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Variables and 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 Color labels

A roll contains $r$ red labels and $b$ blue labels. Write an expression in $r$ and $b$ for the total number of labels.
1. Hint 1
  The total includes both colors.
2. Hint 2
  Picture a roll with $4$ red and $6$ blue labels. What would you do with $4$ and $6$ to get the total? Do the same with the letters.
Answer

$r+b$ , or $b+r$ , labels.

Full solution

The letter $r$ stands for the number of red labels, and $b$ stands for the number of blue labels.
Adding the two counts gives the total:
$r+b$

The expression $b+r$ gives the same total because changing the order of addition does not change its value.
Neither count needs to be known before the expression is written.

Answer

$r+b$ , or $b+r$ , labels.

Key idea

A letter lets one expression describe a quantity even when its component counts are unknown.
Problem 1: Mark complete
Problem 2 An expression to build

Write an expression with two terms in which the coefficient of $k$ is $-\frac{1}{5}$ and the constant is $12$ , with $k$ written in the numerator of a fraction.
1. Hint 1
  The coefficient is the number multiplying the variable, including its sign.
2. Hint 2
  Dividing by $5$ is multiplying by $\frac{1}{5}$ .
Answer

$-\frac{k}{5}+12$ , or $12-\frac{k}{5}$ (also $\frac{-k}{5}+12$ ).

Full solution

A coefficient of $-\frac{1}{5}$ means negative one fifth of $k$ :
$-\frac{1}{5}k=-\frac{k}{5}$

Adding the constant term $12$ gives the two-term expression $-\frac{k}{5}+12$ .
Writing the same two terms in the other order gives $12-\frac{k}{5}$ .

Answer

$-\frac{k}{5}+12$ , or $12-\frac{k}{5}$ (also $\frac{-k}{5}+12$ ).

Key idea

A variable in a fraction can have a fractional coefficient.
Problem 2: Mark complete
Problem 3 A numerator to finish

Find the value of $\frac{2-3v}{5}$ when $v=-6$ .
1. Hint 1
  The fraction bar groups the entire numerator.
2. Hint 2
  Replace $v$ with $(-6)$ , multiply before subtracting in the numerator, then divide.
Answer

$4$

Full solution

Substitute the given value in parentheses.
In the numerator, multiply before subtracting, and subtracting a negative means adding its opposite:
$\begin{aligned}2-3(-6)&=2-(-18)\\2-(-18)&=20\end{aligned}$

The whole numerator is now finished, so divide it by the denominator:
$\frac{20}{5}=4$

Answer

$4$

Key idea

Inside a numerator the order of operations still applies, and the whole numerator is finished before it is divided by the denominator.
Problem 3: Mark complete
Problem 4 Two letters and a phrase

Write "three less than half the sum of $a$ and $b$ " as an expression. Then find its value when $a=-10$ and $b=4$ .
1. Hint 1
  First identify the amount from which three is taken away.
2. Hint 2
  Half the sum means add the two inputs before dividing by two.
3. Hint 3
  Put each given value in parentheses and finish the grouped sum first.
Answer

$\frac{a+b}{2}-3$ , or $\frac{1}{2}(a+b)-3$ (either order of $a$ and $b$ ); value $-6$ .

Full solution

The sum must stay together when it is halved.
Three less than that amount means starting from the half and subtracting $3$ afterward, giving $\frac{a+b}{2}-3$ rather than $3-\frac{a+b}{2}$ .
Halving is multiplying by $\frac{1}{2}$ , so $\frac{1}{2}(a+b)-3$ records the same amount.
Reversing the order of the two addends in either form gives the same amount too.

For the given inputs, the sum is $-6$ and half of it is $-3$ :
$\frac{(-10)+(4)}{2}-3=-6$

Three less than $-3$ is $-6$ , so the number matches the phrase.

Answer

$\frac{a+b}{2}-3$ , or $\frac{1}{2}(a+b)-3$ (either order of $a$ and $b$ ); value $-6$ .

Key idea

Read a grouped phrase as a whole before carrying out an operation on it.
Problem 4: Mark complete
Problem 5 Pages in a pad

Tom's pad starts with $p$ pages, and he uses $8$ of them. Write the number of pages Tom has left, using only $p$ and $8$ . Ana's pad started with $q$ pages; she has used $6$ and has exactly $13$ left. Write one line, using $q$ , $6$ and $13$ , about the pages Ana has left. For each of your two lines, say whether it is an expression or an equation and whether you would evaluate it or solve it; do not solve.
1. Hint 1
  An amount can be described without claiming it equals a particular number.
2. Hint 2
  Ana's pad has a known final amount; how do you write that one amount is equal to another?
Answer

$p-8$ : expression, evaluate; $q-6=13$ (or $13=q-6$ ): equation, solve.

Full solution

For Tom's pad, subtract the pages used from the starting number:
$p-8$

That line has no equals sign, so it is an expression, and it can be evaluated after a value of $p$ is chosen.
For Ana's pad, the pages left are $q-6$ , and that amount is known to be $13$ , so the line is
$q-6=13$

That line has an equals sign, so it is an equation.
Solving it would find the starting number that makes the equality true, while the first line names the pages left for any starting number $p$ .

Answer

$p-8$ : expression, evaluate; $q-6=13$ (or $13=q-6$ ): equation, solve.

