Practice · Linear means the square term's coefficient is zero · Sensim Math

Variant 1 easy

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$a^2x^2+6x+3-x^2+ax-7x-8$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $a^2x^2+6x+3-x^2+ax-7x-8$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(a^2-1)x^2+(6+a-7)x+(3-8)=(a^2-1)x^2+(a-1)x-5

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

1

is

1

or

-1

, since

(-1)\times(-1)=1

.

a^2-1=0\ \Rightarrow\ a^2=1\ \Rightarrow\ a=1\ \text{or}\ a=-1

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-1

must not be

0

. With

a=1

it would be

0

and the expression would shrink to the constant

-5

, which is not linear. So

a=1

is out.

a-1\neq0\ \Rightarrow\ a\neq1\ \Rightarrow\ a=-1

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-1

into the simplified form.

(1-1)x^2+(-1-1)x-5=-2x-5

The leftover expression has degree

1

.

Answer:

a=-1

,

-2x-5

4 · Reviewdoes it hold up?

Substitute $a=-1$ into the original: $x^2+6x+3-x^2-x-7x-8=-2x-5$ . With $a=1$ you get $x^2+6x+3-x^2+x-7x-8=-5$ , a constant, confirming that $a=1$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-1$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=1$ to get both $a=1$ and $a=-1$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 2 easy

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$2a^2x^2+x+3-8x^2+ax+x-7$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $2a^2x^2+x+3-8x^2+ax+x-7$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(2a^2-8)x^2+(1+a+1)x+(3-7)=(2a^2-8)x^2+(a+2)x-4

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

4

is

2

or

-2

, since

(-2)\times(-2)=4

.

2a^2-8=0\ \Rightarrow\ a^2=4\ \Rightarrow\ a=2\ \text{or}\ a=-2

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+2

must not be

0

. With

a=-2

it would be

0

and the expression would shrink to the constant

-4

, which is not linear. So

a=-2

is out.

a+2\neq0\ \Rightarrow\ a\neq-2\ \Rightarrow\ a=2

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=2

into the simplified form.

(8-8)x^2+(2+2)x-4=4x-4

The leftover expression has degree

1

.

Answer:

a=2

,

4x-4

4 · Reviewdoes it hold up?

Substitute $a=2$ into the original: $8x^2+x+3-8x^2+2x+x-7=4x-4$ . With $a=-2$ you get $8x^2+x+3-8x^2-2x+x-7=-4$ , a constant, confirming that $a=-2$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=2$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=4$ to get both $a=2$ and $a=-2$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 3 easy

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$a^2x^2+4x-2-9x^2+ax-x+6$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $a^2x^2+4x-2-9x^2+ax-x+6$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(a^2-9)x^2+(4+a-1)x+(-2+6)=(a^2-9)x^2+(a+3)x+4

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

9

is

3

or

-3

, since

(-3)\times(-3)=9

.

a^2-9=0\ \Rightarrow\ a^2=9\ \Rightarrow\ a=3\ \text{or}\ a=-3

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+3

must not be

0

. With

a=-3

it would be

0

and the expression would shrink to the constant

4

, which is not linear. So

a=-3

is out.

a+3\neq0\ \Rightarrow\ a\neq-3\ \Rightarrow\ a=3

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=3

into the simplified form.

(9-9)x^2+(3+3)x+4=6x+4

The leftover expression has degree

1

.

Answer:

a=3

,

6x+4

4 · Reviewdoes it hold up?

Substitute $a=3$ into the original: $9x^2+4x-2-9x^2+3x-x+6=6x+4$ . With $a=-3$ you get $9x^2+4x-2-9x^2-3x-x+6=4$ , a constant, confirming that $a=-3$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=3$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=9$ to get both $a=3$ and $a=-3$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 4 easy

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$a^2x^2-5x+4-4x^2+ax+3x-1$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $a^2x^2-5x+4-4x^2+ax+3x-1$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(a^2-4)x^2+(-5+a+3)x+(4-1)=(a^2-4)x^2+(a-2)x+3

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

4

is

2

or

-2

, since

(-2)\times(-2)=4

.

a^2-4=0\ \Rightarrow\ a^2=4\ \Rightarrow\ a=2\ \text{or}\ a=-2

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-2

must not be

0

. With

a=2

it would be

0

and the expression would shrink to the constant

3

, which is not linear. So

a=2

is out.

a-2\neq0\ \Rightarrow\ a\neq2\ \Rightarrow\ a=-2

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-2

into the simplified form.

