Math · 100 questions in our bank

SAT Linear Equations in Two Variables: Practice and Worked Solutions

A linear equation in two variables describes a straight line, and the SAT tests whether you can move between its three forms fluently. Slope-intercept form, $y = mx + b$ , is the one worth defaulting to: it exposes the slope and the $y$ -intercept without any work. Point-slope form, $y - y_1 = m(x - x_1)$ , is fastest when you are handed a point and a slope. Standard form, $Ax + By = C$ , is what the test gives you when it wants you to do the rearranging yourself.

Nearly every question reduces to finding two things: a slope and one point. Slope between two points is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ , and this is where most lost points originate, because a single sign error in the subtraction flips the line's direction and usually still produces one of the four answer choices.

The other half of this skill is interpretation. When a line models something real, the slope is a rate and the $y$ -intercept is a starting value. A question asking what the number 47 represents in a cost equation is testing whether you can read those two roles, not whether you can compute anything.

Where students lose the point

Sign errors in the slope formula. Subtracting the coordinates in inconsistent order flips the sign, and the wrong-sign answer is almost always one of the choices.
Swapping slope and $y$ -intercept in word problems. The per-unit rate is the slope; the flat starting amount is the intercept. The question is usually phrased to make you second-guess which is which.
Forgetting that parallel lines share a slope while perpendicular slopes are negative reciprocals. The negative and the reciprocal are separate steps, and doing only one produces a trap answer.
Solving for $y$ incorrectly when rearranging from standard form. Dividing only some terms by the coefficient of $y$ is the single most common algebra slip on this skill.

10 practice questions with worked answers

Work each one before opening the explanation. The answer is hidden until you ask for it, which is the only way practice tells you anything useful.

Question 1

easy

$5p = q + 9r$

The given equation relates the distinct positive numbers $p$ , $q$ , and $r$ . Which equation correctly expresses $p$ in terms of $q$ and $r$ ?

A
$p = 5(q + 9r)$
B
$p = \dfrac{q}{5} + 9r$
C
$p = q + \dfrac{9r}{5}$
D
$p = \dfrac{q + 9r}{5}$

Show the worked answer

Divide both sides of $5p = q + 9r$ by $5$ :

$p = \dfrac{q + 9r}{5}.$

That is choice D. Choice A multiplies by $5$ instead of dividing. Choice B divides only the $q$ term. Choice C divides only the $9r$ term.

Question 2

easy

A potter’s wheel forms wide bowls or narrow cups, one at a time, for a total of $540$ minutes each shift. A wide bowl takes $15$ minutes and a narrow cup takes $6$ minutes. Which equation represents the possible numbers of wide bowls, $x$ , and narrow cups, $y$ , the wheel can form in one shift?

A
$6x + 15y = 540$
B
$15x + 6y = 540$
C
$(x + y)(15 + 6) = 540$
D
$(15 + x)(6 + y) = 540$

Show the worked answer

Each wide bowl uses $15$ minutes, so $x$ bowls use $15x$ minutes. Each narrow cup uses $6$ minutes, so $y$ cups use $6y$ minutes. The times add to the full shift:

$15x + 6y = 540,$

choice B. Choice A swaps the two times. Choice C multiplies the total number of pieces by the sum of the times. Choice D adds each time to the corresponding count before multiplying.

Question 3

easy

Each $5.0$ -ounce serving of lentils and each $2.0$ -ounce serving of sunflower seeds provides about $3$ milligrams of zinc. If a total of $4.5$ milligrams of zinc are consumed from eating $x$ ounces of lentils and $y$ ounces of sunflower seeds, which equation best represents this situation?

A
$5.0x + 2.0y = 4.5$
B
$0.60x + 1.50y = 4.5$
C
$1.50x + 0.60y = 4.5$
D
$2.0x + 5.0y = 4.5$

Show the worked answer

Lentils provide $\dfrac{3}{5.0} = 0.60$ milligrams of zinc per ounce. Sunflower seeds provide $\dfrac{3}{2.0} = 1.50$ milligrams of zinc per ounce. Then $x$ ounces of lentils and $y$ ounces of seeds give

$0.60x + 1.50y = 4.5,$

choice B. Choice C swaps the two rates. Choices A and D use the serving sizes as coefficients instead of milligrams per ounce.

Question 4

medium

A line in the $xy$ -plane contains the points $(0, 10)$ and $(4, 0)$ . An equation of the line is $ax + by = 1$ , where $a$ and $b$ are constants. What is the value of $a$ ?

Student-produced response: type your answer rather than choosing.

Show the worked answer

Substitute each given point into $ax + by = 1$ .

The intercept $(4, 0)$ gives $4a = 1$ , so $a = \dfrac{1}{4}$ .

The intercept $(0, 10)$ gives $10b = 1$ , so $b = \dfrac{1}{10}$ . That value is not needed to find $a$ .

