Algebra Domain Test

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Question 1

What is the value of \(x\) if \(\dfrac{x}{3} - 4 = 2\)?




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Add 4 to both sides:

\[\frac{x}{3} = 6\]

Multiply both sides by 3:

\[x = 18\]

    1. \(2\): solved \(\frac{x}{3} = 2\) (subtracted instead of added 4 to the right side): \(x = 6\)… or divided both sides of \(x = 6\) by \(3\)
    1. \(6\): solved correctly for \(\frac{x}{3} = 6\) but stopped without multiplying by 3
    1. \(12\): computed \(x = 3 \times 2 = 6\)… or added 4 to 2 to get 6, then multiplied by 2 instead of 3

Answer: D

Question 2

Five less than twice a number is 19. What is the number?




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Translate: “five less than twice a number” \(= 2n - 5\).

\[2n - 5 = 19 \implies 2n = 24 \implies n = 12\]

    1. \(7\): set up \(5 - 2n = 19\): \(-2n = 14\), \(n = -7\)… or \(2n + 5 = 19\): \(n = 7\)
    1. \(14\): divided \(19 + 5 = 24\) but then divided by \(\sqrt{24}\)… or solved \(2n = 19 + 5 = 24\) but reported \(n = 24/1.7 \approx 14\); most likely \(n - 5 = 19 \div 2 - 3\)… or simply \(19 - 5 = 14\) (subtracted instead of adding, then didn’t divide)
    1. \(22\): added instead of subtracted: set up \(2n + 5 = 19\)… wait, that gives \(7\); or computed \(n = 19 + 5 = 24\)… then subtracted 2 to get 22; or used \(n + 5 = 19 \times 2 = 38 - 16 = 22\)

Answer: G

Question 3

What is the slope of the line passing through \((-1, 4)\) and \((3, -8)\)?




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\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-8 - 4}{3 - (-1)} = \frac{-12}{4} = -3\]

    1. \(-4\): computed \(\frac{-12}{3} = -4\) (used only \(x_2 = 3\) in the denominator, ignoring \(x_1 = -1\))
    1. \(3\): dropped the negative sign from the correct computation
    1. \(4\): flipped the sign and inverted: \(\frac{4}{-12} \to -\frac{1}{3}\)… or computed \(\frac{4 - (-8)}{-1 - 3} = \frac{12}{-4} = -3\) but reported \(4\)

Answer: B

Question 4

Which of the following is the \(y\)-intercept of the line \(5x - 2y = 10\)?




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Set \(x = 0\) and solve for \(y\):

\[5(0) - 2y = 10 \implies -2y = 10 \implies y = -5\]

The \(y\)-intercept is \(-5\).

Alternatively, solve for \(y\): \(y = \frac{5}{2}x - 5\), confirming \(b = -5\).

    1. \(2\): read the coefficient of \(y\) from the standard form (the \(-2\)) and reported \(2\)
    1. \(5\): read the coefficient of \(x\) directly as the \(y\)-intercept
    1. \(10\): reported the constant on the right-hand side without solving for \(y\)

Answer: F

Question 5

Which of the following is equivalent to \(6x^3 - 9x^2 + 3x\)?




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The GCF of \(6x^3\), \(9x^2\), and \(3x\) is \(3x\):

\[6x^3 - 9x^2 + 3x = 3x(2x^2 - 3x + 1)\]

Verify: \(3x \cdot 2x^2 = 6x^3\) ✓, \(3x \cdot (-3x) = -9x^2\) ✓, \(3x \cdot 1 = 3x\) ✓

    1. \(3x(2x^2 - 3x - 1)\): correct leading terms but wrong sign on the constant: \(3x(-1) = -3x \neq +3x\)
    1. \(3(2x^3 - 3x^2 + x)\): factored out only \(3\) (not \(3x\)); this is equivalent but not fully factored since \(x\) remains as a factor inside
    1. \(x(6x^2 - 9x)\): factored out \(x\) but dropped the \(+3x\) term entirely: \(x \cdot 6x^2 - x \cdot 9x = 6x^3 - 9x^2 \neq 6x^3 - 9x^2 + 3x\)

Answer: A

Question 6

What is the solution set of \(4 - 3x > 13\)?




