PSAT Math Quiz: Equations With Two Variables

20 questions · exam conditions

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Equations With Two VariablesQuestion 1 of 20

A company's cost to produce $x$ items is $C=150+7x$ . What is the value of $x$ when $C=381$ ?

x=21

x=23

x=31

x=33

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PSAT Math Quiz

PSAT Math Quiz: Equations With Two Variables

Practice Equations With Two Variables in PSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Equations With Two Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A company's cost to produce $x$ items is $C=150+7x$ . What is the value of $x$ when $C=381$ ?

$x=21$
$x=23$
$x=31$
$x=33$ (correct answer)

Explanation: This problem gives C = 150 + 7x and asks for x when C = 381. Substituting C = 381: 381 = 150 + 7x. Solving for x: 381 - 150 = 7x, which gives 231 = 7x, so x = 231 ÷ 7 = 33. Let me verify: C = 150 + 7(33) = 150 + 231 = 381 ✓. The company has fixed costs of $150 plus $7 per item produced. A common error is making arithmetic mistakes in subtraction or division. When solving linear cost equations, subtract the fixed cost first, then divide by the variable cost per unit.

Question 2

A recipe uses cups of flour $f$ and cups of sugar $s$ with the constraint $2f+s=11$ . What is the value of $s$ when $f=4$ ?

$1$
$3$ (correct answer)
$7$
$19$

Explanation: The question asks for the value of s when f=4 in the equation 2f + s = 11. This is a linear equation with two variables, constraining the amounts of flour f and sugar s. Substitute f=4: 24 + s = 11, which is 8 + s = 11. Then, subtract 8 from both sides: s = 11 - 8 = 3. A common error is to multiply instead of adding or solving for f instead. Verify by plugging back: 24 + 3 = 8+3=11, correct. When given a value for one variable, substitute directly to find the other in two-variable equations.

Question 3

A landscaper's total charge $T$ (in dollars) is related to the number of labor hours $h$ by $T = 120 + 35h$ . What is the value of $T$ when $h=6$ ?

$210$
$240$
$330$ (correct answer)
$390$

Explanation: We need to find T when h = 6 in the equation T = 120 + 35h. Substituting h = 6 gives T = 120 + 35(6) = 120 + 210 = 330 dollars. The equation represents a linear relationship where 120 is the base charge and 35 is the hourly rate. A common error is arithmetic mistakes when multiplying 35 × 6. When evaluating linear functions, carefully perform the multiplication before adding the constant term.

Question 4

For the equation $6x+9y=54$ , what is the value of $y$ when $x=3$ ?

3
4 (correct answer)
6
12

Explanation: This question tests your ability to substitute a given value into a linear equation and solve for the unknown variable. When you see a linear equation with two variables and are given the value of one variable, you simply substitute and solve. Given the equation

6x + 9y = 54

and

x = 3

, substitute 3 for

x

6(3) + 9y = 54

18 + 9y = 54

To isolate

y

, subtract 18 from both sides:

9y = 54 - 18

9y = 36

Finally, divide both sides by 9:

y = \frac{36}{9} = 4

So the answer is B) 4. Let's examine why the other choices are incorrect. Choice A) 3 might tempt you if you confused the given value of

x

with the answer for

y

, or if you made an arithmetic error during substitution. Choice C) 6 could result from incorrectly dividing 54 by 9 instead of 36 by 9, essentially forgetting to subtract the 18 first. Choice D) 12 might occur if you divided 36 by 3 instead of 36 by 9, mixing up the coefficients in your final step. When solving substitution problems like this, work methodically: substitute the given value, perform arithmetic carefully, and isolate the unknown variable step by step. Double-check your work by substituting both values back into the original equation—

6(3) + 9(4) = 18 + 36 = 54

✓.

Question 5

If $2(x+y)=x-3y$ , what is $y$ in terms of $x$ ?

