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Arithmetic with Expressions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Collecting like terms

Simplify $9k - 4 + 2k^2 - 11k + 7 - 5k^2 + k$ , and check your result by substituting $k = 2$ into both the original expression and your simplified one.
1. Hint 1
  Only like terms can be merged, so sort the terms into families by their variable part, keeping each sign with the term that follows it.
2. Hint 2
  There are three families: the $k^2$ terms, the $k$ terms and the constants. The lone $k$ at the end has coefficient $1$ .
3. Hint 3
  Add the coefficients within each family, so the $k$ terms give $(9 - 11 + 1)k$ . At $k = 2$ , square before multiplying: $2k^2$ is $2 \cdot 4$ .
Answer

$-3k^2 - k + 3$ ; at $k = 2$ both expressions equal $-11$ .

Full solution

Read off the terms with their signs attached: $9k$ , $-4$ , $2k^2$ , $-11k$ , $7$ , $-5k^2$ and $k$ .
Sort them by variable part into three families: the $k^2$ terms, the $k$ terms and the constants.

The $k^2$ terms are $2k^2$ and $-5k^2$ .
Adding their coefficients gives
$\begin{aligned} 2k^2 - 5k^2 &= (2 - 5)k^2 \\ &= -3k^2. \end{aligned}$

The $k$ terms are $9k$ , $-11k$ and $k$ , and the lone $k$ has coefficient $1$ :
$\begin{aligned} 9k - 11k + k &= (9 - 11 + 1)k \\ &= -k. \end{aligned}$
A coefficient of $-1$ is written as the minus sign alone.

The constants $-4$ and $7$ add to $3$ .
So the simplified expression is
$-3k^2 - k + 3.$
Its three terms have different variable parts, so nothing more combines.

Check at $k = 2$ .
The terms of the original become $18$ , $-4$ , $8$ , $-22$ , $7$ , $-20$ and $2$ , and these add to $-11$ .

The simplified form gives $-3(4) - 2 + 3$ , which is also $-11$ .
The two agree, which is good evidence that no term was lost.

Answer

$-3k^2 - k + 3$ ; at $k = 2$ both expressions equal $-11$ .

Key idea

Combine each family of like terms by adding its coefficients, counting a lone variable as having coefficient $1$ .
Problem 1: Mark complete
Problem 2 Subtracting a whole expression

Subtract $(5y^2 - 3y - 6)$ from $(2y^2 - 7y + 4)$ and simplify the result. Check it by substituting $y = -1$ into both the original difference and your answer.
1. Hint 1
  Subtracting a whole expression means adding its opposite, so the minus sign reaches every term of the expression being subtracted. Take care over which expression comes first.
2. Hint 2
  Subtracting $A$ from $B$ means $B - A$ , so start from $(2y^2 - 7y + 4) - (5y^2 - 3y - 6)$ and drop the second parentheses by flipping the sign of each of its three terms.
3. Hint 3
  After the flips, the second group contributes $-5y^2$ , $+3y$ and $+6$ . Collect the $y^2$ terms, the $y$ terms and the constants. At $y = -1$ , $y^2$ is $1$ .
Answer

$-3y^2 - 4y + 10$ ; at $y = -1$ the original difference and the result both equal $11$ .

Full solution

Subtracting $(5y^2 - 3y - 6)$ from $(2y^2 - 7y + 4)$ means starting with $2y^2 - 7y + 4$ and taking the other expression away:
$(2y^2 - 7y + 4) - (5y^2 - 3y - 6).$

The minus sign in front of the second group is a factor of $-1$ , so it flips the sign of every term inside: $5y^2$ becomes $-5y^2$ , $-3y$ becomes $+3y$ , and $-6$ becomes $+6$ .
The difference is now a plain sum:
$2y^2 - 7y + 4 - 5y^2 + 3y + 6.$

Collect each family.
The $y^2$ terms give $2y^2 - 5y^2 = -3y^2$ , the $y$ terms give $-7y + 3y = -4y$ , and the constants $4$ and $6$ give $10$ .
So the result is
$-3y^2 - 4y + 10.$

Check at $y = -1$ , where $y^2 = 1$ .
The first expression is $2 + 7 + 4$ , which is $13$ , and the second is $5 + 3 - 6$ , which is $2$ , so the original difference is $11$ .

