Introduction to Algebra: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Which of these lines is an equation rather than an expression?

Answer choices for question 1 A $11j - 6$ B $5c + 2 = 27$ C $\dfrac{v}{9} + 4$ D $3(k + 7)$
2

Solve $-12u = 54$ for $u$ .

Answer choices for question 2 A $4.5$ B $66$ C $-648$ D $-4.5$
3

Which expression means "twelve less than a number $h$ "?

Answer choices for question 3 A $h - 12$ B $12 - h$ C $12h$ D $h + 12$
4

Simplify $6b - 15 - 9b + 4$ .

Answer choices for question 4 A $3b - 11$ B $-14b$ C $-3b - 11$ D $-3b - 19$
5

Evaluate $9p - 4$ when $p = 6$ .

Answer choices for question 5 A $50$ B $92$ C $58$ D $30$
6

Solve $4x > -28$ for $x$ .

Answer choices for question 6 A $x < -7$ B $x > 7$ C $x > -7$ D $x > -32$
7

A parcel is within a courier's limit when its weight is at most $18$ kilograms. Writing $w$ for the weight in kilograms, which number line shows every weight that is within the limit?

Answer choices for question 7 A An open circle on $18$ , shaded left B An open circle on $18$ , shaded right C A closed circle on $18$ , shaded right D A closed circle on $18$ , shaded left
8

Solve $9x - 14 = 40$ for $x$ .

Answer choices for question 8 A $\dfrac{26}{9}$ B $6$ C $54$ D $45$
9

What is the value of $5(2m - 3) - 4m$ when $m = -2$ ?

Answer choices for question 9 A $-15$ B $18$ C $-27$ D $-3$
10

Solve $\dfrac{3}{8}w = 21$ for $w$ .

Answer choices for question 10 A $56$ B $\dfrac{63}{8}$ C $7$ D $168$
11

How many terms does $7a^2 - a + 3b - 8$ have, and what is its constant term?

Answer choices for question 11 A Four terms, and the constant is $8$ . B Three terms, and the constant is $-8$ . C Four terms, and the constant is $3b$ . D Four terms, and the constant is $-8$ .
12

Solve $-6x \ge 48$ , and say how the solution is graphed.

Answer choices for question 12 A $x \ge -8$ , a closed circle on $-8$ with the shading running right B $x \le -8$ , a closed circle on $-8$ with the shading running left C $x \le 8$ , a closed circle on $8$ with the shading running left D $x < -8$ , an open circle on $-8$ with the shading running left
13

A jar already held $130$ grams of rice. Four identical tins were then emptied into it, and the jar held $530$ grams. How many grams does one tin hold?

Answer choices for question 13 A $100$ grams B $165$ grams C $400$ grams D $2.5$ grams
14

Solve $6(2x - 5) - 4x = 34$ for $x$ .

Answer choices for question 14 A $\dfrac{39}{8}$ B $\dfrac{16}{3}$ C $\dfrac{1}{2}$ D $8$
15

Evaluate $-n^2 + 5n$ when $n = -4$ .

Answer choices for question 15 A $-4$ B $-36$ C $4$ D $-15$
16

Solve $17 - 4x < 5$ for $x$ .

Answer choices for question 16 A $x < 3$ B $x > -\dfrac{11}{2}$ C $x > 3$ D $x > -3$
17

Which value satisfies both $\dfrac{k}{6} = -4$ and $2k + 19 = -29$ ?

Answer choices for question 17 A $-\dfrac{2}{3}$ B $-5$ C $-48$ D $-24$
18

The rule "five less than the square of a number $z$ " is applied at $z = -6$ . What number does it give?

Answer choices for question 18 A $31$ B $-31$ C $121$ D $-41$
19

Solve $3(2x - 7) - 4x \le 9$ for $x$ .

Answer choices for question 19 A $x \le 8$ B $x \le 15$ C $x \ge 15$ D $x \le -6$
20

A solution is graphed on a number line as a filled circle on $-2$ with the shading running left. Exactly one of these statements about that solution is false. Which one?

Answer choices for question 20 A $-2$ is one of the solutions. B The solutions have a greatest member, and it is $-2$ . C The solutions have a least member. D Every solution is less than $-1$ .

