Solving Systems of Equations by Elimination Worksheets - Math Monks - Free Printable

We are given 8 systems of linear equations to solve using the elimination method. The goal is to eliminate one variable by adding or subtracting the equations (possibly after multiplying one or both equations by a constant), then solve for the remaining variable, and finally back-substitute to find the other.

Let’s solve each system one by one.

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①

> 2x + y = 7
> x - 2y = 6

Step 1: Eliminate one variable. Let’s eliminate y.

Multiply the first equation by 2 so that the coefficients of y become opposites:

→ 4x + 2y = 14
→ x - 2y = 6

Step 2: Add the two equations:

(4x + 2y) + (x - 2y) = 14 + 6
→ 5x = 20
→ x = 4

Step 3: Plug x = 4 into the first original equation:

2(4) + y = 7 → 8 + y = 7 → y = -1

✔ Solution: (4, -1)

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②

> x - y = -6
> x + y = 8

Step 1: Add the two equations directly — y will cancel:

(x - y) + (x + y) = -6 + 8
→ 2x = 2
→ x = 1

Step 2: Plug x = 1 into second equation:

1 + y = 8 → y = 7

✔ Solution: (1, 7)

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③

> 3x + 2y = -18
> 13x - 2y = -14

Step 1: Add the equations — y terms cancel:

(3x + 2y) + (13x - 2y) = -18 + (-14)
→ 16x = -32
→ x = -2

Step 2: Plug x = -2 into first equation:

3(-2) + 2y = -18 → -6 + 2y = -18 → 2y = -12 → y = -6

✔ Solution: (-2, -6)

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④

> x + 2y = 4
> 13x + 2y = 28

Step 1: Subtract the first equation from the second to eliminate y:

(13x + 2y) - (x + 2y) = 28 - 4
→ 12x = 24
→ x = 2

Step 2: Plug x = 2 into first equation:

2 + 2y = 4 → 2y = 2 → y = 1

✔ Solution: (2, 1)

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⑤

> 4x + 2y = 14
> 5x + 2y = 16

Step 1: Subtract first equation from second:

(5x + 2y) - (4x + 2y) = 16 - 14
→ x = 2

Step 2: Plug x = 2 into first equation:

4(2) + 2y = 14 → 8 + 2y = 14 → 2y = 6 → y = 3

✔ Solution: (2, 3)

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⑥

> 9x - 2y = 16
> 9x - 2y = -14

Wait — these are the same left-hand side, but different right-hand sides.

That means we have:

9x - 2y = 16
9x - 2y = -14

Subtracting them: 0 = 30 → Contradiction!

✔ No solution — the lines are parallel and never intersect.

Answer: No solution

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⑦

> -2x + 7y = 3
> -4x + 14y = 6

Notice: The second equation is exactly 2 times the first equation:

2*(-2x + 7y) = 2*3 → -4x + 14y = 6 — matches!

So these are dependent equations — same line.

That means infinitely many solutions.

But the problem asks for “the solution as an ordered pair” — which implies a unique solution.

However, since they are identical, any point on the line is a solution. But if forced to write one, we can pick any point.

Let’s solve for y in terms of x from first equation:

-2x + 7y = 3 → 7y = 2x + 3 → y = (2x + 3)/7

Pick x = 0 → y = 3/7 → (0, 3/7)

But technically, this system has infinitely many solutions.

✔ Answer: Infinitely many solutions (or parametrically: (x, (2x+3)/7))

But since the worksheet likely expects a single ordered pair, perhaps it's a trick question — no unique solution.

Actually, let’s check: if you try elimination:

Multiply first equation by -2: 4x - 14y = -6
Add to second: (-4x + 14y) + (4x - 14y) = 6 + (-6) → 0=0 → always true.

So yes — infinitely many solutions.

✔ Answer: Infinitely many solutions

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⑧

> 3x + 4y = 52
> 5x + y = 30

Step 1: Eliminate y. Multiply second equation by 4:

→ 3x + 4y = 52
→ 20x + 4y = 120

Step 2: Subtract first from second:

(20x + 4y) - (3x + 4y) = 120 - 52
→ 17x = 68
→ x = 4

Step 3: Plug x = 4 into second original equation:

5(4) + y = 30 → 20 + y = 30 → y = 10

✔ Solution: (4, 10)

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## ✔ Final Answers:

① (4, -1)
② (1, 7)
③ (-2, -6)
④ (2, 1)
⑤ (2, 3)
⑥ No solution
⑦ Infinitely many solutions
⑧ (4, 10)

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Let me know if you want to see graph interpretations or alternative methods!