Key idea

An expression names an amount, while an equation states that two amounts are equal.
Problem 5: Mark complete
Problem 6 An expression to read

For $-a^2+2b-c+9$ , name each term with its sign, give the coefficient of each variable term, and name the constant.
1. Hint 1
  A term carries the sign written in front of it.
2. Hint 2
  A missing visible multiplier can still mean one copy, or negative one copy, of a variable part.
Answer

Terms: $-a^2$ , $2b$ , $-c$ , $9$ ; coefficients: $-1$ , $2$ , $-1$ respectively; constant: $9$ .

Full solution

The additions and subtractions separate four terms.
Keeping each sign with its term gives $-a^2$ , $2b$ , $-c$ , and $9$ .

The two invisible multipliers become visible when the terms are written as
$-a^2=(-1)a^2$
and
$-c=(-1)c$

Thus the coefficients are $-1$ , $2$ , and $-1$ .
The term $9$ contains no variable, so it is the constant.

Answer

Terms: $-a^2$ , $2b$ , $-c$ , $9$ ; coefficients: $-1$ , $2$ , $-1$ respectively; constant: $9$ .

Key idea

Terms keep their signs, including when their coefficients are not written explicitly.
Problem 6: Mark complete
Problem 7 A storage panel

A panel has $48$ empty slots. A morning group fills $q$ slots and an afternoon group fills $t$ different slots. Write the number of empty slots left as an expression in $q$ and $t$ . Find that number when $q=14$ and $t=18$ .
1. Hint 1
  Both groups reduce the number of empty slots.
2. Hint 2
  Take away each group from the starting total without taking any slot away twice.
Answer

$48-q-t$ , $48-t-q$ , or $48-(q+t)$ ; $16$ slots left.

Full solution

The occupied groups do not overlap, so each must be removed from the $48$ starting slots.
Either $48-q-t$ or $48-t-q$ records that.
Taking away the two groups' total at once, $48-(q+t)$ , records the same count.

Substitute the counts and subtract from left to right:
$\begin{aligned}48-(14)-(18)&=34-18\\34-18&=16\end{aligned}$

The occupied slots total $14+18=32$ , and $32+16=48$ , checking the remaining count.

Answer

$48-q-t$ , $48-t-q$ , or $48-(q+t)$ ; $16$ slots left.

Key idea

Defined variables can record separate contributions to one total.
Problem 7: Mark complete
Problem 8 Noah's written work

Noah is finding $n-\frac{n}{4}$ when $n=-8$ . He writes $-8-\frac{8}{4}$ . Is his substitution correct? Explain and give the correct value.
1. Hint 1
  Every occurrence of the same letter has the same value within one evaluation.
2. Hint 2
  Compare each occurrence of $n$ with the number in its place, sign included.
Answer

No; the correct value is $-6$ .

Full solution

Noah wrote $-8$ for the first $n$ but $8$ for the second, dropping its sign.
Both occurrences must receive the stated value $-8$ , and parentheses keep its sign.

The correct computation divides before subtracting, and subtracting $-2$ means adding $2$ :
$\begin{aligned}(-8)-\frac{(-8)}{4}&=-8-(-2)\\-8-(-2)&=-6\end{aligned}$

Noah's line gives $-8-2=-10$ instead, so his substitution is not correct.

Answer

No; the correct value is $-6$ .

Key idea

Use the given value, sign included, at every occurrence of the same variable.
Problem 8: Mark complete
Problem 9 Jules's swap

For the expression $ab-4$ , Jules claims that exchanging the values assigned to $a$ and $b$ leaves the result unchanged. Is this true for every pair of numbers? Explain.
1. Hint 1
  Only the order of the two factors changes when the inputs are exchanged.
2. Hint 2
  Compare the products $ab$ and $ba$ before considering the subtraction.
Answer

Yes, for every pair of numbers.

Full solution

Exchanging the inputs changes the product from $ab$ to $ba$ .
The commutative property of multiplication gives
$ab=ba$

Subtracting the same $4$ from equal products preserves equality:
$ab-4=ba-4$

Thus Jules is correct for every pair of numbers, including negative numbers and zero.

Answer

Yes, for every pair of numbers.

Key idea

Exchanging two factors leaves their product unchanged, so subtracting the same number from it gives the same result.
Problem 9: Mark complete
Problem 10 Lena's claim

Lena says that choosing different values of $r$ must give different values of $r^2-7$ . Test $r=-4$ and $r=4$ , then explain whether her claim is true.
1. Hint 1
  Think about what squaring does to the sign of a negative input.
2. Hint 2
  Use parentheses around the negative input, and compare the two final values.
Answer

Both values are $9$ ; Lena's claim is false.

Full solution

For the negative input, the square comes before the subtraction, and the square of a negative number is positive:
$\begin{aligned}(-4)^2-7&=16-7\\16-7&=9\end{aligned}$

For the positive input,
$\begin{aligned}(4)^2-7&=16-7\\16-7&=9\end{aligned}$

The inputs differ but the outputs agree, so the claim is false.
The inputs $-4$ and $4$ are opposites, so they have the same square, and subtracting $7$ from equal squares gives equal values.

Answer

Both values are $9$ ; Lena's claim is false.

Key idea

An expression containing a variable can still have the same value for different inputs.
Problem 10: Mark complete

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