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Where you stand

Translate between words and algebraic notation: expressions, equations and inequalities, including the reversal in "less than", the grouping a phrase such as "twice the sum" demands, and the boundary-including wordings "at most" and "at least". Review: Variables and Expressions , Inequalities
Name the parts of an expression (its terms with the signs they carry, the coefficient including the invisible $1$ , the constant) and tell an expression from an equation. Review: Variables and Expressions
Evaluate an expression by substitution under the full order of operations, including negative inputs, powers, and more than one letter. Review: Variables and Expressions , Evaluating and Simplifying Expressions
Simplify an expression by combining like terms and by distributing, including a leading minus read as a factor of $-1$ . Review: Evaluating and Simplifying Expressions
Solve a one-step equation by applying the inverse of the operation attached to the letter, justified by the balance principle, and confirm the value by substitution. Review: Solving One-Step Equations
Solve a two-step or longer equation, clearing the constant before the coefficient, cleaning up a side by combining like terms or distributing when it is needed. Review: Solving Two-Step Equations
Solve an inequality with the balance moves, reversing the symbol exactly when both sides are multiplied or divided by a negative number and not otherwise. Review: Inequalities
Express an inequality's solution as a range and graph it, choosing the circle from the symbol's strictness and the shading from its direction. Review: Inequalities

Core practice

10 problems from across the chapter. 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 Inputs on a card

Find the value of $(-p)^2+\frac{3q}{p-q}$ when $p=-2$ and $q=4$ .
1. Hint 1
  Keep the signs and grouping around each substituted input.
2. Hint 2
  Calculate the square, numerator, and denominator before adding the two terms.
Answer

$2$

Full solution

For the squared term,
$(-(-2))^2=4$
For the fraction, the numerator is $3(4)=12$ and the denominator is $(-2)-(4)=-6$ , which is nonzero.

Finish the division and addition:
$\begin{aligned}\frac{12}{-6}&=-2\\4+(-2)&=2\end{aligned}$

Answer

$2$

Key idea

Substitute into every part of an expression while preserving the original grouping.
Problem 1: Mark complete
Problem 2 A written equality

Find $x$ if $-\frac{5}{12}=\frac{3}{8}+x$ . Explain why your change to the equation is valid and check the value.
1. Hint 1
  The variable, on the right side, has a fraction added to it.
2. Hint 2
  Apply the inverse operation to both sides, writing the fractions in twenty-fourths.
Answer

$x=-\frac{19}{24}$ ; both sides equal $-\frac{5}{12}$ .

Full solution

Subtract $\frac{3}{8}$ from both sides.
Subtracting the same number from both sides keeps the solution and removes the term added to $x$ :
$-\frac{5}{12}-\frac{3}{8}=x$

In twenty-fourths, $-\frac{5}{12}=-\frac{10}{24}$ and $\frac{3}{8}=\frac{9}{24}$ , so
$x=-\frac{19}{24}$

Check in the original: $\frac{3}{8}+(-\frac{19}{24})$ is $\frac{9}{24}-\frac{19}{24}$ , which is $-\frac{10}{24}$ , or $-\frac{5}{12}$ .
The sides agree, so the value is a solution.

Answer

$x=-\frac{19}{24}$ ; both sides equal $-\frac{5}{12}$ .

Key idea

Subtracting the number added to the variable from both sides isolates it and keeps the same solution.
Problem 2: Mark complete
Problem 3 An expression in ink

Write $-(2a-b)+3(a-2b)+4$ without parentheses and with no like terms remaining.
1. Hint 1
  Each outside factor acts on every term in its group.
2. Hint 2
  After removing the groups, collect the $a$ terms separately from the $b$ terms.
Answer

$a-5b+4$ , in any term order.

Full solution

The first group gives $-2a+b$ and the second gives $3a-6b$ .
Together with the constant, the expression is $-2a+b+3a-6b+4$ .

Combine the matching terms:
$\begin{aligned}-2a+3a&=a\\b-6b&=-5b\end{aligned}$

The constant $4$ remains, so the result is $a-5b+4$ .
The terms have different variable parts or are constant, so no more like terms remain.

Answer

$a-5b+4$ , in any term order.

Key idea

Distribute signed factors before collecting each type of term.
Problem 3: Mark complete
Problem 4 A value for $r$

Find $r$ if $\frac{5r+2}{4}-3=1$ . Give an exact answer, as a fraction in lowest terms if it is not a whole number, and check it.
1. Hint 1
  The fraction bar groups its numerator as one amount.
2. Hint 2
  Undo the operations surrounding that numerator before isolating the variable inside it.
Answer

$r=\frac{14}{5}$ ; both sides equal $1$ .

Full solution

Add $3$ to both sides, then multiply by $4$ :
$\begin{aligned}\frac{5r+2}{4}&=4\\5r+2&=16\end{aligned}$

Subtract $2$ from both sides and divide by nonzero $5$ :
$\begin{aligned}5r&=14\\r&=\frac{14}{5}\end{aligned}$

The fraction is in lowest terms.
For the check, the numerator becomes $5(\frac{14}{5})+2=16$ , so the original left side is
$16\div4-3=1.$

Answer

$r=\frac{14}{5}$ ; both sides equal $1$ .

