Practice · Pair each monomial with a term by degree · Sensim Math

Variant 1 easy

$A$ and $B$ are monomials in $x$ , and $\frac{A-4x^2}{2x}=x^2+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-4x^2}{2x}=x^2+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

2x

.

A-4x^2=(x^2+B)\times 2x=2x^3+2xB

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

2xB

to the left and

4x^2

to the right.

A-2xB=2x^3+4x^2

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=2x^3,\ -2xB=4x^2 \qquad (\mathrm{ii})\ A=4x^2,\ -2xB=2x^3

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{4x^2}{-2x}=-2x

, degree

1

, smaller than

3

. Second case:

B=\frac{2x^3}{-2x}=-x^2

, degree

2

, the same as

A

, so it is ruled out.

A=2x^3,\ B=-2x

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{2x^3}{-2x}=-x^2

3-1=2

.

Answer:

-x^2

4 · Reviewdoes it hold up?

Put $A=2x^3$ back in: $\frac{2x^3-4x^2}{2x}=x^2-2x$ , which is $x^2+B$ with $B=-2x$ .

Another way: Read the original equation term by term: the $x^2$ on the right can only come from $\frac{A}{2x}$ , so $A=2x^3$ ; then $\frac{-4x^2}{2x}=-2x$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 2 easy

$A$ and $B$ are monomials in $x$ , and $\frac{A-3x^3}{3x^2}=x^4+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-3x^3}{3x^2}=x^4+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3x^2

.

A-3x^3=(x^4+B)\times 3x^2=3x^6+3x^2B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

3x^2B

to the left and

3x^3

to the right.

A-3x^2B=3x^6+3x^3

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=3x^6,\ -3x^2B=3x^3 \qquad (\mathrm{ii})\ A=3x^3,\ -3x^2B=3x^6

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{3x^3}{-3x^2}=-x

, degree

1

, smaller than

6

. Second case:

B=\frac{3x^6}{-3x^2}=-x^4

, degree

4

, more than

A

's degree

3

, so it is ruled out.

A=3x^6,\ B=-x

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{3x^6}{-x}=-3x^5

6-1=5

.

Answer:

-3x^5

4 · Reviewdoes it hold up?

Put $A=3x^6$ back in: $\frac{3x^6-3x^3}{3x^2}=x^4-x$ , which is $x^4+B$ with $B=-x$ .

Another way: Read the original equation term by term: the $x^4$ on the right can only come from $\frac{A}{3x^2}$ , so $A=3x^6$ ; then $\frac{-3x^3}{3x^2}=-x$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 3 easy

$A$ and $B$ are monomials in $x$ , and $\frac{A+6x^3}{2x}=3x^4+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A+6x^3}{2x}=3x^4+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

2x

.

A+6x^3=(3x^4+B)\times 2x=6x^5+2xB

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

2xB

to the left and

6x^3

to the right.

A-2xB=6x^5-6x^3

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=6x^5,\ -2xB=-6x^3 \qquad (\mathrm{ii})\ A=-6x^3,\ -2xB=6x^5

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{-6x^3}{-2x}=3x^2

, degree

2

, smaller than

5

. Second case:

B=\frac{6x^5}{-2x}=-3x^4

, degree

4

, more than

A

's degree

3

, so it is ruled out.

A=6x^5,\ B=3x^2

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{6x^5}{3x^2}=2x^3

5-2=3

.

Answer:

2x^3

4 · Reviewdoes it hold up?

Put $A=6x^5$ back in: $\frac{6x^5+6x^3}{2x}=3x^4+3x^2$ , which is $3x^4+B$ with $B=3x^2$ .

Another way: Read the original equation term by term: the $3x^4$ on the right can only come from $\frac{A}{2x}$ , so $A=6x^5$ ; then $\frac{6x^3}{2x}=3x^2$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 4 easy

$A$ and $B$ are monomials in $x$ , and $\frac{A+8x^4}{4x^2}=x^5+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A+8x^4}{4x^2}=x^5+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

4x^2

.

A+8x^4=(x^5+B)\times 4x^2=4x^7+4x^2B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

4x^2B

to the left and

8x^4

to the right.

