Practice · Peel a nested defined operation from the outside · Sensim Math

Variant 1 easy

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-2)\circledcirc 4\}\circledcirc 2=4$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $4$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-2)\circledcirc 4\}\circledcirc 2=4$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-2)\circledcirc 4

, so the equation reads

P\circledcirc 2=4

.

P\circledcirc 2=4

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

4

.

2-P=4 \;\Rightarrow\; P=-2

So the inside must equal

-2

.

3Write the inner layer

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

Use the definition with

a=x-2

and

b=4

.

(x-2)\circledcirc 4=(x-2)+4-4(x-2)=4-3(x-2)

Collecting the two

(x-2)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-2

and solve.

4-3(x-2)=-2 \;\Rightarrow\; 3(x-2)=6 \;\Rightarrow\; x-2=2 \;\Rightarrow\; x=4

A linear equation finishes it.

Answer:

4

4 · Reviewdoes it hold up?

At $x=4$ : $2\circledcirc 4=2+4-8=-2$ , then $(-2)\circledcirc 2=-2+2-(-4)=4$ . The result is $4$ , as required.

Another way: Expand everything in terms of $x$ : $(x-2)\circledcirc 4=-3x+10$ , and $(-3x+10)\circledcirc 2=-3x+12-2(-3x+10)=3x-8$ . Then $3x-8=4$ gives $x=4$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-4=(1-P)(1-2)$ , giving $P=-2$ , and the inner layer says $1-(-2)=(1-(x-2))(1-4)$ , giving $x-2=2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-2)+4-4(x-2)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 2 easy

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-3)\circledcirc 3\}\circledcirc 3=5$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $5$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-3)\circledcirc 3\}\circledcirc 3=5$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-3)\circledcirc 3

, so the equation reads

P\circledcirc 3=5

.

P\circledcirc 3=5

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 3=P+3-3P=3-2P

. Set it equal to

5

.

3-2P=5 \;\Rightarrow\; 2P=-2 \;\Rightarrow\; P=-1

So the inside must equal

-1

.

3Write the inner layer

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

Use the definition with

a=x-3

and

b=3

.

(x-3)\circledcirc 3=(x-3)+3-3(x-3)=3-2(x-3)

Collecting the two

(x-3)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-1

and solve.

3-2(x-3)=-1 \;\Rightarrow\; 2(x-3)=4 \;\Rightarrow\; x-3=2 \;\Rightarrow\; x=5

A linear equation finishes it.

Answer:

5

4 · Reviewdoes it hold up?

At $x=5$ : $2\circledcirc 3=2+3-6=-1$ , then $(-1)\circledcirc 3=-1+3-(-3)=5$ . The result is $5$ , as required.

Another way: Expand everything in terms of $x$ : $(x-3)\circledcirc 3=-2x+9$ , and $(-2x+9)\circledcirc 3=-2x+12-3(-2x+9)=4x-15$ . Then $4x-15=5$ gives $x=5$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-5=(1-P)(1-3)$ , giving $P=-1$ , and the inner layer says $1-(-1)=(1-(x-3))(1-3)$ , giving $x-3=2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-3)+3-3(x-3)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 3 easy

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-3)\circledcirc 4\}\circledcirc 2=-5$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $-5$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-3)\circledcirc 4\}\circledcirc 2=-5$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-3)\circledcirc 4

, so the equation reads

P\circledcirc 2=-5

.

P\circledcirc 2=-5

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

-5

.

2-P=-5 \;\Rightarrow\; P=7

So the inside must equal

7

.

3Write the inner layer

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

Use the definition with

a=x-3

and

b=4

.

(x-3)\circledcirc 4=(x-3)+4-4(x-3)=4-3(x-3)

Collecting the two

(x-3)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

7

and solve.

4-3(x-3)=7 \;\Rightarrow\; 3(x-3)=-3 \;\Rightarrow\; x-3=-1 \;\Rightarrow\; x=2

A linear equation finishes it.

