In the equation $x-3=\frac{1}{4}(x-a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-3=\frac{1}{4}(x-a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $3$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

4

.

4x-12=x-a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

3

.

3x=12-a \;\Rightarrow\; x=4-\frac{a}{3}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=4-\frac{a}{3}

is a whole number only when

\frac{a}{3}

is whole, so

a

must be a multiple of

3

.

A number over

3

is whole only when the top is a multiple of

3

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

4-\frac{a}{3}>0

, so

\frac{a}{3}<4

, that is,

a<12

.

\frac{a}{3}<4 \;\Rightarrow\; a<12

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

3

less than

12

, with the matching

x

for each.

a=3,6,9 \;\to\; x=3,2,1

Each pair checks out as natural numbers.

Answer:

3

,

6

,

9

4 · Reviewdoes it hold up?

Take $a=9$ , $x=1$ : left side $1-3=-2$ ; right side $\frac{1}{4}(1-9)=-2$ . Equal. And $a=12$ would give $x=0$ , which is not natural, so the list stops at $9$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=4-\frac{a}{3}$ : $3\frac{2}{3}, 3\frac{1}{3}, 3, 2\frac{2}{3}, 2\frac{1}{3}, 2, 1\frac{2}{3}, 1\frac{1}{3}, 1, \frac{2}{3}, \frac{1}{3}, 0$ . The whole positive entries sit at $a=3,6,9$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $4-\frac{a}{3}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 2 easy

In the equation $x-4=\frac{1}{3}(x-a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-4=\frac{1}{3}(x-a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be even) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3

.

3x-12=x-a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

2

.

2x=12-a \;\Rightarrow\; x=6-\frac{a}{2}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=6-\frac{a}{2}

is a whole number only when

\frac{a}{2}

is whole, so odd values of

a

are out:

a

must be even.

Half of an odd number is never whole.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

6-\frac{a}{2}>0

, so

\frac{a}{2}<6

, that is,

a<12

.

\frac{a}{2}<6 \;\Rightarrow\; a<12

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Even natural numbers less than

12

, with the matching

x

for each.

a=2,4,6,8,10 \;\to\; x=5,4,3,2,1

Each pair checks out as natural numbers.

Answer:

2

,

4

,

6

,

8

,

10

4 · Reviewdoes it hold up?

Take $a=10$ , $x=1$ : left side $1-4=-3$ ; right side $\frac{1}{3}(1-10)=-3$ . Equal. And $a=12$ would give $x=0$ , which is not natural, so the list stops at $10$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=6-\frac{a}{2}$ : $5.5, 5, 4.5, 4, 3.5, 3, 2.5, 2, 1.5, 1, 0.5, 0$ . The whole positive entries sit at $a=2,4,6,8,10$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $6-\frac{a}{2}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 3 easy

In the equation $x-4=\frac{1}{5}(x-a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-4=\frac{1}{5}(x-a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $4$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5

.

5x-20=x-a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

4

.

4x=20-a \;\Rightarrow\; x=5-\frac{a}{4}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=5-\frac{a}{4}

is a whole number only when

\frac{a}{4}

is whole, so

a

must be a multiple of

4

.

A number over

4

is whole only when the top is a multiple of

4

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

5-\frac{a}{4}>0

, so

\frac{a}{4}<5

, that is,

a<20

.

\frac{a}{4}<5 \;\Rightarrow\; a<20

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

4

less than

20

, with the matching

x

for each.

a=4,8,12,16 \;\to\; x=4,3,2,1

Each pair checks out as natural numbers.

Answer:

4

,

8

,

12

,

16

4 · Reviewdoes it hold up?

Take $a=16$ , $x=1$ : left side $1-4=-3$ ; right side $\frac{1}{5}(1-16)=-3$ . Equal. And $a=20$ would give $x=0$ , which is not natural, so the list stops at $16$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=5-\frac{a}{4}$ : $4.75, 4.5, 4.25, 4, 3.75, 3.5, 3.25, 3, 2.75, 2.5, 2.25, 2, 1.75, 1.5, 1.25, 1, 0.75, 0.5, 0.25, 0$ . The whole positive entries sit at $a=4,8,12,16$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $5-\frac{a}{4}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 4 easy

In the equation $x-4=\frac{1}{5}(x-2a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-4=\frac{1}{5}(x-2a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be even) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5

.

5x-20=x-2a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

4

.

4x=20-2a \;\Rightarrow\; x=5-\frac{a}{2}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=5-\frac{a}{2}

is a whole number only when

\frac{a}{2}

is whole, so odd values of

a

are out:

a

must be even.