A-4x^2B=4x^7-8x^4

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=4x^7,\ -4x^2B=-8x^4 \qquad (\mathrm{ii})\ A=-8x^4,\ -4x^2B=4x^7

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{-8x^4}{-4x^2}=2x^2

, degree

2

, smaller than

7

. Second case:

B=\frac{4x^7}{-4x^2}=-x^5

, degree

5

, more than

A

's degree

4

, so it is ruled out.

A=4x^7,\ B=2x^2

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{4x^7}{2x^2}=2x^5

7-2=5

.

Answer:

2x^5

4 · Reviewdoes it hold up?

Put $A=4x^7$ back in: $\frac{4x^7+8x^4}{4x^2}=x^5+2x^2$ , which is $x^5+B$ with $B=2x^2$ .

Another way: Read the original equation term by term: the $x^5$ on the right can only come from $\frac{A}{4x^2}$ , so $A=4x^7$ ; then $\frac{8x^4}{4x^2}=2x^2$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 5 medium

$A$ and $B$ are monomials in $x$ , and $\frac{A+9x^4}{3x^3}=x^4+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A+9x^4}{3x^3}=x^4+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3x^3

.

A+9x^4=(x^4+B)\times 3x^3=3x^7+3x^3B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

3x^3B

to the left and

9x^4

to the right.

A-3x^3B=3x^7-9x^4

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=3x^7,\ -3x^3B=-9x^4 \qquad (\mathrm{ii})\ A=-9x^4,\ -3x^3B=3x^7

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{-9x^4}{-3x^3}=3x

, degree

1

, smaller than

7

. Second case:

B=\frac{3x^7}{-3x^3}=-x^4

, degree

4

, the same as

A

, so it is ruled out.

A=3x^7,\ B=3x

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{3x^7}{3x}=x^6

7-1=6

.

Answer:

x^6

4 · Reviewdoes it hold up?

Put $A=3x^7$ back in: $\frac{3x^7+9x^4}{3x^3}=x^4+3x$ , which is $x^4+B$ with $B=3x$ .

Another way: Read the original equation term by term: the $x^4$ on the right can only come from $\frac{A}{3x^3}$ , so $A=3x^7$ ; then $\frac{9x^4}{3x^3}=3x$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 6 medium

$A$ and $B$ are monomials in $x$ , and $\frac{A-9x^3}{3x}=2x^3+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-9x^3}{3x}=2x^3+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3x

.

A-9x^3=(2x^3+B)\times 3x=6x^4+3xB

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

3xB

to the left and

9x^3

to the right.

A-3xB=6x^4+9x^3

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=6x^4,\ -3xB=9x^3 \qquad (\mathrm{ii})\ A=9x^3,\ -3xB=6x^4

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{9x^3}{-3x}=-3x^2

, degree

2

, smaller than

4

. Second case:

B=\frac{6x^4}{-3x}=-2x^3

, degree

3

, the same as

A

, so it is ruled out.

A=6x^4,\ B=-3x^2

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{6x^4}{-3x^2}=-2x^2

4-2=2

.

Answer:

-2x^2

4 · Reviewdoes it hold up?

Put $A=6x^4$ back in: $\frac{6x^4-9x^3}{3x}=2x^3-3x^2$ , which is $2x^3+B$ with $B=-3x^2$ .

Another way: Read the original equation term by term: the $2x^3$ on the right can only come from $\frac{A}{3x}$ , so $A=6x^4$ ; then $\frac{-9x^3}{3x}=-3x^2$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 7 medium

$A$ and $B$ are monomials in $x$ , and $\frac{A-8x^4}{2x^3}=2x^4+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-8x^4}{2x^3}=2x^4+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

2x^3

.

A-8x^4=(2x^4+B)\times 2x^3=4x^7+2x^3B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

2x^3B

to the left and

8x^4

to the right.

A-2x^3B=4x^7+8x^4

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=4x^7,\ -2x^3B=8x^4 \qquad (\mathrm{ii})\ A=8x^4,\ -2x^3B=4x^7

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{8x^4}{-2x^3}=-4x

, degree

1

, smaller than

7

. Second case:

B=\frac{4x^7}{-2x^3}=-2x^4

, degree

4

, the same as

A

, so it is ruled out.

A=4x^7,\ B=-4x

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{4x^7}{-4x}=-x^6

7-1=6

.

Answer:

-x^6

4 · Reviewdoes it hold up?

Put $A=4x^7$ back in: $\frac{4x^7-8x^4}{2x^3}=2x^4-4x$ , which is $2x^4+B$ with $B=-4x$ .