Answer:

2

4 · Reviewdoes it hold up?

At $x=2$ : $(-1)\circledcirc 4=-1+4-(-4)=7$ , then $7\circledcirc 2=7+2-14=-5$ . The result is $-5$ , as required.

Another way: Expand everything in terms of $x$ : $(x-3)\circledcirc 4=-3x+13$ , and $(-3x+13)\circledcirc 2=-3x+15-2(-3x+13)=3x-11$ . Then $3x-11=-5$ gives $x=2$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-(-5)=(1-P)(1-2)$ , giving $P=7$ , and the inner layer says $1-7=(1-(x-3))(1-4)$ , giving $x-3=-1$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-3)+4-4(x-3)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 4 easy

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-5)\circledcirc 3\}\circledcirc 2=-5$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $-5$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-5)\circledcirc 3\}\circledcirc 2=-5$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-5)\circledcirc 3

, so the equation reads

P\circledcirc 2=-5

.

P\circledcirc 2=-5

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

-5

.

2-P=-5 \;\Rightarrow\; P=7

So the inside must equal

7

.

3Write the inner layer

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

Use the definition with

a=x-5

and

b=3

.

(x-5)\circledcirc 3=(x-5)+3-3(x-5)=3-2(x-5)

Collecting the two

(x-5)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

7

and solve.

3-2(x-5)=7 \;\Rightarrow\; 2(x-5)=-4 \;\Rightarrow\; x-5=-2 \;\Rightarrow\; x=3

A linear equation finishes it.

Answer:

3

4 · Reviewdoes it hold up?

At $x=3$ : $(-2)\circledcirc 3=-2+3-(-6)=7$ , then $7\circledcirc 2=7+2-14=-5$ . The result is $-5$ , as required.

Another way: Expand everything in terms of $x$ : $(x-5)\circledcirc 3=-2x+13$ , and $(-2x+13)\circledcirc 2=-2x+15-2(-2x+13)=2x-11$ . Then $2x-11=-5$ gives $x=3$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-(-5)=(1-P)(1-2)$ , giving $P=7$ , and the inner layer says $1-7=(1-(x-5))(1-3)$ , giving $x-5=-2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-5)+3-3(x-5)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 5 medium

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-5)\circledcirc 6\}\circledcirc 4=1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $1$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-5)\circledcirc 6\}\circledcirc 4=1$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-5)\circledcirc 6

, so the equation reads

P\circledcirc 4=1

.

P\circledcirc 4=1

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 4=P+4-4P=4-3P

. Set it equal to

1

.

4-3P=1 \;\Rightarrow\; 3P=3 \;\Rightarrow\; P=1

So the inside must equal

1

.

3Write the inner layer

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

Use the definition with

a=x-5

and

b=6

.

(x-5)\circledcirc 6=(x-5)+6-6(x-5)=6-5(x-5)

Collecting the two

(x-5)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

1

and solve.

6-5(x-5)=1 \;\Rightarrow\; 5(x-5)=5 \;\Rightarrow\; x-5=1 \;\Rightarrow\; x=6

A linear equation finishes it.

Answer:

6

4 · Reviewdoes it hold up?

At $x=6$ : $1\circledcirc 6=1+6-6=1$ , then $1\circledcirc 4=1+4-4=1$ . The result is $1$ , as required.

Another way: Expand everything in terms of $x$ : $(x-5)\circledcirc 6=-5x+31$ , and $(-5x+31)\circledcirc 4=-5x+35-4(-5x+31)=15x-89$ . Then $15x-89=1$ gives $x=6$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-1=(1-P)(1-4)$ , giving $P=1$ , and the inner layer says $1-1=(1-(x-5))(1-6)$ , giving $x-5=1$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-5)+6-6(x-5)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 6 medium

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-3)\circledcirc 7\}\circledcirc 2=1$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $1$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-3)\circledcirc 7\}\circledcirc 2=1$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-3)\circledcirc 7

, so the equation reads

P\circledcirc 2=1

.