Key idea

Undo surrounding operations before solving for a variable within a grouped numerator.
Problem 4: Mark complete
Problem 5 A rectangle's border

A rectangle's length is $2$ cm less than five times its width of $w$ cm, where $w$ is greater than $1$ . Write the rectangle's perimeter in centimeters as an expression in $w$ , in any equivalent form, then find the perimeter when $w=3$ .
1. Hint 1
  Write the length as an expression in $w$ before building the perimeter from it.
2. Hint 2
  In " $2$ less than five times the width", start from five times the width and take $2$ away.
3. Hint 3
  The perimeter is twice the sum of the length and the width, so the whole sum must stay grouped when it is doubled.
Answer

$2(w+5w-2)$ , or $12w-4$ , or any equivalent form; $32$ cm when $w=3$ .

Full solution

Five times the width is $5w$ , and $2$ less than that starts from $5w$ and takes $2$ away, so the length is $5w-2$ cm, not $2-5w$ .
Since $w$ is greater than $1$ , this length is positive.

The perimeter adds the length and the width and then doubles the whole sum, so the sum stays in parentheses:
$2(w+5w-2)$
Inside the parentheses $w+5w=6w$ , so an equivalent form is
$2(6w-2)=12w-4$

At $w=3$ the length is $5(3)-2=13$ cm, so the perimeter is
$2(3+13)=32$
The other form agrees, since
$12(3)-4=32.$

Answer

$2(w+5w-2)$ , or $12w-4$ , or any equivalent form; $32$ cm when $w=3$ .

Key idea

Build each quantity from its phrase first, then keep grouped any sum that is doubled as a whole.
Problem 5: Mark complete
Problem 6 A condition on $x$

Find every number $x$ for which $-11<5-1.6x$ . Describe the solution range in words and its number-line graph, then check $x=0$ and $x=15$ .
1. Hint 1
  The variable term sits on the right, so isolate it there, keeping track of what each balance move does to the symbol.
2. Hint 2
  After clearing the constant, look at the sign of the number you divide both sides by.
3. Hint 3
  Once $x$ stands alone, read the comparison as a statement about $x$ before choosing the circle and the shading.
Answer

$x<10$ (or $10>x$ ), every number less than $10$ ; open circle at $10$ , shade left; $0$ works and $15$ does not.

Full solution

Subtract $5$ from both sides.
A subtraction leaves the symbol alone:
$-16<-1.6x$

Divide both sides by $-1.6$ .
The divisor is negative, so the symbol reverses:
$10>x$

Read from the side of $x$ , this says $x<10$ , every number less than $10$ .
The symbol is strict, so the circle at $10$ is open, and the smaller numbers lie to the left, so the shading runs left.

At $x=0$ , the right side is $5$ , and $-11<5$ is true.
At $x=15$ , it is $5-24=-19$ , and $-11<-19$ is false.

Answer

$x<10$ (or $10>x$ ), every number less than $10$ ; open circle at $10$ , shade left; $0$ works and $15$ does not.

Key idea

Read a solved comparison from the variable's side before choosing its circle and shading.
Problem 6: Mark complete
Problem 7 A bag of apples

Apples cost $3.20$ dollars per kilogram, and a bag of them costs $8$ dollars. Let the bag hold $k$ kilograms. Find $k$ , explain why each step you take keeps the equation true, and check the result.
1. Hint 1
  The cost of the bag is the price of one kilogram multiplied by the number of kilograms in the bag.
2. Hint 2
  Undo the multiplication by the same nonzero number on both sides.
Answer

$k=2.5$ kilograms; both sides equal $8$ .

Full solution

The cost of the bag is the price per kilogram times the number of kilograms, and $3.20$ is the same number as $3.2$ , so
$3.2k=8$

Divide both sides by $3.2$ .
It is nonzero, so this balance move keeps the two sides equal and leaves $k$ alone on the left:
$k=\frac{8}{3.2}$

Multiplying the top and bottom by $10$ gives $\frac{80}{32}$ , which reduces to $\frac{5}{2}$ , so $k=2.5$ kilograms.
Multiplying by the price checks the result:
$3.2(2.5)=8$
This agrees with the cost of the bag.

Answer

$k=2.5$ kilograms; both sides equal $8$ .

