Algebra Domain Test 1

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Question 1

Which of the following is equivalent to $4(3x - 2) - 3(x + 5)$?

A) $9x - 23$
B) $9x + 7$
C) $15x - 23$
D) $15x - 7$

Show solution

Distribute each term:

\[4(3x - 2) - 3(x + 5) = 12x - 8 - 3x - 15 = 9x - 23\]

1. $9x + 7$: sign error on the second distribution — used $-3(x+5) = -3x + 15$ instead of $-3x - 15$
1. $15x - 23$: added $4x$ and $3x$ instead of subtracting ($4 \cdot 3 + 3 = 15$)
1. $15x - 7$: combined both errors above

Answer: A

Question 2

If $\dfrac{3x}{4} = 9$, what is the value of $x$?

A) $\dfrac{3}{4}$
B) $6$
C) $12$
D) $36$

Show solution

Multiply both sides by $\dfrac{4}{3}$:

\[x = 9 \times \frac{4}{3} = \frac{36}{3} = 12\]

1. $\dfrac{3}{4}$: divided $\frac{3}{4}$ by 9 instead of solving for $x$
1. $6$: divided $9$ by $\frac{3}{2}$ (used $\frac{4}{3} \div 2$) or made a fraction error
1. $36$: multiplied $9 \times 4$ without dividing by 3

Answer: C

Question 3

The first term of an arithmetic sequence is $7$, and the common difference is $-3$. What is the 10th term of the sequence?

A) $-27$
B) $-20$
C) $-17$
D) $34$

Show solution

The $n$th term of an arithmetic sequence is $a_n = a_1 + (n-1)d$.

\[a_{10} = 7 + (10 - 1)(-3) = 7 + 9(-3) = 7 - 27 = -20\]

1. $-27$: computed $(10)(-3) = -30$ and added $3$ (off by one in the $(n-1)$ factor: used $n$ instead of $n-1$, then adjusted incorrectly)
1. $-17$: used $n - 1 = 8$ instead of $9$ (subtracted 2 from $n$ instead of 1)
1. $34$: added instead of subtracted: $7 + 27 = 34$

Answer: B

Question 4

If $|2x - 5| = 11$, what is the positive value of $x$?

Enter your answer:

Show solution

An absolute value equation $|A| = k$ gives two cases:

Case 1: $2x - 5 = 11 \Rightarrow 2x = 16 \Rightarrow x = 8$

Case 2: $2x - 5 = -11 \Rightarrow 2x = -6 \Rightarrow x = -3$

The positive value is $x = 8$.

Answer: 8

Question 5

Which of the following is an equation of the line that passes through $(-3, 1)$ and $(1, 9)$?

A) $y = 2x - 5$
B) $y = 2x + 7$
C) $y = \dfrac{1}{2}x + 7$
D) $y = -2x + 5$

Show solution

Slope: $m = \dfrac{9 - 1}{1 - (-3)} = \dfrac{8}{4} = 2$.

Using point-slope with $(1, 9)$: $y - 9 = 2(x - 1) \Rightarrow y = 2x + 7$.

Verify with $(-3, 1)$: $2(-3) + 7 = -6 + 7 = 1$ ✓

1. $y = 2x - 5$: correct slope but wrong $y$-intercept — used point $(1, -3)$ instead of $(1, 9)$
1. $y = \frac{1}{2}x + 7$: inverted the slope ($\Delta x / \Delta y$ instead of $\Delta y / \Delta x$)
1. $y = -2x + 5$: negated the slope

Answer: B

Question 6

Which of the following inequalities has the same solution set as $-3x + 6 > 15$?

A) $x > -3$
B) $x < -3$
C) $x > 3$
D) $x < 3$

Show solution

\[-3x + 6 > 15 \Rightarrow -3x > 9 \Rightarrow x < -3\]

When dividing by a negative number, the inequality sign flips.

