SAT Algebra Domain Test

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Question 1

What is the value of $4 + 2(3 - 1)^2$?

A) 8
B) 9
C) 12
D) 24

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Evaluate inside the parentheses first: $3 - 1 = 2$. Apply the exponent: $2^2 = 4$. Multiply: $2 \times 4 = 8$. Add: $4 + 8 = 12$.

1. 8: Forgot to apply the exponent, computing $4 + 2(2)$ instead of $4 + 2(2)^2$.
1. 9: Ignored the parentheses and evaluated strictly left to right, computing $4 + 2 \times 3 - 1$.
1. 24: Added $4 + 2$ before evaluating the rest, computing $(4+2)(3-1)^2$.

Answer: C

Question 2

If $|2x + 3| = 11$, what is the least possible value of $x$?

A) -7
B) -4
C) 4
D) 7

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$|2x+3|=11$ means $2x+3=11$ or $2x+3=-11$, giving $x=4$ or $x=-7$. The least value is $-7$.

1. -4: Sign error, solving $2x - 3 = -11$ instead of $2x + 3 = -11$, giving $2x=-8$.
1. 4: Reported the greater solution instead of the least one.
1. 7: Sign error treating the equation as $2x - 3 = 11$ instead of $2x + 3 = 11$, giving $2x=14$.

Answer: A

Question 3

What is the value of $(3^2)(3^4)$?

Enter your answer:

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Using the product rule for exponents: $(3^2)(3^4) = 3^{2+4} = 3^6 = 729$.

Answer: 729

Question 4

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

A) $8x - 12$
B) $8x - 7$
C) $8x + 2$
D) $8x + 17$

Show solution

Distribute: $4(2x-3) = 8x - 12$. Combine like terms: $8x - 12 + 5 = 8x - 7$.

1. $8x - 12$: Distributed correctly but forgot to add the 5 at the end.
1. $8x + 2$: Forgot to distribute the 4 into $-3$, keeping it as $-3$: $4(2x) - 3 + 5 = 8x + 2$.
1. $8x + 17$: Sign error distributing, treating $4(2x-3)$ as $8x+12$ instead of $8x-12$, then adding 5.

Answer: B

Question 5

If $5x - 8 = 27$, what is the value of $x$?

A) 3.8
B) 5.4
C) 7
D) 35

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Add 8 to both sides: $5x = 35$. Divide by 5: $x = 7$.

1. 3.8: Sign error, solving $5x + 8 = 27$ instead of $5x - 8 = 27$, giving $5x=19$.
1. 5.4: Forgot the $-8$ entirely, dividing $27$ by $5$ directly.
1. 35: Correctly found $5x=35$ but forgot to divide by 5, reporting the value of $5x$ instead of $x$.

Answer: C

Question 6

Which of the following represents all values of $x$ satisfying $3x + 6 \leq 21$?

A) $x \geq 7$
B) $x \geq 5$
C) $x \leq 7$
D) $x \leq 5$

Show solution

Subtract 6 from both sides: $3x \leq 15$. Divide by 3: $x \leq 5$.

1. $x \geq 7$: Combined both errors: forgot to subtract 6 and flipped the inequality direction.
1. $x \geq 5$: Solved correctly for the boundary value but flipped the inequality direction without justification.
1. $x \leq 7$: Forgot to subtract 6, dividing 21 by 3 directly instead of 15 by 3.

Answer: D

Question 7

What is the slope of the line that passes through the points $(2, 5)$ and $(6, 13)$?

A) -2
B) 0.5
C) 2
D) 2.25

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Slope $= \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2$.

1. -2: Subtracted the coordinates in inconsistent order for $x$ and $y$, computing $\dfrac{13-5}{2-6}$.
1. 0.5: Inverted the slope ratio, computing $\dfrac{6-2}{13-5}$ instead of $\dfrac{13-5}{6-2}$.
1. 2.25: Added the coordinates instead of subtracting, computing $\dfrac{13+5}{6+2}$.

Answer: C

Question 8

The first term of an arithmetic sequence is 7, and each term after that increases by 4. What is the 5th term of the sequence?

Enter your answer:

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The $n$th term of an arithmetic sequence is $a_n = a_1 + (n-1)d$. Here $a_1=7$, $d=4$, $n=5$: $a_5 = 7 + (4)(4) = 7 + 16 = 23$.