The result gives $-3(1) - 4(-1) + 10$ , which is also $11$ .
The two agree, which is good evidence that no sign was dropped.

Answer

$-3y^2 - 4y + 10$ ; at $y = -1$ the original difference and the result both equal $11$ .

Key idea

Subtracting $A$ from $B$ means $B - A$ , and the minus sign in front of $A$ flips the sign of every one of its terms.
Problem 2: Mark complete
Problem 3 Two products, then collect

Simplify $5(2r - 3) - 3(4r - 7)$ , and check your result by substituting $r = -1$ into both the original expression and your simplified one.
1. Hint 1
  The number in front of each pair of parentheses multiplies every term inside, and its sign travels with it to each term.
2. Hint 2
  The second product is multiplied by $-3$ , not by $3$ , so both of its terms change sign. The $-7$ inside contributes $+21$ .
3. Hint 3
  Once both products are expanded, the expression is a plain sum: collect the $r$ terms and the constants. At $r = -1$ , work out each pair of parentheses first.
Answer

$-2r + 6$ (or $6 - 2r$ ); at $r = -1$ both expressions equal $8$ .

Full solution

Distribute each factor across every term of its parentheses, keeping the signs.
The first product gives
$5(2r - 3) = 10r - 15.$

The second factor is $-3$ , so it multiplies both $4r$ and $-7$ by $-3$ , and $(-3)(-7) = 21$ :
$-3(4r - 7) = -12r + 21.$

The expression is now $10r - 15 - 12r + 21$ .
The $r$ terms give $10r - 12r = -2r$ , and the constants give $-15 + 21 = 6$ , so the result is
$-2r + 6.$

Check at $r = -1$ .
The original is $5(-5) - 3(-11)$ , which is $-25 + 33$ , or $8$ .

The result gives $-2(-1) + 6$ , which is also $8$ .
The two agree, which is good evidence that both products were distributed correctly.

Answer

$-2r + 6$ (or $6 - 2r$ ); at $r = -1$ both expressions equal $8$ .

Key idea

A factor in front of parentheses multiplies every term inside, and a negative factor changes the sign of each one.
Problem 3: Mark complete
Problem 4 Reading the variable parts

Give the coefficient and the variable part of each term of $7ab^2 - a^2b + 3 - ab^2 + 4ba^2 - 8$ , and use them to simplify the expression.
1. Hint 1
  Like terms need exactly the same letters raised to exactly the same powers, while the order the letters are written in does not matter. A term keeps the sign in front of it.
2. Hint 2
  A term that shows no number has coefficient $1$ or $-1$ . Write each variable part with its letters in alphabetical order, so the part of $5yx^2$ is $x^2y$ .
3. Hint 3
  Compare $ab^2$ and $a^2b$ letter by letter: which letter is squared in each? Then add the coefficients within each family of like terms.
Answer

Coefficients, in order: $7$ , $-1$ , $3$ , $-1$ , $4$ , $-8$ . Variable parts: $ab^2$ , $a^2b$ , none, $ab^2$ , $a^2b$ , none. Simplified: $6ab^2 + 3a^2b - 5$ .

Full solution

Read the terms with their signs attached: $7ab^2$ , $-a^2b$ , $3$ , $-ab^2$ , $4ba^2$ and $-8$ .
The coefficient of each is its numerical factor, sign included.

So $7ab^2$ has coefficient $7$ and variable part $ab^2$ .
The terms $-a^2b$ and $-ab^2$ show no number, so each has coefficient $-1$ ; their variable parts are $a^2b$ and $ab^2$ .

In $4ba^2$ the coefficient is $4$ .
Multiplication can be done in any order, so $ba^2$ is the same product as $a^2b$ , and the variable part is $a^2b$ .
The numbers $3$ and $-8$ are constant terms: each is its own coefficient, and neither has a variable part.

The parts $ab^2$ and $a^2b$ are different.
The first is $a \cdot b \cdot b$ , one $a$ and two $b$ 's, while the second is $a \cdot a \cdot b$ , two $a$ 's and one $b$ .
So the terms fall into three families: the $ab^2$ terms, the $a^2b$ terms and the constants.