1. $x > -3$: divided by $-3$ without flipping the inequality sign
1. $x > 3$: divided by $-3$ without flipping AND used $+3$ instead of $-3$
1. $x < 3$: forgot the negative sign — solved as if dividing by $+3$

Answer: B

Question 7

A plumber charges a $\$75$ service fee plus $\$50$ per hour of work. A customer wants to spend no more than $\$275$ total. Which inequality represents the maximum number of hours, $h$, the plumber can work?

A) $50h \leq 275$
B) $75h + 50 \leq 275$
C) $50h + 75 \leq 275$
D) $75h + 50h \leq 275$

Show solution

Total cost $=$ (hourly rate $\times$ hours) $+$ service fee $= 50h + 75$.

The customer wants this to be no more than $\$275$:

\[50h + 75 \leq 275\]

1. $50h \leq 275$: omits the $\$75$ service fee
1. $75h + 50 \leq 275$: swaps the rate and the flat fee
1. $75h + 50h \leq 275$: treats both the fee and rate as per-hour charges

Answer: C

Question 8

In the $xy$-plane, the line $\ell$ passes through the points $(0, -4)$ and $(5, 0)$. What is the slope of $\ell$?

Enter your answer:

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\[m = \frac{0 - (-4)}{5 - 0} = \frac{4}{5} = 0.8\]

Answer: 0.8 (also accepted as 4/5)

Question 9

The equation $ax + b = cx + d$ has no solution. Which of the following must be true?

A) $a = c$ and $b = d$
B) $a = c$ and $b \neq d$
C) $a \neq c$ and $b = d$
D) $a \neq c$ and $b \neq d$

Show solution

Rearrange: $(a - c)x = d - b$.

If $a \neq c$: there is exactly one solution $x = \dfrac{d-b}{a-c}$.
If $a = c$ and $b = d$: the equation becomes $0 = 0$ → infinitely many solutions.
If $a = c$ and $b \neq d$: the equation becomes $0 = d - b$ (a nonzero constant) → no solution.

For no solution: $a = c$ and $b \neq d$.

1. $a = c$ and $b = d$: this gives infinitely many solutions, not no solution
1. $a \neq c$: this always produces exactly one solution
1. $a \neq c$ and $b \neq d$: $a \neq c$ is still enough to guarantee a unique solution regardless of $b$ and $d$

Answer: B

Question 10

Line $p$ has equation $y = 4x - 1$. Line $q$ is parallel to line $p$ and passes through the point $(2, 3)$. Which of the following is an equation of line $q$?

A) $y = \dfrac{1}{4}x + \dfrac{5}{2}$
B) $y = -\dfrac{1}{4}x + \dfrac{7}{2}$
C) $y = 4x - 5$
D) $y = 4x - 1$

Show solution

Parallel lines have equal slopes. Line $q$ has slope $4$.

Using point-slope with $(2, 3)$:

\[y - 3 = 4(x - 2) \Rightarrow y = 4x - 8 + 3 \Rightarrow y = 4x - 5\]

1. Used the perpendicular slope $\frac{1}{4}$ instead of $4$
1. Used the negative perpendicular slope $-\frac{1}{4}$
1. $y = 4x - 1$ is the original line $p$, not a parallel line through $(2, 3)$; verify: $4(2) - 1 = 7 \neq 3$

Answer: C

Question 11

A tank contains 200 gallons of water and is being drained at a constant rate of 15 gallons per minute. A second tank contains 80 gallons and is being filled at a constant rate of 5 gallons per minute. After how many minutes will both tanks contain the same amount of water?

A) $4$
B) $6$
C) $8$
D) $12$

Show solution

Tank 1 after $t$ minutes: $200 - 15t$

Tank 2 after $t$ minutes: $80 + 5t$

Set equal:

\[200 - 15t = 80 + 5t\] \[120 = 20t\] \[t = 6\]

Check: Tank 1 $= 200 - 90 = 110$. Tank 2 $= 80 + 30 = 110$ ✓

1. $4$: set up $200 - 15t = 80$ (ignored the filling of tank 2)
1. $8$: arithmetic error — solved $120 = 15t$ instead of $120 = 20t$
1. $12$: added the rates instead of subtracting ($200 - 80 = 120$, divided by 10 instead of 20)

Answer: B

Question 12

The graph below shows a line in the $xy$-plane.