$y=\dfrac{-x}{5}$ (correct answer)
$y=\dfrac{x}{5}$
$y=\dfrac{-5}{x}$
$y=-5x$

Explanation: This question tests your ability to solve linear equations for one variable in terms of another—a fundamental algebra skill that appears frequently on standardized tests. To isolate

y

in terms of

x

, start by distributing the 2 on the left side:

2x + 2y = x - 3y

. Now collect all terms containing

y

on one side and all terms with

x

on the other. Subtract

2x

from both sides:

2y = x - 3y - 2x

, which simplifies to

2y = -x - 3y

. Next, add

3y

to both sides:

2y + 3y = -x

, giving you

5y = -x

. Finally, divide both sides by 5:

y = \frac{-x}{5}

. Looking at the wrong answers: Choice B gives

y = \frac{x}{5}

, which would result from forgetting the negative sign when moving terms—a common algebraic error. Choice C,

y = \frac{-5}{x}

, incorrectly places

x

in the denominator, likely from confusing which variable should be isolated. Choice D,

y = -5x

, represents a sign error combined with incorrectly multiplying by 5 instead of dividing. You can verify the correct answer A by substituting back into the original equation: if

y = \frac{-x}{5}

, then

2(x + \frac{-x}{5}) = 2(\frac{4x}{5}) = \frac{8x}{5}

and

x - 3(\frac{-x}{5}) = x + \frac{3x}{5} = \frac{8x}{5}

. ✓ Study tip: When solving for one variable in terms of another, always substitute your answer back into the original equation to check your work—it catches algebraic mistakes quickly.

Question 6

A store sells notebooks for $n$ dollars each and charges a flat shipping fee of $s$ dollars per order. The total cost is $T=4n+s$ for an order of 4 notebooks. Which equation solves for $n$ in terms of $T$ and $s$ ?

$n=\dfrac{T+s}{4}$
$n=\dfrac{T-s}{4}$ (correct answer)
$n=4(T-s)$
$n=T-4s$

Explanation: The question asks for the equation that solves for n in terms of T and s, given T = 4n + s. This equation models the total cost T as 4 times the price per notebook n plus shipping s. To solve for n, subtract s from both sides: T - s = 4n. Then, divide both sides by 4: n = (T - s)/4. A common error is to divide by something else or forget to subtract s first. Check by substituting back; if T=20, s=4, n=4, then 20=4*4 +4, yes, and (20-4)/4=4. Focus on inverse operations to isolate the desired variable in equations with multiple variables.

Question 7

The equation $y=\frac{5}{2}x-1$ describes a relationship. If $x$ increases by 4, by how much does $y$ change?

$-1$
$5$
$10$ (correct answer)
$20$

Explanation: This problem involves y = (5/2)x - 1, where if x increases by 4, we need to find how y changes. The change in y equals the coefficient of x times the change in x: Δy = (5/2) × 4 = 5 × 2 = 10. The coefficient 5/2 represents the slope or rate of change. When x increases by 4, y increases by 10. A common error is making mistakes with fraction arithmetic or forgetting to multiply by the coefficient. In linear relationships, changes in the dependent variable equal the slope times changes in the independent variable.

Question 8

A line passes through the points $(2,1)$ and $(6,9)$ on a coordinate plane, where $x$ is the horizontal coordinate and $y$ is the vertical coordinate. What is the equation of the line in the form $y=mx+b$ ?

$y=2x-3$ (correct answer)
$y=\frac{1}{2}x+0$
$y=2x+3$
$y=\frac{1}{2}x-3$

Explanation: This problem asks for the equation of a line passing through (2,1) and (6,9). First, find the slope: m = (9-1)/(6-2) = 8/4 = 2. Using point-slope form with point (2,1): y - 1 = 2(x - 2), which gives y - 1 = 2x - 4, so y = 2x - 3. Let me verify with both points: at (2,1): y = 2(2) - 3 = 1 ✓; at (6,9): y = 2(6) - 3 = 9 ✓. A common error is making arithmetic mistakes in the slope calculation or when converting to slope-intercept form. When finding line equations, calculate slope first, then use point-slope form to find the equation.

Question 9

A charity buys pens and notebooks. Each pen costs $0.50, and each notebook costs $1.20. The total cost is modeled by $0.50p+1.20n=T$ , where $p$ is the number of pens and $n$ is the number of notebooks. If the charity spends $72 and buys 60 pens, how many notebooks did it buy?