Answer: 23

Question 9

If $\dfrac{2}{3}(x - 5) + 4 = 10$, what is the value of $x$?

A) 4
B) 14
C) 18.5
D) 20

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Subtract 4 from both sides: $\dfrac{2}{3}(x-5) = 6$. Multiply both sides by $\dfrac{3}{2}$: $x - 5 = 9$. Add 5: $x = 14$.

1. 4: Sign error on the 5, treating the equation as $\dfrac{2}{3}(x+5)+4=10$ instead of $\dfrac{2}{3}(x-5)+4=10$, giving $x+5=9$.
1. 18.5: Multiplied by $\dfrac{2}{3}$ again instead of the reciprocal $\dfrac{3}{2}$, computing $\dfrac{4}{9}(x-5)=6$.
1. 20: Forgot to subtract 4 first, solving $\dfrac{2}{3}(x-5)=10$ directly instead of $\dfrac{2}{3}(x-5)=6$.

Answer: B

Question 10

If $5x - 3 = 2x + 18$, what is the value of $x$?

A) -5
B) 2.14
C) 7
D) 21

Show solution

Subtract $2x$ from both sides: $3x - 3 = 18$. Add 3: $3x = 21$. Divide by 3: $x = 7$.

1. -5: Sign error on the constant, solving $5x-3=2x-18$ instead of $5x-3=2x+18$, giving $3x=-15$.
1. 2.14: Combined terms incorrectly, adding $2x$ instead of subtracting it and computing $7x-3=18$.
1. 21: Correctly reduced the equation to $3x=21$ but forgot to divide by 3, reporting the value of $3x$ instead of $x$.

Answer: C

Question 11

The volume of a rectangular prism is given by $V = lwh$, where $l$ is length, $w$ is width, and $h$ is height. Which of the following expresses $h$ in terms of $V$, $l$, and $w$?

A) $h = \dfrac{lw}{V}$
B) $h = V - lw$
C) $h = Vlw$
D) $h = \dfrac{V}{lw}$

Show solution

Divide both sides of $V = lwh$ by $lw$: $h = \dfrac{V}{lw}$.

1. $h = \dfrac{lw}{V}$: Inverted the correct expression, dividing $lw$ by $V$ instead of $V$ by $lw$.
1. $h = V - lw$: Treated the equation as additive rather than multiplicative, subtracting instead of dividing.
1. $h = Vlw$: Multiplied all three quantities together instead of isolating $h$ by division.

Answer: D

Question 12

A line has a slope of 3 and a $y$-intercept of $-7$. What is the value of $y$ when $x = 4$?

Enter your answer:

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In slope-intercept form, $y = mx + b = 3x - 7$. Substitute $x=4$: $y = 3(4) - 7 = 12 - 7 = 5$.

Answer: 5

Question 13

Which equation, in point-slope form, represents the line that passes through $(3, -2)$ with a slope of 4?

A) $y + 2 = 4(x - 3)$
B) $y - 2 = 4(x - 3)$
C) $y + 3 = 4(x - 2)$
D) $y + 2 = 4(x + 3)$

Show solution

Point-slope form is $y - y_1 = m(x - x_1)$. Using $(x_1,y_1)=(3,-2)$ and $m=4$: $y - (-2) = 4(x-3)$, which is $y + 2 = 4(x-3)$.

1. $y - 2 = 4(x - 3)$: Sign error on the $y$-coordinate, using $y-2$ instead of $y+2$.
1. $y + 3 = 4(x - 2)$: Swapped the roles of the $x$- and $y$-coordinates from the given point.
1. $y + 2 = 4(x + 3)$: Sign error on the $x$-coordinate, using $x+3$ instead of $x-3$.

Answer: A

Question 14

Which of the following equations represents a line that is perpendicular to $y = \dfrac{1}{3}x + 2$ and passes through the point $(2, 5)$?

A) $y = \dfrac{1}{3}x + \dfrac{13}{3}$
B) $y = -3x + 11$
C) $y = 3x - 1$
D) $y = -3x - 1$

Show solution

Perpendicular lines have slopes that are negative reciprocals: $m = -3$. Using point $(2,5)$: $5 = -3(2) + b \Rightarrow 5 = -6 + b \Rightarrow b = 11$. Equation: $y = -3x + 11$.