Which of the following is an equation of the line shown?

A) $y = -\dfrac{1}{2}x + 5$
B) $y = -2x + 5$
C) $y = \dfrac{1}{2}x + 5$
D) $y = -\dfrac{1}{2}x + 3$

Show solution

From the graph, the $y$-intercept is $(0, 5)$, so $b = 5$.

Slope using the two labeled points $(0, 5)$ and $(4, 3)$:

\[m = \frac{3 - 5}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}\]

Equation: $y = -\dfrac{1}{2}x + 5$.

1. $y = -2x + 5$: inverted the slope ($\Delta y / \Delta x = -2/1$ instead of $-2/4$)
1. $y = \frac{1}{2}x + 5$: correct magnitude but wrong sign on slope; line rises instead of falls
1. $y = -\frac{1}{2}x + 3$: correct slope but used the $y$-coordinate of $(4,3)$ as the intercept

Answer: A

Question 13

The formula for the perimeter of a rectangle is $P = 2l + 2w$. Solving for $l$, what is $l$ in terms of $P$ and $w$?

Enter your answer as a fraction or expression. If $P = 40$ and $w = 7$, what is the value of $l$?

Enter your answer:

Show solution

Solve the literal equation for $l$:

\[P = 2l + 2w \Rightarrow P - 2w = 2l \Rightarrow l = \frac{P - 2w}{2}\]

Substitute $P = 40$, $w = 7$:

\[l = \frac{40 - 14}{2} = \frac{26}{2} = 13\]

Answer: 13

Question 14

A store sells almonds for $\$6$ per pound and cashews for $\$10$ per pound. A customer buys a total of 8 pounds of nuts for $\$60$. How many pounds of almonds did the customer buy?

A) $3$
B) $4$
C) $5$
D) $6$

Show solution

Let $a$ = pounds of almonds, $c$ = pounds of cashews.

\[a + c = 8\] \[6a + 10c = 60\]

From the first equation: $c = 8 - a$. Substitute:

\[6a + 10(8 - a) = 60\] \[6a + 80 - 10a = 60\] \[-4a = -20 \Rightarrow a = 5\]

Check: $5$ lb almonds + $3$ lb cashews $= 8$ lb ✓. Cost: $30 + 30 = \$60$ ✓

1. $3$: solved for cashews and reported that value
1. $4$: arithmetic error in solving
1. $6$: set up $6a = 60$ (ignored cashews entirely)

Answer: C

Question 15

Which of the following represents the solution set of $-2 \leq 3x + 1 < 10$?

A) $-1 \leq x < 3$
B) $-3 \leq x < 3$
C) $-1 < x \leq 3$
D) $\dfrac{1}{3} \leq x < \dfrac{11}{3}$

Show solution

Subtract 1 from all three parts:

\[-2 - 1 \leq 3x + 1 - 1 < 10 - 1\] \[-3 \leq 3x < 9\]

Divide all parts by 3:

\[-1 \leq x < 3\]

1. $-3 \leq x < 3$: divided the left bound by 3 from $-3$ but forgot to first subtract 1 (used $-2/3$ region incorrectly), or subtracted only on the right
1. $-1 < x \leq 3$: reversed the strict/non-strict inequality symbols
1. $\frac{1}{3} \leq x < \frac{11}{3}$: added 1 instead of subtracting 1 in the first step

Answer: A

Question 16

In the $xy$-plane, the graph of the line $y = mx + b$ passes through the first and third quadrants only. Which of the following must be true?

A) $m > 0$ and $b > 0$
B) $m > 0$ and $b < 0$
C) $m > 0$ and $b = 0$
D) $m < 0$ and $b = 0$

Show solution

A line through the origin ($b = 0$) passes through Q1 and Q3 if its slope is positive ($m > 0$), since it rises from the lower-left to the upper-right through the origin.

