SAT Math skill page

SAT Linear Equations Practice

Build the SAT Math skill behind one-variable equations, slope-intercept form, systems, and equation word problems.

Practice linear equations in Dolphin Practice questions

12-18 min practice time3 questions on pageHeart of Algebra

Practice time12-18 min

Question bank3 questions

Best forHeart of Algebra

What this tests

What to know for this SAT skill

Solve one-variable linear equations
Translate word problems into equations
Interpret slope and intercepts
Solve simple systems by substitution or elimination

Dolphin question bank

Practice your skill

Question 1 of 3Hard

The equation 9x + 5 = a(x + b), where a and b are constants, has no solutions. Which of the following must be true?

I . a = 9 I I . b = 5 I I I . b \neq = \frac{5}{9}

None
I only
I and II only
I and III only

Show answer and solution

Correct answerD. I and III only

Solution walkthrough

For a linear equation in a form $a x + b = c x + d$ to have no solutions, the x-terms must have equal coefficients and the remaining terms must not be equal. Expanding the right-hand side of the given equation yields $9 x + 5 = a x + ab$ .
Inspecting the x-terms, 9 must equal $a$ , so statement I must be true. Inspecting the remaining terms, 5 can't equal $9 b$ .
Dividing both of these quantities by 9 yields that $b$ can't equal $\frac{5}{9}$ . Therefore, statement III must be true.
Since $b$ can have any value other than $\frac{5}{9}$ , statement II may or may not be true.

Why the other choices miss

Choice A is incorrect. For the given equation to have no solution, both

a = 9

and

b \neq = \frac{5}{9}

must be true. Choice B is incorrect because it must also be true that

b \neq = \frac{5}{9}

. Choice C is incorrect because when

a = 9

, there are many values of

b

that lead to an equation having no solution. That is,

b

might be 5, but

b

isn’t required to be 5.

Question 2 of 3Medium

What value of x is the solution to the given equation?

\frac{1}{4} (x + 5) - \frac{1}{3} (x + 5) = - 7

-12
-5
79
204

Show answer and solution

Correct answerC. 79

Solution walkthrough

For the given equation, $(x + 5)$ is a factor of both terms on the left-hand side.
Therefore, the given equation can be rewritten as $(\frac{1}{4} - \frac{1}{3}) (x + 5) = - 7$ , or $(\frac{3}{12} - \frac{4}{12}) (x + 5) = - 7$ , which is equivalent to $- \frac{1}{12} (x + 5) = - 7$ .
Multiplying both sides of this equation by $- 12$ yields $x + 5 = 84$ .
Subtracting 5 from both sides of this equation yields $x = 79$ .

Why the other choices miss

Choice A is incorrect. This is the value of

x

for which the left-hand side of the given equation equals

\frac{7}{12}

, not

- 7

. Choice B is incorrect. This is the value of

x

for which the left-hand side of the given equation equals

0

, not

- 7

. Choice D is incorrect. This is the value of

x

for which the left-hand side of the given equation equals

- \frac{209}{12}

, not

- 7

Question 3 of 3Medium

Megan's regular wage at her job is p dollars per hour for the first 8 hours of work in a day plus 1.5 times her regular hourly wage for work in excess of 8 hours that day. On a given day, Megan worked for 10 hours, and her total earnings for that day were $137.50. What is Megan's regular hourly wage?

\11.75$
\12.50$
\13.25$
\13.75$

Show answer and solution

Correct answerB. \12.50$

Solution walkthrough

Since $p$ represents Megan’s regular pay per hour, $1.5 p$ represents the pay per hour in excess of 8 hours. Since Megan worked for 10 hours, she must have been paid $p$ dollars per hour for 8 of the hours plus $1.5 p$ dollars per hour for the remaining 2 hours.
Therefore, since Megan earned $137.50 for the 10 hours, the situation can be represented by the equation $137.5 = 8 p + 2 (1.5 p)$ . Distributing the 2 in the equation gives $137.5 = 8 p + 3 p$ , and combining like terms gives $137.5 = 11 p$ .
Dividing both sides by 11 gives $p = 12.5$ . Therefore, Megan’s regular wage is $12.50$ .

Why the other choices miss

Choices A and C are incorrect and may be the result of calculation errors. Choice D is incorrect and may result from finding the average hourly wage that Megan earned for the 10 hours of work.

Avoid these traps

Common mistakes on this skill

Doing operations on only one side

Every addition, subtraction, multiplication, or division step has to preserve equality on both sides of the equation.

Solving for the wrong variable

SAT questions often ask for a value like 2x, x + 3, or the y-value, not always x itself.

Missing negative signs

Negative constants and negative slopes are common traps. Rewrite the equation slowly before solving.

Study plan

How to practice this skill in Dolphin

Start by isolating the variable in one-variable equations.
Move to slope-intercept questions and practice substituting x-values.
Translate word problems into equations before touching the answer choices.
Review missed questions by naming the exact algebra step that went wrong.

Practice linear equations in Dolphin

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Skill cluster

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FAQ

Questions about SAT Linear Equations Practice

How common are linear equations on SAT Math?

Linear equations show up often because they connect to algebra, word problems, graphs, systems, and function notation.

What should I practice after linear equations?

Move into systems of equations, functions, and algebra word problems so the same equation skills transfer into harder formats.

Can Dolphin give me more linear equation practice?

Yes. Dolphin can filter SAT practice by topic, track missed questions, and keep surfacing the algebra work that needs review.