(4-4)x^2+(-2-2)x+3=-4x+3

The leftover expression has degree

1

.

Answer:

a=-2

,

-4x+3

4 · Reviewdoes it hold up?

Substitute $a=-2$ into the original: $4x^2-5x+4-4x^2-2x+3x-1=-4x+3$ . With $a=2$ you get $4x^2-5x+4-4x^2+2x+3x-1=3$ , a constant, confirming that $a=2$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-2$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=4$ to get both $a=2$ and $a=-2$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 5 medium

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$a^2x^2+3x-6-4x^2+ax-5x+1$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $a^2x^2+3x-6-4x^2+ax-5x+1$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(a^2-4)x^2+(3+a-5)x+(-6+1)=(a^2-4)x^2+(a-2)x-5

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

4

is

2

or

-2

, since

(-2)\times(-2)=4

.

a^2-4=0\ \Rightarrow\ a^2=4\ \Rightarrow\ a=2\ \text{or}\ a=-2

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-2

must not be

0

. With

a=2

it would be

0

and the expression would shrink to the constant

-5

, which is not linear. So

a=2

is out.

a-2\neq0\ \Rightarrow\ a\neq2\ \Rightarrow\ a=-2

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-2

into the simplified form.

(4-4)x^2+(-2-2)x-5=-4x-5

The leftover expression has degree

1

.

Answer:

a=-2

,

-4x-5

4 · Reviewdoes it hold up?

Substitute $a=-2$ into the original: $4x^2+3x-6-4x^2-2x-5x+1=-4x-5$ . With $a=2$ you get $4x^2+3x-6-4x^2+2x-5x+1=-5$ , a constant, confirming that $a=2$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-2$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=4$ to get both $a=2$ and $a=-2$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 6 medium

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$4a^2x^2-2x+7-4x^2+ax+x-3$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $4a^2x^2-2x+7-4x^2+ax+x-3$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(4a^2-4)x^2+(-2+a+1)x+(7-3)=(4a^2-4)x^2+(a-1)x+4

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

1

is

1

or

-1

, since

(-1)\times(-1)=1

.

4a^2-4=0\ \Rightarrow\ a^2=1\ \Rightarrow\ a=1\ \text{or}\ a=-1

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-1

must not be

0

. With

a=1

it would be

0

and the expression would shrink to the constant

4

, which is not linear. So

a=1

is out.

a-1\neq0\ \Rightarrow\ a\neq1\ \Rightarrow\ a=-1

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-1

into the simplified form.

(4-4)x^2+(-1-1)x+4=-2x+4

The leftover expression has degree

1

.

Answer:

a=-1

,

-2x+4

4 · Reviewdoes it hold up?

Substitute $a=-1$ into the original: $4x^2-2x+7-4x^2-x+x-3=-2x+4$ . With $a=1$ you get $4x^2-2x+7-4x^2+x+x-3=4$ , a constant, confirming that $a=1$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-1$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=1$ to get both $a=1$ and $a=-1$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 7 medium

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$2a^2x^2-3x+2-2x^2+ax+4x-3$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $2a^2x^2-3x+2-2x^2+ax+4x-3$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(2a^2-2)x^2+(-3+a+4)x+(2-3)=(2a^2-2)x^2+(a+1)x-1

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

1

is

1

or

-1

, since

(-1)\times(-1)=1

.

2a^2-2=0\ \Rightarrow\ a^2=1\ \Rightarrow\ a=1\ \text{or}\ a=-1

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+1

must not be

0

. With

a=-1

it would be

0

and the expression would shrink to the constant

-1

, which is not linear. So

a=-1

is out.

a+1\neq0\ \Rightarrow\ a\neq-1\ \Rightarrow\ a=1

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=1

into the simplified form.

(2-2)x^2+(1+1)x-1=2x-1

The leftover expression has degree

1

.

Answer:

a=1

,

2x-1

4 · Reviewdoes it hold up?