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Subtract 4 from both sides:

\[-3x > 9\]

Divide by \(-3\) and flip the inequality:

\[x < -3\]

    1. \(x > -3\): correctly found \(-3\) but forgot to flip the inequality when dividing by a negative
    1. \(x > 3\): computed \(-3x > 9\) as \(x > -3\) and then dropped the negative: \(x > 3\)
    1. \(x < 3\): flipped the inequality correctly but used \(+3\) instead of \(-3\): divided \(-3x > 9\) as if it were \(3x > -9\)

Answer: G

Question 7

Using substitution, what is the value of \(y\) in the solution to the system below?

\[\begin{cases} y = 3x - 1 \\ 2x + y = 9 \end{cases}\]




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Substitute \(y = 3x - 1\) into \(2x + y = 9\):

\[2x + (3x - 1) = 9 \implies 5x - 1 = 9 \implies 5x = 10 \implies x = 2\]

Back-substitute: \(y = 3(2) - 1 = 5\).

    1. \(2\): reported \(x = 2\) instead of \(y = 5\)
    1. \(4\): substituted incorrectly — e.g. \(2x + 3x - 1 = 9\), \(4x = 8\), \(x = 2\)… then reported \(x - 1\)… or solved \(5x = 9\), \(x = 1.8\), \(y \approx 4.4 \approx 4\)
    1. \(8\): solved for \(y\) from the second equation without substituting: \(y = 9 - 2x = 9 - 2(0) = 9\)… or set \(y = 9 - 2(1) = 7\)… or a different substitution error giving 8

Answer: C

Question 8

Line \(\ell\) has slope \(\dfrac{2}{3}\). What is the slope of a line perpendicular to \(\ell\)?




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The slope of a perpendicular line is the negative reciprocal of the original slope:

\[m_\perp = -\frac{1}{m} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}\]

Verify: \(\dfrac{2}{3} \times \left(-\dfrac{3}{2}\right) = -1\) ✓ (perpendicular slopes multiply to \(-1\))

    1. \(-\dfrac{2}{3}\): negated the slope but did not take the reciprocal
    1. \(\dfrac{3}{2}\): took the reciprocal but forgot the negative sign
    1. \(\dfrac{2}{3}\): this is the slope of a line parallel to \(\ell\), not perpendicular

Answer: F

Question 9

Which of the following are the solutions to \(x^2 - 2x - 15 = 0\)?




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Factor: find two numbers that multiply to \(-15\) and add to \(-2\): those are \(-5\) and \(+3\).

\[x^2 - 2x - 15 = (x - 5)(x + 3) = 0\]

\[x = 5 \quad \text{or} \quad x = -3\]

Verify: \((5)^2 - 2(5) - 15 = 25 - 10 - 15 = 0\) ✓ and \((-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0\) ✓

    1. \(x = 3\) and \(x = -5\): found factors that multiply to \(-15\) but add to \(-2\) incorrectly — swapped the signs: \((x-3)(x+5)\) expands to \(x^2 + 2x - 15\), not \(x^2 - 2x - 15\)
    1. \(x = 3\) and \(x = 5\): both factors positive, product \(= +15\) not \(-15\): \((x-3)(x-5) = x^2 - 8x + 15\)
    1. \(x = -3\) and \(x = -5\): both factors negative, product \(= +15\) not \(-15\): \((x+3)(x+5) = x^2 + 8x + 15\)

Answer: B

Question 10

What is the result of \((4x^3 - 7x + 2) - (3x^3 + 2x^2 - 5x - 4)\)?




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Distribute the negative sign across the second polynomial:

\[(4x^3 - 7x + 2) - (3x^3 + 2x^2 - 5x - 4)\] \[= 4x^3 - 7x + 2 - 3x^3 - 2x^2 + 5x + 4\]

Combine like terms:

\[= (4-3)x^3 + (-2)x^2 + (-7+5)x + (2+4) = x^3 - 2x^2 - 2x + 6\]

    1. \(x^3 + 2x^2 - 12x + 6\): sign error on \(x^2\): used \(+2x^2\) instead of \(-2x^2\); and \(-7 - 5 = -12\) (subtracted instead of adding \(5x\))
    1. \(x^3 - 2x^2 - 2x - 2\): correct on \(x^3\), \(x^2\), \(x\) terms but sign error on constant: \(2 - 4 = -2\) (subtracted instead of adding the \(-(-4)\))
    1. \(7x^3 - 2x^2 - 12x - 2\): added instead of subtracting \(x^3\) terms: \(4 + 3 = 7\); and subtracted \(5x\): \(-7 - 5 = -12\)

Answer: F

Question 11

The formula for the volume of a cone is \(V = \dfrac{1}{3}\pi r^2 h\). Solving for \(h\) in terms of the other variables gives:




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Multiply both sides by 3:

\[3V = \pi r^2 h\]

Divide both sides by \(\pi r^2\):

\[h = \frac{3V}{\pi r^2}\]

    1. \(h = 3V - \pi r^2\): subtracted \(\pi r^2\) instead of dividing — treated the formula as addition rather than multiplication
    1. \(h = \dfrac{V}{3\pi r^2}\): divided \(V\) by the entire \(3\pi r^2\) without first multiplying by 3: treated the \(\frac{1}{3}\) as if it belonged in the denominator with \(\pi r^2\)
    1. \(h = \dfrac{\pi r^2}{3V}\): inverted the result — took the reciprocal of the correct answer

Answer: B

Question 12

For what value of \(c\) does the system \(\begin{cases} 4x - 2y = 8 \\ 6x - 3y = c \end{cases}\) have no solution?




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The second equation is \(\frac{3}{2}\) times the first (multiply \(4x - 2y = 8\) by \(\frac{3}{2}\): \(6x - 3y = 12\)).

  • If \(c = 12\): both equations represent the same line → infinitely many solutions.
  • If \(c \neq 12\): the lines are parallel (same slope, different \(y\)-intercepts) → no solution.

So the system has no solution for all \(c \neq 12\).

    1. \(c = 12\): this value gives infinitely many solutions (the same line), not no solution
    1. \(c = 0\): no special significance; the lines are still parallel for any \(c \neq 12\), including \(c = 0\)
    1. \(c = 4\): no special significance geometrically

Answer: G

Question 13

What are all values of \(x\) that satisfy \(\dfrac{x}{x - 2} = \dfrac{3}{x - 2} + 1\)?




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Multiply both sides by \((x - 2)\), noting \(x \neq 2\):

\[x = 3 + (x - 2) = x + 1\]

This gives \(x = x + 1\), or \(0 = 1\) — a contradiction.

There is no solution: the equation has no value of \(x\) that satisfies it.

    1. \(x = 5\): after multiplying through, gets \(x = x + 1\) and mistakenly plugs in a value; \(x=5\) doesn’t satisfy the original: \(\frac{5}{3} = \frac{3}{3} + 1 = 2 \neq \frac{5}{3}\)
    1. \(x = 5\) and \(x = 2\): includes \(x = 2\) which is excluded from the domain (denominator \(= 0\))
    1. All real numbers except \(x = 2\): the equation \(0 = 1\) is never true, so no \(x\) works

Answer: C

Question 14

Which of the following represents the solution to \(-2 \leq 3x + 1 < 10\)?




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Subtract 1 from all three parts:

\[-3 \leq 3x < 9\]

Divide all three parts by 3:

\[-1 \leq x < 3\]

    1. \(-3 \leq x < 9\): subtracted 1 correctly but forgot to divide by 3
    1. \(-1 < x \leq 3\): correct values but swapped the inequality types (the \(\leq\) should be on the left, \(<\) on the right)
    1. \(1 \leq x < 11\): added 1 instead of subtracting: \(-2 + 1 \leq 3x < 10 + 1\), giving \(-1 \leq 3x < 11\), then didn’t divide correctly

Answer: F

Question 15

What are the solutions to \(2x^2 - 6x + 3 = 0\)?




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With \(a = 2\), \(b = -6\), \(c = 3\), apply the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4}\]

This simplifies to \(x = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}\).

    1. \(\dfrac{6 \pm \sqrt{36}}{4}\): computed the discriminant as \(b^2 = 36\) only, omitting \(-4ac\): \(36 - 0 = 36\)
    1. \(\dfrac{-6 \pm \sqrt{12}}{4}\): used \(-b = -6\) with the wrong sign — since \(b = -6\), \(-b = +6\), not \(-6\)
    1. \(\dfrac{6 \pm \sqrt{60}}{4}\): computed \(4ac = 4(2)(3) = 24\)… but then added instead of subtracted: \(b^2 + 4ac = 36 + 24 = 60\)

Answer: A

Question 16

What is the vertex of the parabola \(y = x^2 - 8x + 3\), found by completing the square?




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Complete the square:

\[y = x^2 - 8x + 3 = (x^2 - 8x + 16) - 16 + 3 = (x - 4)^2 - 13\]

Vertex form is \((x - h)^2 + k\), so the vertex is \((4, -13)\).

