Example 2 demonstrates how to solve a system of equations by adding them together to eliminate a variable.

Let’s solve each system of equations using the elimination method. We’ll add or subtract the equations to eliminate one variable, then solve for the other.

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Problem 1:
2x + 2y = -2
3x – 2y = 12

Notice: The y terms are +2y and -2y → they cancel when we add!

Add the two equations:

(2x + 3x) + (2y – 2y) = -2 + 12
→ 5x + 0 = 10
→ 5x = 10
→ x = 2

Now plug x = 2 into one of the original equations. Let’s use the first one:

2(2) + 2y = -2
→ 4 + 2y = -2
→ 2y = -6
→ y = -3

✔ Solution: (2, -3)

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Problem 2:
4x – 2y = -1
-4x + 4y = -2

Notice: The x terms are +4x and -4x → they cancel when we add!

Add the two equations:

(4x – 4x) + (-2y + 4y) = -1 + (-2)
→ 0 + 2y = -3
→ 2y = -3
→ y = -3/2 or -1.5

Now plug y = -1.5 into one equation. Use the first:

4x – 2(-1.5) = -1
→ 4x + 3 = -1
→ 4x = -4
→ x = -1

✔ Solution: (-1, -1.5) or written as fraction: (-1, -3/2)

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Problem 3:
x – y = 2
x + y = -3

Notice: The y terms are -y and +y → they cancel when we add!

Add the two equations:

(x + x) + (-y + y) = 2 + (-3)
→ 2x + 0 = -1
→ 2x = -1
→ x = -0.5 or -1/2

Plug x = -0.5 into first equation:

-0.5 – y = 2
→ -y = 2.5
→ y = -2.5 or -5/2

✔ Solution: (-0.5, -2.5) or (-1/2, -5/2)

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Problem 4:
6x + 5y = 4
6x – 7y = -20

Here, both have +6x → if we subtract the second from the first, x will cancel.

Subtract equation 2 from equation 1:

(6x – 6x) + (5y – (-7y)) = 4 – (-20)
→ 0 + (5y + 7y) = 4 + 20
→ 12y = 24
→ y = 2

Plug y = 2 into first equation:

6x + 5(2) = 4
→ 6x + 10 = 4
→ 6x = -6
→ x = -1

✔ Solution: (-1, 2)

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Problem 5:
2x – 3y = 12
4x + 3y = 24

Notice: The y terms are -3y and +3y → they cancel when we add!

Add the two equations:

(2x + 4x) + (-3y + 3y) = 12 + 24
→ 6x + 0 = 36
→ 6x = 36
→ x = 6

Plug x = 6 into first equation:

2(6) – 3y = 12
→ 12 – 3y = 12
→ -3y = 0
→ y = 0

✔ Solution: (6, 0)

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Final Answer:
1. (2, -3)
2. (-1, -3/2)
3. (-1/2, -5/2)
4. (-1, 2)
5. (6, 0)