P\circledcirc 2=1

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

1

.

2-P=1 \;\Rightarrow\; P=1

So the inside must equal

1

.

3Write the inner layer

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

Use the definition with

a=x-3

and

b=7

.

(x-3)\circledcirc 7=(x-3)+7-7(x-3)=7-6(x-3)

Collecting the two

(x-3)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

1

and solve.

7-6(x-3)=1 \;\Rightarrow\; 6(x-3)=6 \;\Rightarrow\; x-3=1 \;\Rightarrow\; x=4

A linear equation finishes it.

Answer:

4

4 · Reviewdoes it hold up?

At $x=4$ : $1\circledcirc 7=1+7-7=1$ , then $1\circledcirc 2=1+2-2=1$ . The result is $1$ , as required.

Another way: Expand everything in terms of $x$ : $(x-3)\circledcirc 7=-6x+25$ , and $(-6x+25)\circledcirc 2=-6x+27-2(-6x+25)=6x-23$ . Then $6x-23=1$ gives $x=4$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-1=(1-P)(1-2)$ , giving $P=1$ , and the inner layer says $1-1=(1-(x-3))(1-7)$ , giving $x-3=1$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-3)+7-7(x-3)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 7 medium

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-6)\circledcirc 7\}\circledcirc 2=7$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $7$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-6)\circledcirc 7\}\circledcirc 2=7$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-6)\circledcirc 7

, so the equation reads

P\circledcirc 2=7

.

P\circledcirc 2=7

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

7

.

2-P=7 \;\Rightarrow\; P=-5

So the inside must equal

-5

.

3Write the inner layer

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

Use the definition with

a=x-6

and

b=7

.

(x-6)\circledcirc 7=(x-6)+7-7(x-6)=7-6(x-6)

Collecting the two

(x-6)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-5

and solve.

7-6(x-6)=-5 \;\Rightarrow\; 6(x-6)=12 \;\Rightarrow\; x-6=2 \;\Rightarrow\; x=8

A linear equation finishes it.

Answer:

8

4 · Reviewdoes it hold up?

At $x=8$ : $2\circledcirc 7=2+7-14=-5$ , then $(-5)\circledcirc 2=-5+2-(-10)=7$ . The result is $7$ , as required.

Another way: Expand everything in terms of $x$ : $(x-6)\circledcirc 7=-6x+43$ , and $(-6x+43)\circledcirc 2=-6x+45-2(-6x+43)=6x-41$ . Then $6x-41=7$ gives $x=8$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-7=(1-P)(1-2)$ , giving $P=-5$ , and the inner layer says $1-(-5)=(1-(x-6))(1-7)$ , giving $x-6=2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-6)+7-7(x-6)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 8 medium

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-5)\circledcirc 4\}\circledcirc 3=7$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $7$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-5)\circledcirc 4\}\circledcirc 3=7$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-5)\circledcirc 4

, so the equation reads

P\circledcirc 3=7

.

P\circledcirc 3=7

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 3=P+3-3P=3-2P

. Set it equal to

7

.

3-2P=7 \;\Rightarrow\; 2P=-4 \;\Rightarrow\; P=-2

So the inside must equal

-2

.

3Write the inner layer

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

Use the definition with

a=x-5

and

b=4

.

(x-5)\circledcirc 4=(x-5)+4-4(x-5)=4-3(x-5)

Collecting the two

(x-5)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-2

and solve.

4-3(x-5)=-2 \;\Rightarrow\; 3(x-5)=6 \;\Rightarrow\; x-5=2 \;\Rightarrow\; x=7

A linear equation finishes it.

Answer:

7

4 · Reviewdoes it hold up?

At $x=7$ : $2\circledcirc 4=2+4-8=-2$ , then $(-2)\circledcirc 3=-2+3-(-6)=7$ . The result is $7$ , as required.

