Algebra 1 Systems of Linear Equations Worksheet Solutions (PDF)

Algebra 1 math worksheet solutions: Systems of Linear Equations – Maths PDF

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This systems practice solutions for Algebra 1 covers systems of linear equations (Common Core A-REI.C.5, A-REI.C.6, A-REI.D.11, A-CED.A.3), with a free PDF to print.

Use this key to check your work on the systems of equations worksheet. Every solution shows the steps, and the final answer appears in bold. If your answer differs, find the first line where your work and the key disagree, then redo the problem from there without looking.

Algebra 1 · Common Core State Standards: A-REI.C.5, A-REI.C.6, A-REI.D.11, A-CED.A.3

Systems practice solutions: exercises 1 to 10

Exercise 1 – Check a solution

First equation: $3 = 2 + 1$ is true. Second equation: $3 = -2 + 5$ is true.

Yes, it is a solution

Exercise 2 – Solve by graphing

The lines meet at (2, 3).

The lines cross at $(2, 3)$. Check: $2(2) – 1 = 3$ and $-2 + 5 = 3$.

$(2, 3)$

Exercise 3 – Graph and read

Both lines pass through $(-1, 3)$: $-1 + 4 = 3$ and $-2(-1) + 1 = 3$.

$(-1, 3)$

Exercise 4 – Parallel lines

Both lines have slope 3 but different y-intercepts (2 and -4), so they are parallel.

No solution

Exercise 5 – Same line

Divide the second equation by 2: $y = 2x + 1$. The equations describe the same line.

Infinitely many solutions

Exercise 6 – Substitution with y = 3x

Substitute $y = 3x$: $x + 3x = 12$, so $4x = 12$ and $x = 3$.

Then $y = 3(3) = 9$.

$(3, 9)$

Exercise 7 – Substitution with x

Substitute: $3(2y – 1) + y = 11$. Distribute: $6y – 3 + y = 11$.

$7y = 14$, so $y = 2$. Then $x = 2(2) – 1 = 3$.

$(3, 2)$

Exercise 8 – Two slope-intercept equations

Set the expressions equal: $-x + 7 = 2x + 1$. Then $6 = 3x$, so $x = 2$.

$y = -2 + 7 = 5$.

$(2, 5)$

Exercise 9 – Substitution after rearranging

From the first equation, $y = 10 – 2x$. Substitute: $3x – (10 – 2x) = 5$.

$5x – 10 = 5$, so $x = 3$. Then $y = 10 – 6 = 4$.

$(3, 4)$

Exercise 10 – Elimination by adding

Add the equations: $2x = 20$, so $x = 10$.

Substitute: $10 + y = 14$, so $y = 4$.

$(10, 4)$

Exercises 11 and up

Exercise 11 – Elimination with opposites

Adding cancels y.

Add: $6x = 18$, so $x = 3$.

Substitute: $2(3) + 3y = 12$, so $3y = 6$ and $y = 2$.

$(3, 2)$

Exercise 12 – Elimination by subtracting

Subtract the second equation from the first: $2x = 8$, so $x = 4$.

Substitute: $4 + 2y = 8$, so $y = 2$.

$(4, 2)$

Exercise 13 – Multiply one equation

Multiply the first equation by 2 and the second by 3: $4x + 6y = 26$ and $15x – 6y = 12$.

Add: $19x = 38$, so $x = 2$. Then $2(2) + 3y = 13$ gives $y = 3$.

$(2, 3)$

Exercise 14 – Multiply both equations

Multiply the first by 3 and the second by 4: $9x – 12y = 3$ and $8x + 12y = 48$.

Add: $17x = 51$, so $x = 3$. Then $3(3) – 4y = 1$ gives $y = 2$.

$(3, 2)$

Exercise 15 – Elimination gives no solution

Multiply the second equation by $-2$: $-2x – 4y = -20$.

Add to the first: $0 = -12$, which is false.

No solution

Exercise 16 – Movie tickets

Split of the 8 tickets.

$a + s = 8$ and $9a + 6s = 60$.

Substitute $s = 8 – a$: $9a + 6(8 – a) = 60$, so $3a + 48 = 60$ and $a = 4$.

4 adult tickets and 4 student tickets

Exercise 17 – Coins

$d + q = 18$ and $10d + 25q = 300$ (in cents).

Substitute $d = 18 – q$: $10(18 – q) + 25q = 300$, so $180 + 15q = 300$ and $q = 8$.

$d = 18 – 8 = 10$.

10 dimes and 8 quarters

Exercise 18 – Two gyms

Gym A: $y = 20m + 30$. Gym B: $y = 35m$.

Set equal: $20m + 30 = 35m$, so $30 = 15m$ and $m = 2$.

Cost: $35(2) = 70$.

After 2 months, both cost $70

Exercise 19 – Bake sale

$c + m = 40$ and $1.5c + 2.5m = 70$.

Substitute $c = 40 – m$: $1.5(40 – m) + 2.5m = 70$, so $60 + m = 70$ and $m = 10$.

$c = 30$.

30 cookies and 10 muffins

Exercise 20 – Boat and current

Downstream speed: $b + c = \frac{24}{2} = 12$. Upstream speed: $b – c = \frac{24}{3} = 8$.

Add: $2b = 20$, so $b = 10$. Then $c = 12 – 10 = 2$.

Boat 10 mph, current 2 mph

Exercise 21 – Number problem

Let the numbers be $x$ and $y$: $x + y = 42$ and $x = 2y + 6$.

Substitute: $2y + 6 + y = 42$, so $3y = 36$ and $y = 12$.

$x = 2(12) + 6 = 30$.

30 and 12

Key facts behind this systems practice solutions

Key facts

A solution of a system is an ordered pair that satisfies every equation.
Graphing: the solution is the intersection point of the lines.
Substitution: solve one equation for a variable and substitute into the other.
Elimination: add or subtract equations to cancel a variable.
Parallel lines give no solution; identical lines give infinitely many.

How to use this systems practice solutions

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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