25
35 (correct answer)
42
60

Explanation: This is a linear equation application where you're given specific values and need to solve for an unknown variable. When you see a real-world scenario with a formula already provided, substitute the known values and solve algebraically. You're told the charity spends $72 total and buys 60 pens. Substitute these values into the equation $

0.50p + 1.20n = T

0.50(60) + 1.20n = 72

First, calculate the cost of the pens:

0.50 \times 60 = 30

So the equation becomes:

30 + 1.20n = 72

Subtract 30 from both sides:

1.20n = 42

Divide by 1.20:

n = \frac{42}{1.20} = 35

The charity bought 35 notebooks. Looking at the wrong answers: Choice (A) 25 would result from incorrectly calculating

\frac{30}{1.20}

instead of

\frac{42}{1.20}

, mixing up which dollar amount to divide. Choice (C) 42 comes from forgetting to divide by the notebook price—this is just the remaining dollars after buying pens, not the number of notebooks. Choice (D) 60 assumes they bought the same number of notebooks as pens, ignoring the price difference entirely. Strategy tip: In substitution problems, work methodically through each step and always check your answer makes sense. Here, 35 notebooks at $1.20 each plus 60 pens at $0.50 each should equal $72: $$35(1.20) + 60(0.50) = 42 + 30 = 72$$ ✓

Question 10

The relationship between the number of adult tickets $a$ and student tickets $s$ sold for a show is given by $12a+8s=480$ , where 12 and 8 are the ticket prices in dollars and 480 is the total revenue. Rewrite the equation to solve for $s$ in terms of $a$ .

$s=60-\frac{3}{2}a$
$s=\frac{480-12a}{8}$ (correct answer)
$s=60-\frac{2}{3}a$
$s=\frac{12a-480}{8}$

Explanation: This problem asks to solve

12a + 8s = 480

for s in terms of a. Starting with

12a + 8s = 480

, subtract 12a from both sides:

8s = 480 - 12a

. Then divide both sides by 8:

s = \frac{480 - 12a}{8}

. This can also be written as

s = 60 - 1.5a

, but the fraction form matches choice B exactly. A common error is incorrectly rearranging the terms or making sign errors when moving terms across the equal sign. When isolating a variable, perform the same operation to both sides of the equation systematically.

Question 11

A line on the coordinate plane passes through $(0,-4)$ and $(5,6)$ . What is the y-intercept of the line?

$-4$ (correct answer)
$0$
$5$
$6$

Explanation: This problem asks for the y-intercept of a line passing through (0,-4) and (5,6). The y-intercept is the y-value when x = 0. Since one of the given points is (0,-4), the y-intercept is -4. The y-intercept can be read directly from a point where x = 0, without needing to find the full equation. A common error is confusing the y-intercept with the slope or using the wrong point. The y-intercept is simply the y-coordinate where the line crosses the y-axis (x = 0).

Question 12

A fundraiser has the constraint $x+y=75$ , where $x$ is the number of online donations and $y$ is the number of in-person donations. If $y$ decreases by 12, how must $x$ change to keep the total number of donations the same?

Increase by 12 (correct answer)
Decrease by 12
Increase by 63
Decrease by 63

Explanation: This problem involves the constraint x + y = 75, where if y decreases by 12, we need to find how x must change to maintain the total. If y decreases by 12, then to keep x + y = 75, x must increase by 12. The constraint equation requires that any decrease in one variable be offset by an equal increase in the other variable. A common error is thinking both variables change in the same direction or getting the magnitude wrong. In constraint equations where variables sum to a constant, changes in the variables must be equal and opposite.

Question 13

A gym charges a one-time sign-up fee plus a monthly fee. The total cost $C$ (in dollars) after $m$ months is modeled by $C = 25 + 18m$ . If the number of months $m$ increases by 4, by how much does the total cost $C$ change?

$18$
$43$
$72$ (correct answer)
$100$

Explanation: This question asks how the total cost C changes when the number of months m increases by 4. The cost function is C = 25 + 18m, where 25 is the one-time sign-up fee and 18 is the monthly fee. When m increases by 4, the new cost becomes C_new = 25 + 18(m + 4) = 25 + 18m + 72. The change in cost is C_new - C = (25 + 18m + 72) - (25 + 18m) = 72 dollars. A common error is to only multiply the increase in months by the coefficient without considering the constant term, but since the constant 25 appears in both expressions, it cancels out. When dealing with linear functions, the change in output equals the rate of change times the change in input.

Question 14

A line is modeled by $y= -\frac{3}{4}x+6$ . What is the value of $y$ when $x=8$ ?

$0$ (correct answer)
$2$
$6$
$12$

Explanation: This problem gives y = (-3/4)x + 6 and asks for y when x = 8. Substituting x = 8: y = (-3/4)(8) + 6 = -6 + 6 = 0. The calculation involves multiplying the fraction by 8: (-3/4) × 8 = -3 × 2 = -6, then adding 6 gives 0. A common error is making mistakes with fraction arithmetic or sign errors. When evaluating linear expressions with fractions, multiply carefully and follow the order of operations.

