SAT Linear Equations Worksheet

Solve a one-variable linear equation by performing the same operation on both sides. Distribute across every term, clear fractions carefully, and check the final value in the original equation.

Written by Luke Finigan

$SAT Linear Equations Worksheet, page 1: distribution, fractions, constants, and equations with no solution$

This 12-question worksheet includes word problems, constants, and equations with one, no, or infinitely many solutions. Download a student PDF for an attempt without answers or choose the full explained version.

12 questions · about 25 minutes · Algebra › Linear equations in one variable

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$SAT Linear Equations Worksheet, page 1: distribution, fractions, constants, and equations with no solution$

Simplify before deciding how many solutions exist

Distribute first, then collect the variable terms on one side and the constants on the other. For $3 (2 x - 5) = 21$ , distribution gives $6 x - 15 = 21$ , then $6 x = 36$ and $x = 6$ . Substitution gives $3 (12 - 5) = 21$ .

After simplifying, an equation of the form $a x = b$ has one solution when $a \neq = 0$ . If $a = 0$ and $b \neq = 0$ , it has no solution. If both are zero, every allowed value works. A parameter question asks which coefficient produces one of these cases.

In a word problem, define what $x$ measures and distinguish an amount remaining from an amount used. Reread the requested expression at the end: finding $x$ does not finish a question asking for $2 x + 1$ .

Try it yourself

How many solutions does

5 x + 2 = 5 x - 3

have?

AOneBNoneCInfinitely many

Choose an answer.

Questions

Question 1
If $3 (2 x - 5) = 21$ , what is $x$ ?
B. Distribute: $6 x - 15 = 21$ . Add 15 to get $6 x = 36$ , so $x = 6$ . Checking gives $3 (12 - 5) = 21$ .
Question 2
If $\frac{x}{4} + 3 = 8$ , what is $x$ ?
D. Subtract 3: $\frac{x}{4} = 5$ . Multiply both sides by 4 to obtain $x = 20$ .
Question 3
If $5 x + 7 = 2 x - 11$ , what is $x$ ?
A. Subtract $2 x$ and 7 from both sides: $3 x = - 18$ . Divide by 3 to obtain $x = - 6$ .
Question 4
A repair costs a fixed fee of $35 plus $18 per hour. A customer pays $107 before tax. How many hours of work are charged?
C. Set $35 + 18 h = 107$ . Then $18 h = 72$ , so $h = 4$ hours. The fixed fee is paid once.
Question 5
For which value of $k$ does the equation $k x + 4 = 7 x - 2$ have no solution?
B. No solution occurs when the $x$ coefficients are equal but constants differ. With $k = 7$ , subtracting $7 x$ gives the false statement $4 = - 2$ .
Question 6
The equation $2 (3 x + a) = 6 x + 10$ is true for every real $x$ . What is $a$ ?
A. Expand to $6 x + 2 a = 6 x + 10$ . For every $x$ , the constants must agree: $2 a = 10$ , so $a = 5$ .
Question 7
If $4 (x + 2) - 3 x = 19$ , what is $x$ ?
C. Distribute and combine like terms: $4 x + 8 - 3 x = 19$ becomes $x + 8 = 19$ . Subtract 8 to obtain $x = 11$ .
Question 8
If $\frac{2 x - 3}{5} = 7$ , what is $x$ ?
B. Multiply both sides by 5: $2 x - 3 = 35$ . Adding 3 gives $2 x = 38$ , so $x = 19$ .
Question 9
A tank initially contains 120 liters of water and drains at 8 liters per minute. After how many minutes will 56 liters remain?
B. Set $120 - 8 t = 56$ . The tank loses 64 liters, so $t = \frac{64}{8} = 8$ minutes.
Question 10
If $\frac{x - 2}{3} = \frac{x + 4}{5}$ , what is $x$ ?
C. Multiply by 15 to get $5 (x - 2) = 3 (x + 4)$ . Thus $5 x - 10 = 3 x + 12$ , giving $2 x = 22$ and $x = 11$ .
Question 11
For which value of $a$ does $5 x + 2 = 5 (x + a)$ have infinitely many solutions?
B. The right side expands to $5 x + 5 a$ . The equations agree for every $x$ only if $2 = 5 a$ , so $a = \frac{2}{5}$ .
Question 12
If $7 x - 4 = 3 x + 20$ , what is $2 x + 1$ ?
D. First solve $4 x = 24$ , giving $x = 6$ . The question asks for $2 x + 1$ , which is $12 + 1 = 13$ .

0/12 correct

Check your linear equations answers

Each row is a wrong answer choice from one of the sheet's problems. Find the one you picked, then run the check in the last column.

What your answer looked like	What actually happened	Fix it next time
32 for $\frac{x}{4} + 3 = 8$ (problem 2)	You multiplied 8 by 4 before dealing with the $+ 3$ . That solves $\frac{x}{4} = 8$ , a different equation.	Undo the addition first: $\frac{x}{4} = 5$ , so $x = 20$ . Check 32: $\frac{32}{4} + 3 = 11$ , not 8.
6 for $5 x + 7 = 2 x - 11$ (problem 3)	A sign slipped while you moved the constants. Subtracting 7 from $- 11$ gives $- 18$ , not 18.	From $3 x = - 18$ , $x = - 6$ . Check 6 in the original: the left side is $5 (6) + 7 = 37$ , and the right side is $2 (6) - 11 = 1$ .
10 in problem 6	The 2 outside the parentheses multiplies $a$ too, so the left side is $6 x + 2 a$ , not $6 x + a$ .	Match the constants: $2 a = 10$ , so $a = 5$ . Problem 11 has the same trap: $5 (x + a) = 5 x + 5 a$ , so $5 a = 2$ and $a = \frac{2}{5}$ , not 2.
16 for $\frac{2 x - 3}{5} = 7$ (problem 8)	You subtracted 3 when you needed to add it. After multiplying by 5, the equation is $2 x - 3 = 35$ .	Add 3 to get $2 x = 38$ , so $x = 19$ . Check: $\frac{2 ( 19 ) - 3}{5} = \frac{35}{5} = 7$ .
7 or 15 in problem 9	$\frac{56}{8} = 7$ is how long 56 liters take to drain, and $\frac{120}{8} = 15$ is when the tank is empty. The question says 56 liters remain.	Find what drains: $120 - 56 = 64$ liters, and $\frac{64}{8} = 8$ minutes. Check: $120 - 8 (8) = 56$ .
6 or 12 in problem 12	Those are $x$ and $2 x$ . The question asks for $2 x + 1$ .	Solve $4 x = 24$ for $x = 6$ , then finish the expression: $2 (6) + 1 = 13$ .

Worked example: the equation with no solution

Problem 5 asks for the value of $k$ that gives $k x + 4 = 7 x - 2$ no solution. Collect the $x$ -terms on one side and the constants on the other: $k x - 7 x = - 2 - 4$ , which is $(k - 7) x = - 6$ .

If $k - 7$ isn't 0, you can divide by it and get exactly one solution, $x = \frac{- 6}{k - 7}$ . If $k = 7$ , the left side is $0 \cdot x$ , which is 0 for every $x$ , and $0 = - 6$ is never true. So the answer is 7. With $k = - 7$ , the equation becomes $- 14 x = - 6$ , which has the solution $x = \frac{3}{7}$ . Problem 11 is the other case: when the $x$ -terms and the constants both match, every $x$ works.

Apply the check to fresh questions

On another linear-equations question, finish by substituting into the original equation and checking the requested quantity. The practice link provides fresh problems on the same skill.

Practice linear equations questions

Review the linear equations in one variable method

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