Half of an odd number is never whole.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

5-\frac{a}{2}>0

, so

\frac{a}{2}<5

, that is,

a<10

.

\frac{a}{2}<5 \;\Rightarrow\; a<10

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Even natural numbers less than

10

, with the matching

x

for each.

a=2,4,6,8 \;\to\; x=4,3,2,1

Each pair checks out as natural numbers.

Answer:

2

,

4

,

6

,

8

4 · Reviewdoes it hold up?

Take $a=8$ , $x=1$ : left side $1-4=-3$ ; right side $\frac{1}{5}(1-16)=-3$ . Equal. And $a=10$ would give $x=0$ , which is not natural, so the list stops at $8$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=5-\frac{a}{2}$ : $4.5, 4, 3.5, 3, 2.5, 2, 1.5, 1, 0.5, 0$ . The whole positive entries sit at $a=2,4,6,8$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $5-\frac{a}{2}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 5 medium

In the equation $x-6=\frac{1}{4}(x-2a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-6=\frac{1}{4}(x-2a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $3$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

4

.

4x-24=x-2a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

3

.

3x=24-2a \;\Rightarrow\; x=8-\frac{2a}{3}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=8-\frac{2a}{3}

is a whole number only when

\frac{2a}{3}

is whole, so

a

must be a multiple of

3

.

A number over

3

is whole only when the top is a multiple of

3

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

8-\frac{2a}{3}>0

, so

\frac{2a}{3}<8

, that is,

a<12

.

\frac{2a}{3}<8 \;\Rightarrow\; a<12

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

3

less than

12

, with the matching

x

for each.

a=3,6,9 \;\to\; x=6,4,2

Each pair checks out as natural numbers.

Answer:

3

,

6

,

9

4 · Reviewdoes it hold up?

Take $a=9$ , $x=2$ : left side $2-6=-4$ ; right side $\frac{1}{4}(2-18)=-4$ . Equal. And $a=12$ would give $x=0$ , which is not natural, so the list stops at $9$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=8-\frac{2a}{3}$ : $7\frac{1}{3}, 6\frac{2}{3}, 6, 5\frac{1}{3}, 4\frac{2}{3}, 4, 3\frac{1}{3}, 2\frac{2}{3}, 2, 1\frac{1}{3}, \frac{2}{3}, 0$ . The whole positive entries sit at $a=3,6,9$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $8-\frac{2a}{3}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 6 medium

In the equation $x-6=\frac{1}{7}(x-4a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-6=\frac{1}{7}(x-4a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $3$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

7

.

7x-42=x-4a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

6

.

6x=42-4a \;\Rightarrow\; x=7-\frac{2a}{3}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=7-\frac{2a}{3}

is a whole number only when

\frac{2a}{3}

is whole, so

a

must be a multiple of

3

.

A number over

3

is whole only when the top is a multiple of

3

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

7-\frac{2a}{3}>0

, so

\frac{2a}{3}<7

, that is,

a<10\frac{1}{2}

.

\frac{2a}{3}<7 \;\Rightarrow\; a<10\frac{1}{2}

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

3

less than

10\frac{1}{2}

, with the matching

x

for each.

a=3,6,9 \;\to\; x=5,3,1

Each pair checks out as natural numbers.

Answer:

3

,

6

,

9

4 · Reviewdoes it hold up?

Take $a=9$ , $x=1$ : left side $1-6=-5$ ; right side $\frac{1}{7}(1-36)=-5$ . Equal. And $a=12$ would give $x=-1$ , which is not natural, so the list stops at $9$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=7-\frac{2a}{3}$ : $6\frac{1}{3}, 5\frac{2}{3}, 5, 4\frac{1}{3}, 3\frac{2}{3}, 3, 2\frac{1}{3}, 1\frac{2}{3}, 1, \frac{1}{3}, -\frac{1}{3}, -1$ . The whole positive entries sit at $a=3,6,9$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $7-\frac{2a}{3}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 7 medium

In the equation $x-8=\frac{1}{5}(x-3a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-8=\frac{1}{5}(x-3a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $4$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5

.

5x-40=x-3a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

4

.

4x=40-3a \;\Rightarrow\; x=10-\frac{3a}{4}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=10-\frac{3a}{4}

is a whole number only when

\frac{3a}{4}

is whole, so

a

must be a multiple of

4

.

A number over

4

is whole only when the top is a multiple of

4

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

10-\frac{3a}{4}>0

, so

\frac{3a}{4}<10

, that is,

a<13\frac{1}{3}

.

\frac{3a}{4}<10 \;\Rightarrow\; a<13\frac{1}{3}

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

4

less than

13\frac{1}{3}

, with the matching

x

for each.

a=4,8,12 \;\to\; x=7,4,1

Each pair checks out as natural numbers.