    1. \((4, 3)\): correct \(x\)-coordinate but used the original constant \(3\) instead of \(3 - 16 = -13\)
    1. \((-4, -13)\): correct \(y\)-coordinate but wrong sign on \(h\): \(h = 4\), not \(-4\)
    1. \((8, 3)\): used the coefficient of \(x\) (\(-8\)) directly for the vertex \(x\)-value without halving

Answer: F

Question 17

What is the quotient when \(2x^3 + 3x^2 - 11x - 6\) is divided by \((x + 3)\)?




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Use synthetic division with root \(x = -3\):

\[\begin{array}{r|rrrr} -3 & 2 & 3 & -11 & -6 \\ & & -6 & 9 & 6 \\ \hline & 2 & -3 & -2 & 0 \end{array}\]

The quotient is \(2x^2 - 3x - 2\) with remainder \(0\).

Verify: \((x + 3)(2x^2 - 3x - 2) = 2x^3 - 3x^2 - 2x + 6x^2 - 9x - 6 = 2x^3 + 3x^2 - 11x - 6\) ✓

    1. \(2x^2 + 3x - 2\): sign error on the \(x\) coefficient — brought down \(+3\) instead of computing \(-6 + 3 = -3\)
    1. \(2x^2 - 3x + 2\): sign error on the constant term: \(9 + (-6) = 3\)… or computed \(9 - 6 = 3\) then reported \(+2\)
    1. \(2x^2 + 9x + 16\): added instead of multiplied in synthetic division steps

Answer: A

Question 18

What are all values of \(x\) that satisfy \(|2x - 5| = 9\)?




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An absolute value equation \(|A| = k\) (with \(k > 0\)) has two cases:

Case 1: \(2x - 5 = 9 \implies 2x = 14 \implies x = 7\)

Case 2: \(2x - 5 = -9 \implies 2x = -4 \implies x = -2\)

Verify: \(|2(7)-5| = |9| = 9\) ✓ and \(|2(-2)-5| = |-9| = 9\) ✓

    1. \(x = 7\) only: solved only the positive case and missed the negative case
    1. \(x = -2\) only: solved only the negative case and missed the positive case
    1. \(x = 7\) and \(x = 2\): found the positive case correctly but made an arithmetic error in the negative case: \(2x = -4\) but reported \(x = 2\) instead of \(x = -2\)

Answer: H

Question 19

Two friends, Alicia and Ben, together have 84 trading cards. Alicia has 12 more than twice the number Ben has. How many cards does Ben have?




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Let \(b\) = Ben’s cards. Then Alicia has \(2b + 12\).

\[b + (2b + 12) = 84 \implies 3b + 12 = 84 \implies 3b = 72 \implies b = 24\]

Verify: Ben \(= 24\), Alicia \(= 2(24) + 12 = 60\); total \(= 84\) ✓

    1. \(18\): set up \(3b = 84 - 12 = 72\) correctly but computed \(72/4 = 18\) (divided by 4 instead of 3)
    1. \(28\): arithmetic error — solved \(3b = 84\): \(b = 28\) (forgot to subtract 12)
    1. \(32\): set up \(b + (b + 12) = 84\) (dropped the factor of 2 on Ben’s count): \(2b = 72\), \(b = 36\)… or \(3b = 96\), \(b = 32\)

Answer: B

Question 20

A system of three equations is given below:

\[\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}\]

What is the value of \(x\)?




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Step 1: Add equations (1) and (2) to eliminate \(y\):

\[(x+y+z)+(2x-y+z) = 6+3 \implies 3x+2z = 9 \quad\cdots(4)\]

Step 2: Add equations (2) and (3) to eliminate \(z\):

\[(2x-y+z)+(x+2y-z) = 3+2 \implies 3x+y = 5 \quad\cdots(5)\]

Step 3: From equation (5): \(y = 5 - 3x\). Add equations (1) and (3) to eliminate \(z\):

\[(x+y+z)+(x+2y-z) = 6+2 \implies 2x+3y = 8 \quad\cdots(6)\]

Substitute \(y = 5 - 3x\) into (6):

\[2x + 3(5-3x) = 8 \implies 2x + 15 - 9x = 8 \implies -7x = -7 \implies x = 1\]

Verify: \(x=1 \Rightarrow y=2\); from (1): \(z = 6-1-2=3\). Check (2): \(2-2+3=3\) ✓. Check (3): \(1+4-3=2\) ✓.

    1. \(2\): arithmetic error in the elimination step — solved \(-7x = -14\), giving \(x = 2\)
    1. \(3\): read the value of \(z\) (\(z = 3\)) instead of \(x\)
    1. \(4\): added the right-hand sides of two equations without eliminating a variable correctly

Answer: F