Another way: Read the original equation term by term: the $2x^4$ on the right can only come from $\frac{A}{2x^3}$ , so $A=4x^7$ ; then $\frac{-8x^4}{2x^3}=-4x$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 8 medium

$A$ and $B$ are monomials in $x$ , and $\frac{A-16x^5}{4x^2}=x^5+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-16x^5}{4x^2}=x^5+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

4x^2

.

A-16x^5=(x^5+B)\times 4x^2=4x^7+4x^2B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

4x^2B

to the left and

16x^5

to the right.

A-4x^2B=4x^7+16x^5

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=4x^7,\ -4x^2B=16x^5 \qquad (\mathrm{ii})\ A=16x^5,\ -4x^2B=4x^7

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{16x^5}{-4x^2}=-4x^3

, degree

3

, smaller than

7

. Second case:

B=\frac{4x^7}{-4x^2}=-x^5

, degree

5

, the same as

A

, so it is ruled out.

A=4x^7,\ B=-4x^3

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{4x^7}{-4x^3}=-x^4

7-3=4

.

Answer:

-x^4

4 · Reviewdoes it hold up?

Put $A=4x^7$ back in: $\frac{4x^7-16x^5}{4x^2}=x^5-4x^3$ , which is $x^5+B$ with $B=-4x^3$ .

Another way: Read the original equation term by term: the $x^5$ on the right can only come from $\frac{A}{4x^2}$ , so $A=4x^7$ ; then $\frac{-16x^5}{4x^2}=-4x^3$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 9 hard

$A$ and $B$ are monomials in $x$ , and $\frac{A-10x^5}{5x^3}=2x^6+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-10x^5}{5x^3}=2x^6+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5x^3

.

A-10x^5=(2x^6+B)\times 5x^3=10x^9+5x^3B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

5x^3B

to the left and

10x^5

to the right.

A-5x^3B=10x^9+10x^5

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=10x^9,\ -5x^3B=10x^5 \qquad (\mathrm{ii})\ A=10x^5,\ -5x^3B=10x^9

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{10x^5}{-5x^3}=-2x^2

, degree

2

, smaller than

9

. Second case:

B=\frac{10x^9}{-5x^3}=-2x^6

, degree

6

, more than

A

's degree

5

, so it is ruled out.

A=10x^9,\ B=-2x^2

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{10x^9}{-2x^2}=-5x^7

9-2=7

.

Answer:

-5x^7

4 · Reviewdoes it hold up?

Put $A=10x^9$ back in: $\frac{10x^9-10x^5}{5x^3}=2x^6-2x^2$ , which is $2x^6+B$ with $B=-2x^2$ .

Another way: Read the original equation term by term: the $2x^6$ on the right can only come from $\frac{A}{5x^3}$ , so $A=10x^9$ ; then $\frac{-10x^5}{5x^3}=-2x^2$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 10 hard

$A$ and $B$ are monomials in $x$ , and $\frac{A-12x^4}{3x^2}=4x^4+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A-12x^4}{3x^2}=4x^4+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

3x^2

.

A-12x^4=(4x^4+B)\times 3x^2=12x^6+3x^2B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

3x^2B

to the left and

12x^4

to the right.

A-3x^2B=12x^6+12x^4

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=12x^6,\ -3x^2B=12x^4 \qquad (\mathrm{ii})\ A=12x^4,\ -3x^2B=12x^6

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{12x^4}{-3x^2}=-4x^2

, degree

2

, smaller than

6

. Second case:

B=\frac{12x^6}{-3x^2}=-4x^4

, degree

4

, the same as

A

, so it is ruled out.

A=12x^6,\ B=-4x^2

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{12x^6}{-4x^2}=-3x^4

6-2=4

.

Answer:

-3x^4

4 · Reviewdoes it hold up?

Put $A=12x^6$ back in: $\frac{12x^6-12x^4}{3x^2}=4x^4-4x^2$ , which is $4x^4+B$ with $B=-4x^2$ .

Another way: Read the original equation term by term: the $4x^4$ on the right can only come from $\frac{A}{3x^2}$ , so $A=12x^6$ ; then $\frac{-12x^4}{3x^2}=-4x^2$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 11 hard

$A$ and $B$ are monomials in $x$ , and $\frac{A+12x^5}{2x^2}=3x^5+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A+12x^5}{2x^2}=3x^5+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

2x^2

.