Answer:

4

,

8

,

12

4 · Reviewdoes it hold up?

Take $a=12$ , $x=1$ : left side $1-8=-7$ ; right side $\frac{1}{5}(1-36)=-7$ . Equal. And $a=16$ would give $x=-2$ , which is not natural, so the list stops at $12$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=10-\frac{3a}{4}$ : $9.25, 8.5, 7.75, 7, 6.25, 5.5, 4.75, 4, 3.25, 2.5, 1.75, 1, 0.25, -0.5, -1.25, -2$ . The whole positive entries sit at $a=4,8,12$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $10-\frac{3a}{4}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 8 medium

In the equation $x-8=\frac{1}{3}(x-3a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-8=\frac{1}{3}(x-3a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be even) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3

.

3x-24=x-3a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

2

.

2x=24-3a \;\Rightarrow\; x=12-\frac{3a}{2}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=12-\frac{3a}{2}

is a whole number only when

\frac{3a}{2}

is whole, so odd values of

a

are out:

a

must be even.

Half of an odd number is never whole.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

12-\frac{3a}{2}>0

, so

\frac{3a}{2}<12

, that is,

a<8

.

\frac{3a}{2}<12 \;\Rightarrow\; a<8

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Even natural numbers less than

8

, with the matching

x

for each.

a=2,4,6 \;\to\; x=9,6,3

Each pair checks out as natural numbers.

Answer:

2

,

4

,

6

4 · Reviewdoes it hold up?

Take $a=6$ , $x=3$ : left side $3-8=-5$ ; right side $\frac{1}{3}(3-18)=-5$ . Equal. And $a=8$ would give $x=0$ , which is not natural, so the list stops at $6$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=12-\frac{3a}{2}$ : $10.5, 9, 7.5, 6, 4.5, 3, 1.5, 0$ . The whole positive entries sit at $a=2,4,6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $12-\frac{3a}{2}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 9 hard

In the equation $x-10=\frac{1}{6}(x-3a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-10=\frac{1}{6}(x-3a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $5$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

6

.

6x-60=x-3a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

5

.

5x=60-3a \;\Rightarrow\; x=12-\frac{3a}{5}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=12-\frac{3a}{5}

is a whole number only when

\frac{3a}{5}

is whole, so

a

must be a multiple of

5

.

A number over

5

is whole only when the top is a multiple of

5

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

12-\frac{3a}{5}>0

, so

\frac{3a}{5}<12

, that is,

a<20

.

\frac{3a}{5}<12 \;\Rightarrow\; a<20

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

5

less than

20

, with the matching

x

for each.

a=5,10,15 \;\to\; x=9,6,3

Each pair checks out as natural numbers.

Answer:

5

,

10

,

15

4 · Reviewdoes it hold up?

Take $a=15$ , $x=3$ : left side $3-10=-7$ ; right side $\frac{1}{6}(3-45)=-7$ . Equal. And $a=20$ would give $x=0$ , which is not natural, so the list stops at $15$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=12-\frac{3a}{5}$ : $11.4, 10.8, 10.2, 9.6, 9, 8.4, 7.8, 7.2, 6.6, 6, 5.4, 4.8, 4.2, 3.6, 3, 2.4, 1.8, 1.2, 0.6, 0$ . The whole positive entries sit at $a=5,10,15$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $12-\frac{3a}{5}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 10 hard

In the equation $x-10=\frac{1}{3}(x-3a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-10=\frac{1}{3}(x-3a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be even) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3

.

3x-30=x-3a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

2

.

2x=30-3a \;\Rightarrow\; x=15-\frac{3a}{2}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=15-\frac{3a}{2}

is a whole number only when

\frac{3a}{2}

is whole, so odd values of

a

are out:

a

must be even.

Half of an odd number is never whole.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

15-\frac{3a}{2}>0

, so

\frac{3a}{2}<15

, that is,

a<10

.

\frac{3a}{2}<15 \;\Rightarrow\; a<10

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Even natural numbers less than

10

, with the matching

x

for each.

a=2,4,6,8 \;\to\; x=12,9,6,3

Each pair checks out as natural numbers.

Answer:

2

,

4

,

6

,

8

4 · Reviewdoes it hold up?

Take $a=8$ , $x=3$ : left side $3-10=-7$ ; right side $\frac{1}{3}(3-24)=-7$ . Equal. And $a=10$ would give $x=0$ , which is not natural, so the list stops at $8$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=15-\frac{3a}{2}$ : $13.5, 12, 10.5, 9, 7.5, 6, 4.5, 3, 1.5, 0$ . The whole positive entries sit at $a=2,4,6,8$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $15-\frac{3a}{2}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 11 hard

In the equation $x-12=\frac{1}{5}(x-6a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-12=\frac{1}{5}(x-6a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be even) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5

.