Add the coefficients within each family.
The $ab^2$ terms give $7 - 1$ , which is $6$ ; the $a^2b$ terms give $-1 + 4$ , which is $3$ ; and the constants $3$ and $-8$ give $-5$ .
So the simplified expression is
$6ab^2 + 3a^2b - 5.$

Check at $a = -1$ and $b = 2$ , not at $a = 1$ , where $a^2$ and $a$ are equal and a slip in a power of $a$ would not show.
The terms of the original become $-28$ , $-2$ , $3$ , $4$ , $8$ and $-8$ , which add to $-23$ , and the simplified form gives $-24 + 6 - 5$ , also $-23$ .

Answer

Coefficients, in order: $7$ , $-1$ , $3$ , $-1$ , $4$ , $-8$ . Variable parts: $ab^2$ , $a^2b$ , none, $ab^2$ , $a^2b$ , none. Simplified: $6ab^2 + 3a^2b - 5$ .

Key idea

Like terms need the same letters to the same powers; the order the letters are written in does not matter, but the powers do.
Problem 4: Mark complete
Problem 5 Brackets inside brackets

Simplify $3t - 4[2t - 3(t - 2)]$ .
1. Hint 1
  Nested grouping symbols are cleared from the inside out: finish the innermost parentheses completely before touching the brackets around them.
2. Hint 2
  Inside the brackets, distribute the $-3$ across $(t - 2)$ and combine the result with the $2t$ . Keep what you get inside the brackets for now.
3. Hint 3
  The $-4$ in front of the brackets multiplies every term inside, sign included, so a negative term inside becomes positive. Then combine with the $3t$ outside.
Answer

$7t - 24$

Full solution

Work from the inside out, starting with the parentheses.
The $-3$ multiplies both terms of $(t - 2)$ , and $(-3)(-2) = 6$ :
$2t - 3(t - 2) = 2t - 3t + 6.$

Combining $2t$ and $-3t$ gives $-t$ , so the brackets now hold $-t + 6$ .

Next, distribute the $-4$ across both terms in the brackets.
It turns $-t$ into $4t$ and $6$ into $-24$ :
$-4(-t + 6) = 4t - 24.$

The whole expression is now $3t + 4t - 24$ .
Combining the $t$ terms gives
$7t - 24.$

Check at $t = 3$ .
Inside the brackets, $2(3) - 3(3 - 2)$ is $6 - 3$ , or $3$ , so the original is $9 - 4(3)$ , which is $-3$ .
The result gives $21 - 24$ , also $-3$ .

Answer

$7t - 24$

Key idea

Clear nested grouping from the innermost layer outward, and let each negative factor change the sign of every term it multiplies.
Problem 5: Mark complete
Problem 6 What was taken away?

Find the expression that must be subtracted from $3x^2 - 4x + 1$ to leave $x^2 + 2x - 6$ , and check your answer by carrying out the subtraction.
1. Hint 1
  Think of the same question with numbers first: what must be subtracted from $10$ to leave $4$ ? The relationship you use there holds for expressions too.
2. Hint 2
  If subtracting $E$ from $A$ leaves $B$ , then $E = A - B$ . Here $A$ is $3x^2 - 4x + 1$ and $B$ is $x^2 + 2x - 6$ ; put $B$ in parentheses before you subtract it.
3. Hint 3
  Subtracting $B$ flips the sign of each of its terms. For the check, subtract your expression from $3x^2 - 4x + 1$ in the same way and compare with $x^2 + 2x - 6$ .
Answer

$2x^2 - 6x + 7$ ; the check: $(3x^2 - 4x + 1) - (2x^2 - 6x + 7)$ is $x^2 + 2x - 6$ .

Full solution

With numbers, the amount subtracted from $10$ to leave $4$ is $6$ , found as $10 - 4$ .
In the same way, if subtracting an expression $E$ from $3x^2 - 4x + 1$ leaves $x^2 + 2x - 6$ , then $E$ is the first expression minus the second, so $E$ equals
$(3x^2 - 4x + 1) - (x^2 + 2x - 6).$

Drop the second parentheses, flipping the sign of each of its terms:
$E = 3x^2 - 4x + 1 - x^2 - 2x + 6.$

Collect each family.
The $x^2$ terms give $3x^2 - x^2 = 2x^2$ , the $x$ terms give $-4x - 2x = -6x$ , and the constants $1$ and $6$ give $7$ .
So
$E = 2x^2 - 6x + 7.$

Check by carrying out the subtraction, flipping each sign of $E$ :
$\begin{aligned} &(3x^2 - 4x + 1) - (2x^2 - 6x + 7) \\ &= 3x^2 - 4x + 1 - 2x^2 + 6x - 7 \\ &= x^2 + 2x - 6. \end{aligned}$
That is exactly what was to be left, so the answer checks.