Substitute $a=1$ into the original: $2x^2-3x+2-2x^2+x+4x-3=2x-1$ . With $a=-1$ you get $2x^2-3x+2-2x^2-x+4x-3=-1$ , a constant, confirming that $a=-1$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=1$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=1$ to get both $a=1$ and $a=-1$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 8 hard

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$2a^2x^2+5x-4-2x^2+ax-4x-2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $2a^2x^2+5x-4-2x^2+ax-4x-2$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(2a^2-2)x^2+(5+a-4)x+(-4-2)=(2a^2-2)x^2+(a+1)x-6

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

1

is

1

or

-1

, since

(-1)\times(-1)=1

.

2a^2-2=0\ \Rightarrow\ a^2=1\ \Rightarrow\ a=1\ \text{or}\ a=-1

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+1

must not be

0

. With

a=-1

it would be

0

and the expression would shrink to the constant

-6

, which is not linear. So

a=-1

is out.

a+1\neq0\ \Rightarrow\ a\neq-1\ \Rightarrow\ a=1

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=1

into the simplified form.

(2-2)x^2+(1+1)x-6=2x-6

The leftover expression has degree

1

.

Answer:

a=1

,

2x-6

4 · Reviewdoes it hold up?

Substitute $a=1$ into the original: $2x^2+5x-4-2x^2+x-4x-2=2x-6$ . With $a=-1$ you get $2x^2+5x-4-2x^2-x-4x-2=-6$ , a constant, confirming that $a=-1$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=1$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=1$ to get both $a=1$ and $a=-1$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 9 medium

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$3a^2x^2+2x-5-3x^2+ax-3x+1$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $3a^2x^2+2x-5-3x^2+ax-3x+1$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(3a^2-3)x^2+(2+a-3)x+(-5+1)=(3a^2-3)x^2+(a-1)x-4

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

1

is

1

or

-1

, since

(-1)\times(-1)=1

.

3a^2-3=0\ \Rightarrow\ a^2=1\ \Rightarrow\ a=1\ \text{or}\ a=-1

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-1

must not be

0

. With

a=1

it would be

0

and the expression would shrink to the constant

-4

, which is not linear. So

a=1

is out.

a-1\neq0\ \Rightarrow\ a\neq1\ \Rightarrow\ a=-1

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-1

into the simplified form.

(3-3)x^2+(-1-1)x-4=-2x-4

The leftover expression has degree

1

.

Answer:

a=-1

,

-2x-4

4 · Reviewdoes it hold up?

Substitute $a=-1$ into the original: $3x^2+2x-5-3x^2-x-3x+1=-2x-4$ . With $a=1$ you get $3x^2+2x-5-3x^2+x-3x+1=-4$ , a constant, confirming that $a=1$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-1$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=1$ to get both $a=1$ and $a=-1$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 10 hard

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$3a^2x^2-6x+5-12x^2+ax+8x+2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $3a^2x^2-6x+5-12x^2+ax+8x+2$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(3a^2-12)x^2+(-6+a+8)x+(5+2)=(3a^2-12)x^2+(a+2)x+7

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

4

is

2

or

-2

, since

(-2)\times(-2)=4

.

3a^2-12=0\ \Rightarrow\ a^2=4\ \Rightarrow\ a=2\ \text{or}\ a=-2

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+2

must not be

0

. With

a=-2

it would be

0

and the expression would shrink to the constant

7

, which is not linear. So

a=-2

is out.

a+2\neq0\ \Rightarrow\ a\neq-2\ \Rightarrow\ a=2

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=2

into the simplified form.

(12-12)x^2+(2+2)x+7=4x+7

The leftover expression has degree

1

.

Answer:

a=2

,

4x+7

4 · Reviewdoes it hold up?

Substitute $a=2$ into the original: $12x^2-6x+5-12x^2+2x+8x+2=4x+7$ . With $a=-2$ you get $12x^2-6x+5-12x^2-2x+8x+2=7$ , a constant, confirming that $a=-2$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=2$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=4$ to get both $a=2$ and $a=-2$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 11 hard

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$2a^2x^2-7x+1-18x^2+ax+4x+4$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $2a^2x^2-7x+1-18x^2+ax+4x+4$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(2a^2-18)x^2+(-7+a+4)x+(1+4)=(2a^2-18)x^2+(a-3)x+5

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

9

is

3

or

-3

, since

(-3)\times(-3)=9

.