5x-60=x-6a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

4

.

4x=60-6a \;\Rightarrow\; x=15-\frac{3a}{2}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=15-\frac{3a}{2}

is a whole number only when

\frac{3a}{2}

is whole, so odd values of

a

are out:

a

must be even.

Half of an odd number is never whole.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

15-\frac{3a}{2}>0

, so

\frac{3a}{2}<15

, that is,

a<10

.

\frac{3a}{2}<15 \;\Rightarrow\; a<10

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Even natural numbers less than

10

, with the matching

x

for each.

a=2,4,6,8 \;\to\; x=12,9,6,3

Each pair checks out as natural numbers.

Answer:

2

,

4

,

6

,

8

4 · Reviewdoes it hold up?

Take $a=8$ , $x=3$ : left side $3-12=-9$ ; right side $\frac{1}{5}(3-48)=-9$ . Equal. And $a=10$ would give $x=0$ , which is not natural, so the list stops at $8$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=15-\frac{3a}{2}$ : $13.5, 12, 10.5, 9, 7.5, 6, 4.5, 3, 1.5, 0$ . The whole positive entries sit at $a=2,4,6,8$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $15-\frac{3a}{2}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Variant 12 hard

In the equation $x-12=\frac{1}{4}(x-5a)$ in $x$ , both the solution $x$ and the constant $a$ are natural numbers (positive whole numbers). Find all possible values of $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Solve the equation for $x$ in terms of $a$ . Then find every natural number $a$ that also makes $x$ a natural number.

Givens

The equation is $x-12=\frac{1}{4}(x-5a)$ .
$a$ is a natural number.
The solution $x$ is a natural number.

Unknowns

Every possible value of $a$ .

Constraints

$x$ must be a whole number and must be positive.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

First get $x$ as a formula in $a$ . Two filters then cut the natural numbers down: $x$ must be whole (so $a$ must be a multiple of $3$ ) and $x$ must be positive (so $a$ must be small). A short list of the survivors is the answer.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

4

.

4x-48=x-5a

Multiplying both sides by the same number keeps the equation true.

2Solve for x in terms of a

#13 Convert to Algebra 7.EE.A.1

Collect the

x

terms on the left and the rest on the right, then divide by

3

.

3x=48-5a \;\Rightarrow\; x=16-\frac{5a}{3}

Now

x

is a formula in

a

.

3x must be whole

#3 Eliminate Possibilities 7.EE.B.4

x=16-\frac{5a}{3}

is a whole number only when

\frac{5a}{3}

is whole, so

a

must be a multiple of

3

.

A number over

3

is whole only when the top is a multiple of

3

.

4x must be positive

#3 Eliminate Possibilities 7.EE.B.4

x>0

means

16-\frac{5a}{3}>0

, so

\frac{5a}{3}<16

, that is,

a<9\frac{3}{5}

.

\frac{5a}{3}<16 \;\Rightarrow\; a<9\frac{3}{5}

Too large an

a

pushes

x

to zero or below.

5List the survivors

#2 Make a Systematic List 7.EE.B.4

Multiples of

3

less than

9\frac{3}{5}

, with the matching

x

for each.

a=3,6,9 \;\to\; x=11,6,1

Each pair checks out as natural numbers.

Answer:

3

,

6

,

9

4 · Reviewdoes it hold up?

Take $a=9$ , $x=1$ : left side $1-12=-11$ ; right side $\frac{1}{4}(1-45)=-11$ . Equal. And $a=12$ would give $x=-4$ , which is not natural, so the list stops at $9$ .

Another way: Make a table (Make a Systematic List) of $a=1,2,3,\dots$ with $x=16-\frac{5a}{3}$ : $14\frac{1}{3}, 12\frac{2}{3}, 11, 9\frac{1}{3}, 7\frac{2}{3}, 6, 4\frac{1}{3}, 2\frac{2}{3}, 1, -\frac{2}{3}, -2\frac{1}{3}, -4$ . The whole positive entries sit at $a=3,6,9$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the linear equation and reasoning about which values of $a$ give a valid solution.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Rearranging to write $x$ as $16-\frac{5a}{3}$.

💡Takeaway. Solve for

x

with the letter still in it, then let "whole" and "positive" do the sorting.

Write the solution in a, then filter by wholeness · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

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Constraints

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Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

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Givens

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Where next?