Answer

$2x^2 - 6x + 7$ ; the check: $(3x^2 - 4x + 1) - (2x^2 - 6x + 7)$ is $x^2 + 2x - 6$ .

Key idea

The expression that must be subtracted from $A$ to leave $B$ is $A - B$ , and subtracting $B$ flips the sign of each of its terms.
Problem 6: Mark complete
Problem 7 Trimming a poster

A rectangular poster is $w$ cm wide and $(3w + 4)$ cm tall, where $w$ is at least $1$ . A strip $5$ cm tall is cut off along its bottom edge, across the full width, so the poster keeps its width and becomes $5$ cm shorter.

Write the area of the trimmed poster as a simplified expression in square centimeters. Then find that area when $w = 12$ , and check it by multiplying the trimmed poster's width by its height.
1. Hint 1
  The area of a rectangle is its width times its height, and that stays true when a length is written with a letter. The strip that is cut off is a rectangle too.
2. Hint 2
  The whole poster has area $w(3w + 4)$ and the strip has area $5w$ . Distribute the $w$ across both terms, remembering that $w \cdot 3w$ is $3w^2$ .
3. Hint 3
  Subtract the strip and collect the two $w$ terms. For the check, the trimmed poster is $12$ cm wide, and its height is $3w + 4$ with $5$ taken off.
Answer

$3w^2 - w$ square centimeters. At $w = 12$ the area is $420$ square centimeters, matching the trimmed poster's $12$ cm by $35$ cm.

Full solution

The whole poster is a rectangle $w$ cm wide and $(3w + 4)$ cm tall, so its area is $w(3w + 4)$ square centimeters.

Distribute the $w$ across both terms.
Since $w \cdot 3w$ is $3w^2$ , this gives
$w(3w + 4) = 3w^2 + 4w.$

The strip cut off is a rectangle $w$ cm wide and $5$ cm tall, so its area is $5w$ square centimeters.

The trimmed poster is what remains, so subtract the strip and collect the $w$ terms:
$\begin{aligned} 3w^2 + 4w - 5w &= 3w^2 + (4 - 5)w \\ &= 3w^2 - w. \end{aligned}$

The same area can be found directly.
The trimmed poster is $(3w + 4) - 5$ cm tall, which is $3w - 1$ cm, and $w(3w - 1)$ is also $3w^2 - w$ .

At $w = 12$ the area is $3(144) - 12$ , which is $420$ square centimeters.

Check: the trimmed poster is $12$ cm wide and $3(12) - 1 = 35$ cm tall, and $12 \cdot 35$ is $420$ .
The two results agree.

Answer

$3w^2 - w$ square centimeters. At $w = 12$ the area is $420$ square centimeters, matching the trimmed poster's $12$ cm by $35$ cm.

Key idea

An area can be found as the whole minus the piece removed, or as the new width times the new height, and both routes simplify to the same expression.
Problem 7: Mark complete
Problem 8 Two days of ticket sales

A cinema charges $11$ dollars for an adult ticket and $6$ dollars for a child ticket. On Saturday it sold $a$ adult tickets and $c$ child tickets, where $c$ is at least $4$ . On Sunday it sold twice as many adult tickets as on Saturday, and $4$ fewer child tickets than on Saturday.

Write a simplified expression for Sunday's takings minus Saturday's takings, in dollars. Then say which of $a$ and $c$ your expression depends on, and explain why.
1. Hint 1
  Each day's takings are the number of adult tickets times their price plus the number of child tickets times theirs. Write Sunday's ticket numbers in terms of $a$ and $c$ first.
2. Hint 2
  Sunday sold $2a$ adult tickets and $c - 4$ child tickets, so its takings are $11(2a) + 6(c - 4)$ . Subtract Saturday's takings as a whole, in parentheses.
3. Hint 3
  Once the products are distributed and Saturday's signs are flipped, look at the two terms that contain $c$ . What does each day's $6c$ stand for?
Answer

$11a - 24$ dollars. It depends on $a$ only: the child-ticket terms $6c$ and $-6c$ cancel, and Sunday's child tickets bring in $24$ dollars less than Saturday's, whatever $c$ is.