2a^2-18=0\ \Rightarrow\ a^2=9\ \Rightarrow\ a=3\ \text{or}\ a=-3

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a-3

must not be

0

. With

a=3

it would be

0

and the expression would shrink to the constant

5

, which is not linear. So

a=3

is out.

a-3\neq0\ \Rightarrow\ a\neq3\ \Rightarrow\ a=-3

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=-3

into the simplified form.

(18-18)x^2+(-3-3)x+5=-6x+5

The leftover expression has degree

1

.

Answer:

a=-3

,

-6x+5

4 · Reviewdoes it hold up?

Substitute $a=-3$ into the original: $18x^2-7x+1-18x^2-3x+4x+4=-6x+5$ . With $a=3$ you get $18x^2-7x+1-18x^2+3x+4x+4=5$ , a constant, confirming that $a=3$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=-3$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=9$ to get both $a=3$ and $a=-3$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Variant 12 hard

When the expression below is simplified, find the value of the constant $a$ that makes it a linear expression in $x$ , and find that linear expression.

$3a^2x^2+5x-1-27x^2+ax-2x-4$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The expression mixes $x^2$ terms, $x$ terms and constants, with an unknown constant $a$ in two coefficients. Choose $a$ so that after simplifying, the expression is linear in $x$ , then write it.

Givens

The expression $3a^2x^2+5x-1-27x^2+ax-2x-4$ .
$a$ is a constant.

Unknowns

The value of $a$ .
The linear expression it produces.

Constraints

Linear in $x$ : the $x^2$ term must vanish, but the $x$ term must not.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Collect like terms by degree so each coefficient is visible. A linear expression needs a zero $x^2$ coefficient and a nonzero $x$ coefficient; the first gives two candidates for $a$ , and the second rules one out.

3 · Execute4 carry out the plan

1Group by degree

#13 Convert to Algebra A-APR.A.1

Collect the

x^2

terms, the

x

terms and the constants.

(3a^2-27)x^2+(5+a-2)x+(-1-4)=(3a^2-27)x^2+(a+3)x-5

Each power of

x

now has one coefficient.

2Make the square term vanish

#13 Convert to Algebra 8.EE.A.2

For the expression to be linear, the

x^2

coefficient must be

0

. A number whose square is

9

is

3

or

-3

, since

(-3)\times(-3)=9

.

3a^2-27=0\ \Rightarrow\ a^2=9\ \Rightarrow\ a=3\ \text{or}\ a=-3

Don't forget the negative candidate.

3Keep the x term alive

#3 Eliminate Possibilities A-APR.A.1

The

x

coefficient

a+3

must not be

0

. With

a=-3

it would be

0

and the expression would shrink to the constant

-5

, which is not linear. So

a=-3

is out.

a+3\neq0\ \Rightarrow\ a\neq-3\ \Rightarrow\ a=3

A linear expression must actually contain

x

.

4Write the linear expression

#7 Identify Subproblems A-APR.A.1

Put

a=3

into the simplified form.

(27-27)x^2+(3+3)x-5=6x-5

The leftover expression has degree

1

.

Answer:

a=3

,

6x-5

4 · Reviewdoes it hold up?

Substitute $a=3$ into the original: $27x^2+5x-1-27x^2+3x-2x-4=6x-5$ . With $a=-3$ you get $27x^2+5x-1-27x^2-3x-2x-4=-5$ , a constant, confirming that $a=-3$ fails.

Another way: Test the two candidates directly in the original expression (guess and check) instead of reasoning about coefficients; only $a=3$ leaves an $x$ term.

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Adding the polynomial's like terms by degree and reading each coefficient.
8.EE.A.2 Use square root and cube root symbols to represent solutions — Solving $a^2=9$ to get both $a=3$ and $a=-3$.

💡Takeaway. Linear means the

x^2

part is gone but the

x

part is still there; check both conditions.

Linear means the square term's coefficient is zero · 12 practice problems

Generated variants — 12

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