Key idea

Dividing both sides by a nonzero price per unit recovers the number of units.
Problem 7: Mark complete
Problem 8 A harbor warning light

A harbor gauge measures the water level $h$ , in meters, relative to a fixed marker, with levels below the marker negative. A warning light stays off while $h$ is at least $-2.5$ , and it comes on at every lower level. Write an inequality for the levels at which the light is off and one for the levels at which it is on, and describe both number-line graphs.
1. Hint 1
  Decide which of the two groups the boundary level itself belongs to.
2. Hint 2
  Determine which condition covers the higher levels and which covers the lower levels.
Answer

Off: $h\ge-2.5$ (or $-2.5\le h$ ), closed circle at $-2.5$ , shade right; on: $h<-2.5$ (or $-2.5>h$ ), open circle at $-2.5$ , shade left.

Full solution

"At least $-2.5$ " allows $-2.5$ itself and every greater level, so the light is off exactly when
$h\ge-2.5$
Its graph has a closed circle at $-2.5$ and shading to the right.

The light comes on at every level lower than $-2.5$ , but not at $-2.5$ itself, since the light is off there:
$h<-2.5$
Its graph has an open circle at $-2.5$ and shading to the left.

The two graphs share the boundary but differ in both the circle and the direction, and together they cover every level once.

Answer

Off: $h\ge-2.5$ (or $-2.5\le h$ ), closed circle at $-2.5$ , shade right; on: $h<-2.5$ (or $-2.5>h$ ), open circle at $-2.5$ , shade left.

Key idea

A condition and its opposite share one boundary, which belongs to exactly one of them.
Problem 8: Mark complete
Problem 9 Dana's rewrite

Dana claims that $4(x-5)-(2x+1)+x$ equals $x-21$ for every $x$ . Judge the claim, rewrite the original expression with no parentheses or like terms remaining, and find the value of $x$ for which the original equals $-9$ .
1. Hint 1
  Check the expression rewrite before relying on it to find an input.
2. Hint 2
  The minus before the second group reaches only the terms inside it, so the final $+x$ keeps its plus sign.
3. Hint 3
  Use the correct shorter expression when finding the input, then check the original.
Answer

The claim is false; correct expression: $3x-21$ ; required value: $x=4$ .

Full solution

Distributing the $4$ gives $4x-20$ , and the minus before $(2x+1)$ multiplies both of its terms by $-1$ , giving $-2x-1$ , so the expression becomes $4x-20-2x-1+x$ .
The $x$ terms combine to $4x-2x+x=3x$ , and the constants total $-21$ , leaving
$3x-21$
Thus the claimed $x-21$ is not a valid rewrite for every input.
For example, at $x=1$ the original is $4(-4)-3+1=-18$ , while $x-21$ gives $-20$ , so the two forms differ.

For the required input, solve
$3x-21=-9$
Add $21$ to both sides, then divide by $3$ :
$\begin{aligned}3x&=12\\x&=4\end{aligned}$

In the original, $4(4-5)=-4$ and $(2(4)+1)=9$ , so the value is $-4-9+4=-9$ , confirming the solution.

Answer

The claim is false; correct expression: $3x-21$ ; required value: $x=4$ .

Key idea

A rewrite can be trusted in an equation only when it gives the same value as the original for every input.
Problem 9: Mark complete
Problem 10 Kai's two lines

Kai's solution of $\frac{x+5}{-3}\le2$ has two lines: $x+5\le-6$ , then $x\le-11$ . Say whether his first line follows validly from the original inequality and whether his second follows validly from his first, solve the original inequality, and test $x=-12$ in it.
1. Hint 1
  Judge each line as a move made on the line before it, not only by where it ends.
2. Hint 2
  Clearing the fraction multiplies both sides by $-3$ ; decide what that does to the symbol.
3. Hint 3
  When testing, finish the whole numerator before dividing it by $-3$ .
Answer

The first line does not follow validly; the second follows validly from the first; the solution is $x\ge-11$ ; $x=-12$ is not a solution.

Full solution

The fraction bar divides the whole numerator by $-3$ .
Multiplying both sides by $-3$ clears it, and because the multiplier is negative the symbol must reverse:
$x+5\ge-6$
Kai kept $\le$ , so the first line does not follow validly from the original.

The second line subtracts $5$ from both sides of the first, which leaves the symbol alone, so it follows validly from the first; it carries the earlier error forward.

Subtracting $5$ from both sides of the correct line gives
$x\ge-11$

At $x=-12$ , the original left side is
$\frac{-12+5}{-3}=\frac{7}{3}$
This is greater than $2$ , so the comparison is false and $-12$ is not a solution.
It does satisfy Kai's $x\le-11$ , which confirms that his range is wrong.

Answer

The first line does not follow validly; the second follows validly from the first; the solution is $x\ge-11$ ; $x=-12$ is not a solution.

Key idea

Judge each step against the line before it, since a valid step can carry an earlier error forward.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Pre-Algebra. You can skip it. → Continue to Chapter 9: Geometry Basics →

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