Full solution

Saturday's takings are $11a$ dollars from adult tickets and $6c$ dollars from child tickets, so $11a + 6c$ dollars in all.

On Sunday the cinema sold $2a$ adult tickets and $c - 4$ child tickets.
Its takings are $11(2a) + 6(c - 4)$ dollars, and distributing gives $22a + 6c - 24$ dollars.

Subtract Saturday's takings as a whole, flipping the sign of each of its terms:
$\begin{aligned} &(22a + 6c - 24) - (11a + 6c) \\ &= 22a + 6c - 24 - 11a - 6c. \end{aligned}$

Collect like terms.
The $a$ terms give $22a - 11a = 11a$ , and the $c$ terms give $6c - 6c = 0$ , which leaves
$11a - 24.$

So the expression depends on $a$ only.
The $c$ terms cancel because both days' takings contain $6c$ .
Sunday's child tickets are Saturday's with $4$ taken away, and those $4$ tickets are worth $6 \cdot 4 = 24$ dollars however large $c$ is.

Each term of $11a - 24$ has a meaning.
Sunday sold $a$ more adult tickets than Saturday, worth $11a$ dollars, and $4$ fewer child tickets, worth $24$ dollars.
Neither depends on $c$ .

Check with $a = 5$ and $c = 10$ : Saturday takes $55 + 60 = 115$ dollars and Sunday takes $110 + 36 = 146$ dollars.
The difference is $31$ dollars, and $11(5) - 24$ is $31$ too.

Answer

$11a - 24$ dollars. It depends on $a$ only: the child-ticket terms $6c$ and $-6c$ cancel, and Sunday's child tickets bring in $24$ dollars less than Saturday's, whatever $c$ is.

Key idea

When a whole expression is subtracted, terms can cancel, and a letter that cancels is a quantity the result does not depend on.
Problem 8: Mark complete
Problem 9 A check that passed

Priya simplifies $6 - 2(x - 3) - (5 - x)$ . Her work reads
$\begin{aligned} &6 - 2(x - 3) - (5 - x) \\ &= 6 - 2x + 6 - 5 - x \\ &= -3x + 7. \end{aligned}$
She substitutes $x = 0$ into the original expression and into $-3x + 7$ , gets $7$ both times, and decides that her answer is right.

Find her mistake and write the correct simplified expression. Then give a value of $x$ at which the original expression and $-3x + 7$ disagree, and explain why her check at $x = 0$ could not catch the mistake.
1. Hint 1
  A match at one value is evidence, not proof, while a mismatch at any value proves that something went wrong. Look for a term whose sign was not handled.
2. Hint 2
  Expand $-2(x - 3)$ and $-(5 - x)$ yourself, one group at a time, and compare each with what Priya wrote in her second line.
3. Hint 3
  At $x = 0$ every term that contains $x$ is worth $0$ . Compare the $x$ terms of her answer with those of the correct one, and ask what each is worth there.
Answer

Mistake: $-(5 - x)$ became $-5 - x$ ; it is $-5 + x$ . Correct form: $-x + 7$ . At $x = 2$ the original is $5$ but $-3x + 7$ is $1$ (any $x \neq 0$ works). At $x = 0$ every $x$ term is $0$ , so a slip in an $x$ term cannot show.

Full solution

Expand one group at a time.
The $-2$ reaches both terms of $(x - 3)$ and gives $-2x + 6$ , which is what Priya wrote.

The minus sign in front of $(5 - x)$ is a factor of $-1$ , so it flips both terms:
$-(5 - x) = -5 + x.$
Priya wrote $-5 - x$ , flipping the $5$ but not the $-x$ .
That is her mistake.

With the correct signs the expression is $6 - 2x + 6 - 5 + x$ .
The $x$ terms give $-2x + x = -x$ , and the constants $6$ , $6$ and $-5$ give $7$ , so the correct simplified expression is
$-x + 7.$

Try $x = 2$ .
The original is $6 - 2(-1) - 3$ , which is $5$ , and the correct form gives $-2 + 7 = 5$ as well.
Priya's answer gives $-6 + 7$ , which is $1$ .