Another way: Expand everything in terms of $x$ : $(x-5)\circledcirc 4=-3x+19$ , and $(-3x+19)\circledcirc 3=-3x+22-3(-3x+19)=6x-35$ . Then $6x-35=7$ gives $x=7$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-7=(1-P)(1-3)$ , giving $P=-2$ , and the inner layer says $1-(-2)=(1-(x-5))(1-4)$ , giving $x-5=2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-5)+4-4(x-5)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 9 hard

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-1)\circledcirc 3\}\circledcirc 2=-9$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $-9$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-1)\circledcirc 3\}\circledcirc 2=-9$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-1)\circledcirc 3

, so the equation reads

P\circledcirc 2=-9

.

P\circledcirc 2=-9

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

-9

.

2-P=-9 \;\Rightarrow\; P=11

So the inside must equal

11

.

3Write the inner layer

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

Use the definition with

a=x-1

and

b=3

.

(x-1)\circledcirc 3=(x-1)+3-3(x-1)=3-2(x-1)

Collecting the two

(x-1)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

11

and solve.

3-2(x-1)=11 \;\Rightarrow\; 2(x-1)=-8 \;\Rightarrow\; x-1=-4 \;\Rightarrow\; x=-3

A linear equation finishes it.

Answer:

-3

4 · Reviewdoes it hold up?

At $x=-3$ : $(-4)\circledcirc 3=-4+3-(-12)=11$ , then $11\circledcirc 2=11+2-22=-9$ . The result is $-9$ , as required.

Another way: Expand everything in terms of $x$ : $(x-1)\circledcirc 3=-2x+5$ , and $(-2x+5)\circledcirc 2=-2x+7-2(-2x+5)=2x-3$ . Then $2x-3=-9$ gives $x=-3$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-(-9)=(1-P)(1-2)$ , giving $P=11$ , and the inner layer says $1-11=(1-(x-1))(1-3)$ , giving $x-1=-4$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-1)+3-3(x-1)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 10 hard

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-3)\circledcirc 5\}\circledcirc 3=9$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $9$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-3)\circledcirc 5\}\circledcirc 3=9$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-3)\circledcirc 5

, so the equation reads

P\circledcirc 3=9

.

P\circledcirc 3=9

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 3=P+3-3P=3-2P

. Set it equal to

9

.

3-2P=9 \;\Rightarrow\; 2P=-6 \;\Rightarrow\; P=-3

So the inside must equal

-3

.

3Write the inner layer

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

Use the definition with

a=x-3

and

b=5

.

(x-3)\circledcirc 5=(x-3)+5-5(x-3)=5-4(x-3)

Collecting the two

(x-3)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-3

and solve.

5-4(x-3)=-3 \;\Rightarrow\; 4(x-3)=8 \;\Rightarrow\; x-3=2 \;\Rightarrow\; x=5

A linear equation finishes it.

Answer:

5

4 · Reviewdoes it hold up?

At $x=5$ : $2\circledcirc 5=2+5-10=-3$ , then $(-3)\circledcirc 3=-3+3-(-9)=9$ . The result is $9$ , as required.

Another way: Expand everything in terms of $x$ : $(x-3)\circledcirc 5=-4x+17$ , and $(-4x+17)\circledcirc 3=-4x+20-3(-4x+17)=8x-31$ . Then $8x-31=9$ gives $x=5$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-9=(1-P)(1-3)$ , giving $P=-3$ , and the inner layer says $1-(-3)=(1-(x-3))(1-5)$ , giving $x-3=2$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-3)+5-5(x-3)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 11 hard

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-2)\circledcirc 3\}\circledcirc 3=9$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $9$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-2)\circledcirc 3\}\circledcirc 3=9$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-2)\circledcirc 3

, so the equation reads

P\circledcirc 3=9

.

P\circledcirc 3=9

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 3=P+3-3P=3-2P

. Set it equal to

9

.

3-2P=9 \;\Rightarrow\; 2P=-6 \;\Rightarrow\; P=-3

So the inside must equal

-3

.

3Write the inner layer

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

Use the definition with

a=x-2

and

b=3

.