1. $y = \dfrac{1}{3}x + \dfrac{13}{3}$: Used the same slope as the given line (parallel) instead of the negative reciprocal (perpendicular).
1. $y = 3x - 1$: Used the reciprocal of the slope but forgot to negate it.
1. $y = -3x - 1$: Used the correct perpendicular slope but made a sign error solving for the intercept: $5 - (-3)(2)$ computed as $5-6$ instead of $5+6$.

Answer: B

Question 15

A taxi company charges a $\$3$ base fee plus $\$2.50$ per mile. If a ride costs $\$20.50$, how many miles was the ride?

A) 5.83
B) 7
C) 8.2
D) 9.4

Show solution

Let $m$ = miles. $3 + 2.50m = 20.50$, so $2.50m = 17.50$, $m = 7$.

1. 5.83: Subtracted the base fee but divided by the wrong rate (3 instead of 2.50).
1. 8.2: Forgot the base fee entirely, dividing the full cost by the per-mile rate: $20.50/2.50$.
1. 9.4: Added the base fee back on instead of subtracting it: $(20.50+3)/2.50$.

Answer: B

Question 16

If $y = 2x + 1$ and $3x + y = 16$, what is the value of $x$?

Enter your answer:

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Substitute $y=2x+1$ into $3x+y=16$: $3x+(2x+1)=16$, so $5x+1=16$, $5x=15$, $x=3$.

Answer: 3

Question 17

If $4x - 3y = 7$ and $2x + 3y = 17$, what is the value of $xy$?

A) -12
B) 7
C) 12
D) 16

Show solution

Add the two equations: $(4x-3y)+(2x+3y)=7+17$, giving $6x=24$, so $x=4$. Substitute into $2x+3y=17$: $8+3y=17$, so $y=3$. Then $xy = 4 \times 3 = 12$.

1. -12: Sign error solving for $y$, obtaining $y=-3$ instead of $3$, giving $xy=4(-3)=-12$.
1. 7: Added $x$ and $y$ instead of multiplying them, computing $4+3$.
1. 16: Reported $x^2$ instead of $xy$, computing $4^2$.

Answer: C

Question 18

A choir sells adult tickets for $20 and student tickets for $12. They sold 160 tickets total for $2,600. How many more adult tickets were sold than student tickets?

A) 10
B) 30
C) 75
D) 85

Show solution

Let $a$ = adult tickets, $s$ = student tickets. $a+s=160$ and $20a+12s=2{,}600$. Substituting $s=160-a$: $20a+12(160-a)=2{,}600$, so $8a+1{,}920=2{,}600$, $8a=680$, $a=85$. Then $s=160-85=75$. The difference is $85-75=10$.

1. 30: Arithmetic slip in the elimination step, computing $8a=760$ instead of $8a=680$, giving $a=95$, $s=65$, and a difference of 30.
1. 75: Reported the number of student tickets instead of the difference.
1. 85: Reported the number of adult tickets instead of the difference.

Answer: A

Question 19

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

\[\begin{aligned} 6x - 8y &= 20 \\ 9x - 12y &= k \end{aligned}\]

Enter your answer:

Show solution

The coefficients of the second equation are $\dfrac{3}{2}$ of the first equation’s coefficients ($9 = \frac{3}{2}(6)$ and $12 = \frac{3}{2}(8)$). For the system to represent the same line, the constant terms must follow the same ratio: $k = \dfrac{3}{2}(20) = 30$.

Answer: 30

Question 20

Which of the following represents the solution set of $-3 < 2x + 5 \leq 11$?

A) $x < -4$ or $x \geq 3$
B) $-1.5 < x \leq 5.5$
C) $-8 < x \leq 6$
D) $-4 < x \leq 3$

Show solution

Subtract 5 from all three parts: $-8 < 2x \leq 6$. Divide all three parts by 2: $-4 < x \leq 3$.

1. $x < -4$ or $x \geq 3$: Treated the compound inequality as two separate conditions joined by “or” instead of a single combined range.
1. $-1.5 < x \leq 5.5$: Forgot to subtract 5 from all three parts, dividing $-3 < 2x + 5 \leq 11$ by 2 directly.
1. $-8 < x \leq 6$: Correctly subtracted 5 from all three parts but forgot to divide by 2.

Answer: D