One disagreement is enough: $-3x + 7$ does not equal the original at $x = 2$ , so it is not equivalent to it, and her answer is wrong.

Her check at $x = 0$ could not see this.
Her answer and the correct one have the same constant, $7$ , and differ only in their $x$ terms, $-3x$ against $-x$ .
At $x = 0$ every term containing $x$ is worth $0$ , so both come to $7$ there.

The correct form minus hers is $(-x + 7) - (-3x + 7)$ , which is $2x$ .
That difference is $0$ at $x = 0$ and nonzero at every other value, so a check at any other value would have exposed the slip.

Answer

Mistake: $-(5 - x)$ became $-5 - x$ ; it is $-5 + x$ . Correct form: $-x + 7$ . At $x = 2$ the original is $5$ but $-3x + 7$ is $1$ (any $x \neq 0$ works). At $x = 0$ every $x$ term is $0$ , so a slip in an $x$ term cannot show.

Key idea

A check at $x = 0$ cannot tell apart two expressions that differ only in their $x$ terms, so pair it with a check at a value such as $x = 2$ , where those terms count.
Problem 9: Mark complete
Problem 10 Four claimed simplifications

Each claim below says that its two sides are equal for every value of the letters in it.

Claim A: $4m + 3m^2 = 7m^3$

Claim B: $6ab - 2ba = 4ab$

Claim C: $7 - (k - 3) = 4 - k$

Claim D: $p(p + 6) - 6p = p^2$

Decide whether each claim is true. Show why each true claim holds for every value, using the laws of arithmetic, and disprove each false claim with a value that makes its two sides differ.
1. Hint 1
  A claim about every value needs a reason that works for every value at once. A claim that fails needs only one value at which its two sides come out different.
2. Hint 2
  Simplify the left side of each claim with the tools of this lesson, reading each variable part carefully, and compare the result with the right side.
3. Hint 3
  When you test a value, do not stop at $0$ or $1$ : they can make different powers of a letter look alike. Try $2$ or $-1$ as well.
Answer

A: false (for example, at $m = 2$ the sides are $20$ and $56$ ). B: true. C: false (for example, at $k = 0$ the sides are $10$ and $4$ ; any $k$ works). D: true.

Full solution

A claim about every value is proved by rewriting one side into the other with laws that hold for every number.
It is disproved by a single value at which the two sides differ.

Claim A is false.
The terms $4m$ and $3m^2$ have different variable parts, $m$ and $m^2$ , so they cannot be combined by adding coefficients.
At $m = 2$ the left side is $8 + 12 = 20$ and the right side is $7 \cdot 8$ , which is $56$ .

The values $m = 0$ and $m = 1$ happen to make the two sides of claim A agree, at $0$ and at $7$ .
Those matches are coincidences, and the one mismatch at $m = 2$ settles the claim.

Claim B is true.
Multiplication can be done in either order, so $ba = ab$ , and the two terms are like terms with variable part $ab$ :
$\begin{aligned} 6ab - 2ba &= 6ab - 2ab \\ &= (6 - 2)ab \\ &= 4ab. \end{aligned}$

Claim C is false.
The minus sign flips both terms of $(k - 3)$ , so the left side is $7 - k + 3$ , which is $10 - k$ , not $4 - k$ .
The claim forgot to flip the $-3$ .

At $k = 0$ the left side is $7 - (-3)$ , which is $10$ , and the right side is $4$ .
In fact the two sides differ by $6$ at every value of $k$ .

Claim D is true.
Distributing the $p$ across both terms gives
$p(p + 6) = p^2 + 6p.$
Then the $6p$ and the $-6p$ cancel:
$p^2 + 6p - 6p = p^2.$

Answer

A: false (for example, at $m = 2$ the sides are $20$ and $56$ ). B: true. C: false (for example, at $k = 0$ the sides are $10$ and $4$ ; any $k$ works). D: true.

Key idea

A claimed simplification is proved by laws that hold for every value and disproved by one value where its sides differ; matching values alone do not prove it.
Problem 10: Mark complete

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