A+12x^5=(3x^5+B)\times 2x^2=6x^7+2x^2B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

2x^2B

to the left and

12x^5

to the right.

A-2x^2B=6x^7-12x^5

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=6x^7,\ -2x^2B=-12x^5 \qquad (\mathrm{ii})\ A=-12x^5,\ -2x^2B=6x^7

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{-12x^5}{-2x^2}=6x^3

, degree

3

, smaller than

7

. Second case:

B=\frac{6x^7}{-2x^2}=-3x^5

, degree

5

, the same as

A

, so it is ruled out.

A=6x^7,\ B=6x^3

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{6x^7}{6x^3}=x^4

7-3=4

.

Answer:

x^4

4 · Reviewdoes it hold up?

Put $A=6x^7$ back in: $\frac{6x^7+12x^5}{2x^2}=3x^5+6x^3$ , which is $3x^5+B$ with $B=6x^3$ .

Another way: Read the original equation term by term: the $3x^5$ on the right can only come from $\frac{A}{2x^2}$ , so $A=6x^7$ ; then $\frac{12x^5}{2x^2}=6x^3$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Variant 12 hard

$A$ and $B$ are monomials in $x$ , and $\frac{A+15x^6}{5x^3}=3x^6+B$ . Simplify $\frac{A}{B}$ . (The degree of $A$ is greater than the degree of $B$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two unknown monomials $A$ and $B$ satisfy one equation. Using that $A$ has the higher degree, find $\frac{A}{B}$ .

Givens

$A$ and $B$ are monomials in $x$ .
$\frac{A+15x^6}{5x^3}=3x^6+B$
The degree of $A$ is greater than the degree of $B$ .

Unknowns

$\frac{A}{B}$ in simplest form.

Constraints

A monomial is a single term, so it can equal only one term of the same degree.

2 · Planchoose the strategy

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

Clear the fraction and put the unknowns on one side. Two single-term unknowns must match the two terms on the other side, so list the two possible pairings and keep the one that fits the degree condition.

3 · Execute5 carry out the plan

1Clear the fraction

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

Multiply both sides by

5x^3

.

A+15x^6=(3x^6+B)\times 5x^3=15x^9+5x^3B

No more fraction to worry about.

2Gather the unknowns

#13 Convert to Algebra 6.EE.A.4

Move

5x^3B

to the left and

15x^6

to the right.

A-5x^3B=15x^9-15x^6

Unknowns on one side, known terms on the other.

3List the pairings

#2 Make a Systematic List 8.EE.A.1

The left is two monomials and the right is two terms of different degree, so each monomial takes one whole term. There are only two ways to pair them.

(\mathrm{i})\ A=15x^9,\ -5x^3B=-15x^6 \qquad (\mathrm{ii})\ A=-15x^6,\ -5x^3B=15x^9

Only two cases to check.

4Keep the case that fits

#3 Eliminate Possibilities 8.EE.A.1

First case:

B=\frac{-15x^6}{-5x^3}=3x^3

, degree

3

, smaller than

9

. Second case:

B=\frac{15x^9}{-5x^3}=-3x^6

, degree

6

, the same as

A

, so it is ruled out.

A=15x^9,\ B=3x^3

The degree condition picks exactly one case.

5Divide

#13 Convert to Algebra 8.EE.A.1

Divide the coefficients and subtract the exponents.

\frac{A}{B}=\frac{15x^9}{3x^3}=5x^6

9-3=6

.

Answer:

5x^6

4 · Reviewdoes it hold up?

Put $A=15x^9$ back in: $\frac{15x^9+15x^6}{5x^3}=3x^6+3x^3$ , which is $3x^6+B$ with $B=3x^3$ .

Another way: Read the original equation term by term: the $3x^6$ on the right can only come from $\frac{A}{5x^3}$ , so $A=15x^9$ ; then $\frac{15x^6}{5x^3}=3x^3$ must be $B$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Multiplying and dividing powers of $x$ with exponent laws.
6.EE.A.4 Identify equivalent expressions — Seeing that equal expressions must match term by term.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the fractional equation without the denominator.

💡Takeaway. A monomial is one term, so it can only match a term of its own degree.

Pair each monomial with a term by degree · 12 practice problems

Generated variants — 12

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Where next?