Solving Systems of Equations Worksheets with Two Variables - Free Printable

You’ve uploaded a worksheet titled “Systems of Equations - Elimination Method”, which contains 8 systems of linear equations to solve using the elimination method.

I’ll solve each system step-by-step and explain the process.

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🔁 General Strategy for Elimination:

1. Align equations so variables are in the same order.
2. Multiply one or both equations by constants to make the coefficients of one variable opposites (or equal).
3. Add or subtract the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.
6. Write the solution as an ordered pair (x, y).

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## ✔ Problem 1:
2x + 5y = 20
6x – 5y = 12

→ Notice: The y-terms are already opposites (+5y and –5y).
Add the equations:

```
2x + 5y = 20
+ 6x – 5y = 12
--------------
8x = 32
```

→ x = 32 / 8 = 4

Substitute x = 4 into first equation:

2(4) + 5y = 20 → 8 + 5y = 20 → 5y = 12 → y = 12/5 = 2.4

✔ Solution: (4, 12/5) or (4, 2.4)

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## ✔ Problem 2:
3p + 4q = -3
–p + 4q = -15

→ q-coefficients are the same. Subtract equations to eliminate q.

Write them:

Eq1: 3p + 4q = -3
Eq2: –p + 4q = -15

Subtract Eq2 from Eq1:

```
3p + 4q = -3
– (–p + 4q = -15)
------------------
4p = 12
```

→ p = 12 / 4 = 3

Plug into Eq2: –3 + 4q = -15 → 4q = -12 → q = -3

✔ Solution: (3, -3)

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## ✔ Problem 3:
–2a + 3b = 15
7a – 6b = –3

→ We want to eliminate one variable. Let’s eliminate b.

Multiply first equation by 2 to match coefficients of b:

Eq1 × 2: –4a + 6b = 30
Eq2: 7a – 6b = –3

Now add:

```
–4a + 6b = 30
+ 7a – 6b = –3
---------------
3a = 27
```

→ a = 9

Plug into original Eq1: –2(9) + 3b = 15 → –18 + 3b = 15 → 3b = 33 → b = 11

✔ Solution: (9, 11)

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## ✔ Problem 4:
u + v = 11
–u + v = –1

→ u-coefficients are opposites. Add equations:

```
u + v = 11
+ –u + v = –1
-------------
2v = 10
```

→ v = 5

Plug into first equation: u + 5 = 11 → u = 6

✔ Solution: (6, 5)

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## ✔ Problem 5:
5r + 8s = 3
4r + 8s = –4

→ s-coefficients are same. Subtract equations:

Eq1: 5r + 8s = 3
Eq2: 4r + 8s = –4

Subtract Eq2 from Eq1:

```
5r + 8s = 3
– 4r + 8s = –4
---------------
r = 7
```

→ r = 7

Plug into Eq2: 4(7) + 8s = –4 → 28 + 8s = –4 → 8s = –32 → s = –4

✔ Solution: (7, –4)

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## ✔ Problem 6:
–6c – 2d = 8
–8c – d = 6

→ Let’s eliminate d. Multiply second equation by 2 to match d-coefficients:

Eq2 × 2: –16c – 2d = 12
Eq1: –6c – 2d = 8

Now subtract Eq1 from Eq2×2:

```
–16c – 2d = 12
– (–6c – 2d = 8)
-----------------
–10c = 4
```

→ c = 4 / (–10) = –2/5

Plug into Eq2: –8(–2/5) – d = 6 → 16/5 – d = 6 → –d = 6 – 16/5 = 30/5 – 16/5 = 14/5 → d = –14/5

✔ Solution: (–2/5, –14/5)

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## ✔ Problem 7:
m – 4n = 13
m + n = 12

→ m-coefficients are same. Subtract equations:

Eq1: m – 4n = 13
Eq2: m + n = 12

Subtract Eq2 from Eq1:

```
m – 4n = 13
– m + n = 12
------------
–5n = 1
```

→ n = –1/5

Plug into Eq2: m + (–1/5) = 12 → m = 12 + 1/5 = 61/5

✔ Solution: (61/5, –1/5)

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## ✔ Problem 8:
3x + 3y = 18
3x – 4y = –48

→ x-coefficients are same. Subtract equations:

Eq1: 3x + 3y = 18
Eq2: 3x – 4y = –48

Subtract Eq2 from Eq1:

```
3x + 3y = 18
– 3x – 4y = –48
---------------
7y = 66
```

→ y = 66/7

Plug into Eq1: 3x + 3(66/7) = 18 → 3x + 198/7 = 18 → 3x = 18 – 198/7 = 126/7 – 198/7 = –72/7 → x = –24/7

✔ Solution: (–24/7, 66/7)

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## 📝 Final Answers Summary:

1. (4, 12/5)
2. (3, -3)
3. (9, 11)
4. (6, 5)
5. (7, -4)
6. (–2/5, –14/5)
7. (61/5, –1/5)
8. (–24/7, 66/7)

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