If $b > 0$: the line crosses the $y$-axis above the origin, passing through Q1, Q2, and part of Q3 or Q4 — it enters Q2 (negative $x$, positive $y$), so it does NOT pass through Q1 and Q3 only.
If $b < 0$: the $y$-intercept is below the origin, so the line also passes through Q4.
If $b = 0$ and $m > 0$: the line passes through exactly Q1 and Q3. ✓
1. $m > 0$, $b > 0$: line passes through Q1, Q2, and Q3
1. $m > 0$, $b < 0$: line passes through Q1, Q3, and Q4
1. $m < 0$, $b = 0$: line passes through Q2 and Q4, not Q1 and Q3

Answer: C

Question 17

The system of equations below has solution $(x, y)$. What is the value of $2x - y$?

\[2x + y = 11\] \[x - y = 1\]

Enter your answer:

Show solution

Add the equations to eliminate $y$:

\[3x = 12 \Rightarrow x = 4\]

Substitute into the second equation:

\[4 - y = 1 \Rightarrow y = 3\]

\[2x - y = 2(4) - 3 = 8 - 3 = 5\]

Check in equation 1: $2(4) + 3 = 11$ ✓

Answer: 5

Question 18

For what value of $c$ does the system below have infinitely many solutions?

\[4x - 2y = 10\] \[-6x + 3y = c\]

A) $-20$
B) $-15$
C) $15$
D) $20$

Show solution

For infinitely many solutions, the two equations must be equivalent (same line).

Multiply the first equation by $-\dfrac{3}{2}$:

\[4x - 2y = 10 \xrightarrow{\times (-3/2)} -6x + 3y = -15\]

So the second equation must have $c = -15$.

Alternatively: the ratio of $x$-coefficients is $\dfrac{-6}{4} = -\dfrac{3}{2}$, and of $y$-coefficients is $\dfrac{3}{-2} = -\dfrac{3}{2}$. For the lines to be identical, the constants must share the same ratio: $c = 10 \times \left(-\dfrac{3}{2}\right) = -15$.

1. $-20$: used ratio $-2$ instead of $-\frac{3}{2}$
1. $15$: correct magnitude but wrong sign (forgot the negative)
1. $20$: multiplied by $+2$ instead of $-\frac{3}{2}$

Answer: B

Question 19

A line in the $xy$-plane is perpendicular to the line $3x + 5y = 15$ and passes through the origin. Which of the following is an equation of the line?

A) $y = -\dfrac{3}{5}x$
B) $y = \dfrac{3}{5}x$
C) $y = \dfrac{5}{3}x$
D) $y = -\dfrac{5}{3}x$

Show solution

Rewrite the given line in slope-intercept form:

\[3x + 5y = 15 \Rightarrow y = -\frac{3}{5}x + 3\]

Slope of the given line: $m = -\dfrac{3}{5}$.

Perpendicular slope: $m_\perp = \dfrac{5}{3}$ (negative reciprocal).

Through the origin ($b = 0$): $y = \dfrac{5}{3}x$.

1. $y = -\frac{3}{5}x$: used the original slope (parallel, not perpendicular)
1. $y = \frac{3}{5}x$: negated the original slope but didn’t take the reciprocal
1. $y = -\frac{5}{3}x$: correct reciprocal but wrong sign

Answer: C

Question 20

A company’s monthly profit $P$, in dollars, is modeled by $P = 12n - 3{,}000$, where $n$ is the number of units sold. If the company increases its monthly sales by 50 units, by how many dollars does monthly profit increase?

Enter your answer:

Show solution

In the model $P = 12n - 3{,}000$, the slope is $12$: each additional unit sold increases profit by $\$12$.

For 50 additional units:

\[\Delta P = 12 \times 50 = 600\]

The $-3{,}000$ constant (fixed costs) does not change, so it has no effect on the increase in profit.

Answer: 600