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

Let’s solve each system of equations using elimination. We’ll go one by one, step by step.

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Problem ①:

Equations:
1) 2x + 2y = 6
2) 3x - 5 = y → Let’s rewrite this as: 3x - y = 5 (move y to left and -5 to right)

Now we have:
A) 2x + 2y = 6
B) 3x - y = 5

We want to eliminate one variable. Let’s eliminate y.

Multiply equation B by 2 so that the y terms become -2y and +2y:

→ 2*(3x - y) = 2*5 → 6x - 2y = 10 (call this C)

Now add equation A and C:

(2x + 2y) + (6x - 2y) = 6 + 10
→ 8x = 16
→ x = 2

Plug x = 2 into equation B: 3x - y = 5
→ 3(2) - y = 5 → 6 - y = 5 → y = 1

✔ Solution: x = 2, y = 1

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Problem ②:

Equations:
1) -20x + 6y = -6
2) -10x - 4y = 4

Let’s make coefficients of x match. Multiply equation 2 by 2:

→ 2*(-10x - 4y) = 2*4 → -20x - 8y = 8 (call this C)

Now subtract equation C from equation 1:

(-20x + 6y) - (-20x - 8y) = -6 - 8
→ -20x + 6y + 20x + 8y = -14
→ 14y = -14
→ y = -1

Plug y = -1 into original equation 2: -10x - 4y = 4
→ -10x - 4(-1) = 4 → -10x + 4 = 4 → -10x = 0 → x = 0

✔ Solution: x = 0, y = -1

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Problem ③:

Equations:
1) 3x + 5y = -9
2) -6x - 5y = -18

Notice: If we add them directly, the y terms cancel!

Add equations:

(3x + 5y) + (-6x - 5y) = -9 + (-18)
→ -3x = -27
→ x = 9

Plug x = 9 into equation 1: 3(9) + 5y = -9 → 27 + 5y = -9 → 5y = -36 → y = -36/5 = -7.2

Wait — let me check if I did that right.

Actually, maybe better to use fractions or double-check.

But let’s verify with equation 2:

-6x -5y = -18 → -6(9) -5y = -18 → -54 -5y = -18 → -5y = 36 → y = -36/5 → same.

So it’s correct, but unusual fraction. Maybe I made a sign error?

Original problem:

③ 3x + 5y = -9
-6x -5y = -18

Adding: 3x -6x = -3x; 5y -5y = 0; -9 -18 = -27 → -3x = -27 → x=9 → yes.

Then 3(9)+5y=-9 → 27+5y=-9 → 5y=-36 → y=-36/5

That’s correct. So answer is x=9, y=-36/5 or -7.2

But let’s write as fraction: y = -36/5

✔ Solution: x = 9, y = -36/5

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Problem ④:

Equations:
1) -2x + 3y = -1
2) 2x + 5y = 25

Add them directly — x terms cancel!

(-2x + 3y) + (2x + 5y) = -1 + 25
→ 8y = 24 → y = 3

Plug y=3 into equation 2: 2x + 5(3)=25 → 2x +15=25 → 2x=10 → x=5

✔ Solution: x = 5, y = 3

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Problem ⑤:

Equations:
1) x + 4y = 7
2) 4x - 3y = 9

Let’s eliminate x. Multiply equation 1 by 4:

→ 4x + 16y = 28 (call this C)

Now subtract equation 2 from C:

(4x + 16y) - (4x - 3y) = 28 - 9
→ 4x + 16y -4x +3y = 19
→ 19y = 19 → y = 1

Plug y=1 into equation 1: x + 4(1)=7 → x=3

✔ Solution: x = 3, y = 1

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Problem ⑥:

Equations:
1) 2x - 5y = 30
2) x + 5y = -45

Add them — y terms cancel!

(2x -5y) + (x +5y) = 30 + (-45)
→ 3x = -15 → x = -5

Plug x=-5 into equation 2: -5 + 5y = -45 → 5y = -40 → y = -8

✔ Solution: x = -5, y = -8

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Problem ⑦:

Equations:
1) -7x -6y = 11
2) -8x -12y = 28

Let’s try to eliminate y. Notice: if we multiply equation 1 by 2, we get:

→ 2*(-7x -6y) = 2*11 → -14x -12y = 22 (call this C)

Now subtract equation 2 from C:

(-14x -12y) - (-8x -12y) = 22 - 28
→ -14x -12y +8x +12y = -6
→ -6x = -6 → x = 1

Plug x=1 into equation 1: -7(1) -6y = 11 → -7 -6y = 11 → -6y = 18 → y = -3

Check in equation 2: -8(1) -12(-3) = -8 + 36 = 28 → matches!

✔ Solution: x = 1, y = -3

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Problem ⑧:

Equations:
1) 3x + 4y = -1
2) 4x - 3y = 7

Let’s eliminate one variable. Let’s eliminate y.

Find LCM of 4 and 3 → 12.

Multiply equation 1 by 3: → 9x + 12y = -3 (C)
Multiply equation 2 by 4: → 16x - 12y = 28 (D)

Now add C and D:

(9x + 12y) + (16x - 12y) = -3 + 28
→ 25x = 25 → x = 1

Plug x=1 into equation 1: 3(1) + 4y = -1 → 3 + 4y = -1 → 4y = -4 → y = -1

Check in equation 2: 4(1) -3(-1) = 4 + 3 = 7 → correct!

✔ Solution: x = 1, y = -1

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Final Answer:

① x = 2, y = 1
② x = 0, y = -1
③ x = 9, y = -36/5
④ x = 5, y = 3
⑤ x = 3, y = 1
⑥ x = -5, y = -8
⑦ x = 1, y = -3
⑧ x = 1, y = -1