Question 15

A movie theater sells tickets with total revenue $R=9a+6c$ , where $a$ is the number of adult tickets and $c$ is the number of child tickets. If $c$ increases by 5 while $a$ stays the same, how does $R$ change?

Increases by $30$ (correct answer)
Increases by $45$
Decreases by $30$
Decreases by $45$

Explanation: This problem involves revenue R = 9a + 6c, where if c increases by 5 while a stays the same, we need to find how R changes. The change in R equals the coefficient of c times the change in c: ΔR = 6 × 5 = 30. Since the coefficient is positive and c increases, R increases by 30. The coefficient 6 represents the revenue per child ticket. A common error is using the wrong coefficient or forgetting that only one variable changes. When one variable in a linear expression changes while others stay constant, multiply the coefficient by the change in that variable.

Question 16

A school club has a budget constraint $5x+2y=64$ , where $x$ is the number of T-shirts and $y$ is the number of stickers purchased. If $x$ increases by 4, by how much must $y$ change to keep the total cost the same?

Increase by 10
Decrease by 10
Increase by 20
Decrease by 20 (correct answer)

Explanation: This problem involves the constraint 5x + 2y = 64, where if x increases by 4, we need to find how y must change to keep the equation satisfied. The change in the left side from increasing x by 4 is 5(4) = 20. To maintain equality, 2y must decrease by 20, so 2Δy = -20, which means Δy = -10. Therefore, y must decrease by 10. A common error is forgetting to account for the coefficients or getting the sign wrong. In constraint equations, when one term increases, another must decrease to maintain the constant sum.

Question 17

A concert venue has $r$ regular seats and $v$ VIP seats with the constraint $r+v=350$ . If the number of VIP seats is $v=120$ , what is $r$ ?

$230$ (correct answer)
$240$
$350$
$470$

Explanation: This problem gives the constraint r + v = 350 with v = 120, asking for r. Substituting v = 120: r + 120 = 350. Solving for r: r = 350 - 120 = 230. This represents a concert venue where the total number of regular and VIP seats is fixed. With 120 VIP seats, there are 230 regular seats. A common error is adding instead of subtracting or making arithmetic mistakes. In constraint equations where variables sum to a constant, one variable equals the constant minus the other.

Question 18

A baker uses the constraint $x+2y=30$ , where $x$ is the number of small boxes and $y$ is the number of large boxes packed, and each large box holds twice as many items. What is $x$ when $y=11$ ?

$x=4$
$x=8$ (correct answer)
$x=11$
$x=19$

Explanation: This problem gives x + 2y = 30 and asks for x when y = 11. Substituting y = 11: x + 2(11) = 30, which gives x + 22 = 30. Solving for x: x = 30 - 22 = 8. Let me verify: 8 + 2(11) = 8 + 22 = 30 ✓. This represents a packing constraint where large boxes hold twice as many items as small boxes. A common error is making arithmetic mistakes in substitution or subtraction. When solving constraint equations, substitute known values carefully and solve systematically.

Question 19

The equation $y=\frac{x-6}{3}$ relates $x$ and $y$ . What is the value of $x$ when $y=5$ ?

$x=9$
$x=11$
$x=15$
$x=21$ (correct answer)

Explanation: This problem gives

y = \frac{x - 6}{3}

and asks for

x

when

y = 5

. Substituting

y = 5

5 = \frac{x - 6}{3}

. Multiplying both sides by 3:

15 = x - 6

. Adding 6 to both sides:

x = 15 + 6 = 21

. Let me verify:

y = \frac{21 - 6}{3} = \frac{15}{3} = 5

✓. The equation can be solved by clearing the fraction first, then isolating the variable. A common error is making mistakes when multiplying by the denominator or combining terms. When solving equations with fractions, multiply both sides by the denominator to eliminate the fraction first.

Question 20

A water tank contains $W=500-20t$ liters after $t$ minutes of draining. What is the value of $W$ when $t=12$ ?

$140$
$240$
$260$ (correct answer)
$740$

Explanation: This problem gives W = 500 - 20t and asks for W when t = 12. Substituting t = 12: W = 500 - 20(12) = 500 - 240 = 260. This represents a water tank that starts with 500 liters and drains at 20 liters per minute. After 12 minutes, 260 liters remain. A common error is making arithmetic mistakes in the multiplication or subtraction. When evaluating linear expressions involving time, substitute the time value carefully and follow the order of operations.