Algebra 1 Solving Linear Equations in One Variable Worksheet Solutions (PDF)

Algebra 1 math worksheet solutions: Solving Linear Equations in One Variable – Maths PDF

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This algebra 1 linear equations answers for Algebra 1 covers solving linear equations in one variable (Common Core A-REI.A.1, A-REI.B.3, A-CED.A.1), with a free PDF to print.

Algebra 1 linear equations answers (Algebra 1) – first page of the printable PDF

Here are full solutions for the linear equations worksheet. Read each step, not just the bold answer, and compare it with your own method. Redo any problem you missed, and note which kind of mistake you made so you can avoid it on the test.

Algebra 1 · Common Core State Standards: A-REI.A.1, A-REI.B.3, A-CED.A.1

Algebra 1 linear equations answers: exercises 1 to 10

Exercise 1 – Undo addition

Subtract 7 from both sides: $x = 15 – 7$.

$x = 8$

Exercise 2 – Undo multiplication

Divide both sides by $-4$: $x = \frac{28}{-4}$.

$x = -7$

Exercise 3 – Undo division

Multiply both sides by 5: $x = -3 \cdot 5$.

$x = -15$

Exercise 4 – Two steps with a tape diagram

Each x equals 5.

Subtract 5 from both sides: $3x = 15$.

Divide by 3: $x = 5$.

Check: $3(5) + 5 = 20$.

$x = 5$

Exercise 5 – Two-step with negatives

Add 9 to both sides: $2x = 8$.

Divide by 2: $x = 4$.

$x = 4$

Exercise 6 – Fraction two-step

Subtract 3: $\frac{x}{4} = 4$.

Multiply by 4: $x = 16$.

$x = 16$

Exercise 7 – Variables on both sides

Steps for the equation.

Subtract $2x$ from both sides: $3x – 3 = 12$.

Add 3: $3x = 15$. Divide by 3: $x = 5$.

$x = 5$

Exercise 8 – Larger coefficient on the right

Subtract $4x$: $7 = 5x – 13$.

Add 13: $20 = 5x$. Divide by 5: $4 = x$.

$x = 4$

Exercise 9 – Distribute once

Distribute: $3x – 6 = 2x + 5$.

Subtract $2x$: $x – 6 = 5$. Add 6: $x = 11$.

$x = 11$

Exercise 10 – Distribute on both sides

Distribute: $6x + 2 = 4x + 20$.

Subtract $4x$: $2x + 2 = 20$. Subtract 2: $2x = 18$.

Divide by 2: $x = 9$.

$x = 9$

Exercises 11 and up

Exercise 11 – Fraction coefficient

Add 4: $\frac{2}{3}x = 12$.

Multiply both sides by the reciprocal $\frac{3}{2}$: $x = 12 \cdot \frac{3}{2} = 18$.

$x = 18$

Exercise 12 – Combine like terms first

Combine $4x – x = 3x$: $3x + 2 = 17$.

Subtract 2: $3x = 15$. Divide by 3: $x = 5$.

$x = 5$

Exercise 13 – No solution

Distribute: $3x + 6 = 3x + 5$.

Subtract $3x$ from both sides: $6 = 5$, which is false for every value of $x$.

No solution

Exercise 14 – Infinitely many solutions

Distribute: $2x + 8 = 2x + 8$.

Subtract $2x$: $8 = 8$, which is always true.

Infinitely many solutions (all real numbers)

Exercise 15 – Check a solution

Substitute: $5(-2) + 4 = -10 + 4 = -6$.

The left side equals the right side.

Yes, $x = -2$ is a solution

Exercise 16 – Concert tickets

Equation: $4t + 6 = 38$.

Subtract 6: $4t = 32$. Divide by 4: $t = 8$.

$8 per ticket

Exercise 17 – Saving for a bike

Equation: $40 + 15w = 190$.

Subtract 40: $15w = 150$. Divide by 15: $w = 10$.

Check: $40 + 15(10) = 190$.

10 weeks

Exercise 18 – Rides on two passes

Costs are equal when $20 + 3r = 5 + 4r$.

Subtract $3r$: $20 = 5 + r$. Subtract 5: $15 = r$.

Cost: $20 + 3(15) = 65$ and $5 + 4(15) = 65$. The table shows the passes switching order between 10 and 20 rides.

15 rides; $65

Exercise 19 – Jacket money

Three-fifths of her money is $27: $\frac{3}{5}m = 27$.

Multiply by $\frac{5}{3}$: $m = 27 \cdot \frac{5}{3} = 45$.

$45

Exercise 20 – Consecutive integers

Let the integers be $n$, $n + 1$, and $n + 2$. Then $n + (n + 1) + (n + 2) = 87$.

Combine: $3n + 3 = 87$. Subtract 3: $3n = 84$. Divide: $n = 28$.

28, 29, 30

Exercise 21 – Garden fence

Width 8 feet, length 19 feet.

Perimeter: $2w + 2(2w + 3) = 54$.

Distribute and combine: $6w + 6 = 54$. Subtract 6: $6w = 48$. Divide: $w = 8$.

Length: $2(8) + 3 = 19$.

Check: $2(8) + 2(19) = 54$.

Width 8 feet, length 19 feet

Key facts behind this algebra 1 linear equations answers

Key facts

Do the same operation to both sides of an equation to keep it balanced.
Undo addition and subtraction first, then multiplication and division.
With variables on both sides, collect variable terms on one side and constants on the other.
A false statement such as 6 = 5 means no solution; a true statement such as 8 = 8 means infinitely many.
Check a solution by substituting it into the original equation.

How to use this algebra 1 linear equations answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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