Using the $y$ -intercept in place of the $x$ -intercept would produce $b$ instead of $a$ .

Question 5

medium

In the $xy$ -plane, line $\ell$ passes through the origin and is parallel to the line $y = 5x + 9$ . If line $\ell$ also contains the point $(7, d)$ , what is the value of $d$ ?

Student-produced response: type your answer rather than choosing.

Show the worked answer

Parallel lines have equal slopes, so line $\ell$ has slope $5$ . Combined with the origin, its equation is $y = 5x$ . The point $(7, d)$ lies on the line, so $d = 5 \cdot 7 = 35$ .

Using the given $y$ -intercept $9$ as if the line through the origin were $y = 5x + 9$ would give $5 \cdot 7 + 9 = 44$ , which is not on the parallel line through $(0, 0)$ .

Question 6

medium

In the $xy$ -plane, line $k$ passes through the points $(3, 5)$ and $(7, 17)$ . If the point $(10, w)$ also lies on line $k$ , what is the value of $w$ ?

Student-produced response: type your answer rather than choosing.

Show the worked answer

Find the slope from the two given points: $m = \dfrac{17 - 5}{7 - 3} = \dfrac{12}{4} = 3$ .

Use point-slope form with $(3, 5)$ : $y - 5 = 3(x - 3)$ , so $y = 3x - 4$ .

Substitute $x = 10$ : $w = 3(10) - 4 = 26$ .

Reversing the slope to $\dfrac{7 - 3}{17 - 5} = \dfrac{1}{3}$ would give $w \approx 7.33$ , and forgetting the $-4$ intercept would give $30$ ; neither point lies on line $k$ .

Question 7

medium

The graph of $9x + 4y = -36$ in the $xy$ -plane has an $x$ -intercept of $(a, 0)$ and a $y$ -intercept of $(0, b)$ , where $a$ and $b$ are constants. What is the value of $\dfrac{b}{a}$ ?

A
$-\dfrac{9}{4}$
B
$-\dfrac{4}{9}$
C
$\dfrac{4}{9}$
D
$\dfrac{9}{4}$

Show the worked answer

The $x$ -intercept occurs when $y = 0$ : $9a = -36$ , so $a = -4$ . The $y$ -intercept occurs when $x = 0$ : $4b = -36$ , so $b = -9$ . Then $\dfrac{b}{a} = \dfrac{-9}{-4} = \dfrac{9}{4}$ , choice D.

In general, for $Ax + By = C$ the ratio of intercepts is $\dfrac{b}{a} = \dfrac{A}{B}$ . Choice A is the opposite of this ratio, and choices B and C are the reciprocal (with and without a sign error).

Question 8

hard

$x$	$y$
$k$	11
$k+5$	$-19$

A line in the $xy$ -plane passes through the two points in the table. This line meets the $y$ -axis at $(k-4, b)$ , where $k$ and $b$ are constants. Find $b$ .

Student-produced response: type your answer rather than choosing.

Show the worked answer

The slope is $\dfrac{-19-11}{(k+5)-k} = \dfrac{-30}{5} = -6$ . Using the point $(k, 11)$ gives $y-11 = -6(x-k)$ . The given $y$ -intercept occurs at $x = k-4$ , so $b = 11-6((k-4)-k) = 11-6(-4) = 35$ .

Question 9

hard

In the $xy$ -plane, the line $9x + 4y = -47$ crosses the axes at $(a, 0)$ and $(0, b)$ . What is $\dfrac{b}{a}$ ?

A
$-\dfrac{9}{4}$
B
$-\dfrac{4}{9}$
C
$\dfrac{4}{9}$
D
$\dfrac{9}{4}$

Show the worked answer

At the $x$ -intercept, $y = 0$ , so $9a = -47$ and $a = -\dfrac{47}{9}$ . At the $y$ -intercept, $x = 0$ , so $4b = -47$ and $b = -\dfrac{47}{4}$ . Then $\dfrac{b}{a} = \dfrac{47}{4} \cdot \dfrac{9}{47} = \dfrac{9}{4}$ . Choice D. A is the opposite sign of the correct ratio, and B and C are the reciprocal (with and without a sign error).

Question 10

hard

In the $xy$ -plane, line $q$ is perpendicular to the graph of $x = -5$ . Which of the following could be an equation of line $q$ ?

A
$y = 3$
B
$x = 3$
C
$y = -5x$
D
$y = x - 5$

Show the worked answer

The graph of $x = -5$ is a vertical line. A line perpendicular to a vertical line is horizontal, and every horizontal line has an equation of the form $y = c$ . Only choice A, $y = 3$ , has that form. Choice B is another vertical line (parallel, not perpendicular), and choices C and D have nonzero slopes, so they are neither horizontal nor vertical.