(x-2)\circledcirc 3=(x-2)+3-3(x-2)=3-2(x-2)

Collecting the two

(x-2)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-3

and solve.

3-2(x-2)=-3 \;\Rightarrow\; 2(x-2)=6 \;\Rightarrow\; x-2=3 \;\Rightarrow\; x=5

A linear equation finishes it.

Answer:

5

4 · Reviewdoes it hold up?

At $x=5$ : $3\circledcirc 3=3+3-9=-3$ , then $(-3)\circledcirc 3=-3+3-(-9)=9$ . The result is $9$ , as required.

Another way: Expand everything in terms of $x$ : $(x-2)\circledcirc 3=-2x+7$ , and $(-2x+7)\circledcirc 3=-2x+10-3(-2x+7)=4x-11$ . Then $4x-11=9$ gives $x=5$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-9=(1-P)(1-3)$ , giving $P=-3$ , and the inner layer says $1-(-3)=(1-(x-2))(1-3)$ , giving $x-2=3$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-2)+3-3(x-2)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Variant 12 hard

For two numbers $a$ and $b$ , define $a \circledcirc b=a+b-ab$ . Find the value of $x$ that satisfies $\{(x-1)\circledcirc 5\}\circledcirc 2=9$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A new operation is defined by $a \circledcirc b=a+b-ab$ . It is used twice, one inside the other, and the result is $9$ . Find $x$ .

Givens

$a \circledcirc b=a+b-ab$ for any numbers $a$ and $b$ .
$\{(x-1)\circledcirc 5\}\circledcirc 2=9$ .

Unknowns

The value of $x$ .

Constraints

Apply the definition exactly, with the left number as $a$ and the right number as $b$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Give the inner block a name, $P$ . Undo the outer operation first to learn what $P$ must be, then undo the inner operation to get $x$ . Each layer is a short linear equation.

3 · Execute4 carry out the plan

1Name the inside

#7 Identify Subproblems 7.EE.A.1

Let

P=(x-1)\circledcirc 5

, so the equation reads

P\circledcirc 2=9

.

P\circledcirc 2=9

One layer at a time.

2Undo the outer layer

#11 Work Backwards 7.EE.A.1

By the definition,

P\circledcirc 2=P+2-2P=2-P

. Set it equal to

9

.

2-P=9 \;\Rightarrow\; P=-7

So the inside must equal

-7

.

3Write the inner layer

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

Use the definition with

a=x-1

and

b=5

.

(x-1)\circledcirc 5=(x-1)+5-5(x-1)=5-4(x-1)

Collecting the two

(x-1)

terms keeps it short.

4Undo the inner layer

#11 Work Backwards 7.EE.B.4

Set it equal to

-7

and solve.

5-4(x-1)=-7 \;\Rightarrow\; 4(x-1)=12 \;\Rightarrow\; x-1=3 \;\Rightarrow\; x=4

A linear equation finishes it.

Answer:

4

4 · Reviewdoes it hold up?

At $x=4$ : $3\circledcirc 5=3+5-15=-7$ , then $(-7)\circledcirc 2=-7+2-(-14)=9$ . The result is $9$ , as required.

Another way: Expand everything in terms of $x$ : $(x-1)\circledcirc 5=-4x+9$ , and $(-4x+9)\circledcirc 2=-4x+11-2(-4x+9)=4x-7$ . Then $4x-7=9$ gives $x=4$ . A slicker view: $1-(a\circledcirc b)=(1-a)(1-b)$ , so the outer layer says $1-9=(1-P)(1-2)$ , giving $P=-7$ , and the inner layer says $1-(-7)=(1-(x-1))(1-5)$ , giving $x-1=3$ .

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Applying the definition and simplifying linear expressions such as $(x-1)+5-5(x-1)$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear equations.

💡Takeaway. For an operation inside an operation, work from the outside in: find what the inside has to be, then solve for

x

.

Peel a nested defined operation from the outside · 12 practice problems

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