Solving Systems of Linear Equations Worksheet | Fun and Engaging ... - Free Printable

Here is the complete solution for all 12 systems of linear equations from Section A and Section B, solved algebraically using either elimination or substitution methods. Each solution includes a clear step-by-step explanation.

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## ✔ Section A

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1)

> 4x + y = 17
> 2x + y = 9

Method: Elimination

Subtract the second equation from the first:

(4x + y) - (2x + y) = 17 - 9
→ 2x = 8
→ x = 4

Plug into second equation:
2(4) + y = 9 → 8 + y = 9 → y = 1

✔ Answer: x = 4, y = 1

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2)

> 2x + y = 7
> 5x - y = 14

Method: Elimination

Add the two equations:

(2x + y) + (5x - y) = 7 + 14
→ 7x = 21
→ x = 3

Plug into first equation:
2(3) + y = 7 → 6 + y = 7 → y = 1

✔ Answer: x = 3, y = 1

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3)

> 3x + 2y = 19
> 2x - 2y = 6

Method: Elimination

Add the two equations:

(3x + 2y) + (2x - 2y) = 19 + 6
→ 5x = 25
→ x = 5

Plug into first equation:
3(5) + 2y = 19 → 15 + 2y = 19 → 2y = 4 → y = 2

✔ Answer: x = 5, y = 2

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4)

> 3x - 4y = 17
> x - 4y = 3

Method: Elimination

Subtract the second equation from the first:

(3x - 4y) - (x - 4y) = 17 - 3
→ 2x = 14
→ x = 7

Plug into second equation:
7 - 4y = 3 → -4y = -4 → y = 1

✔ Answer: x = 7, y = 1

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5)

> 2x + 5y = 24
> 2x + 3y = 16

Method: Elimination

Subtract the second equation from the first:

(2x + 5y) - (2x + 3y) = 24 - 16
→ 2y = 8
→ y = 4

Plug into second equation:
2x + 3(4) = 16 → 2x + 12 = 16 → 2x = 4 → x = 2

✔ Answer: x = 2, y = 4

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6)

> 4x - 3y = 7
> x + 3y = 13

Method: Elimination

Add the two equations:

(4x - 3y) + (x + 3y) = 7 + 13
→ 5x = 20
→ x = 4

Plug into second equation:
4 + 3y = 13 → 3y = 9 → y = 3

✔ Answer: x = 4, y = 3

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## ✔ Section B

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1)

> 2x + y = 4
> 5x + 4y = 7

Method: Substitution

From first equation: y = 4 - 2x

Plug into second equation:
5x + 4(4 - 2x) = 7
→ 5x + 16 - 8x = 7
→ -3x = -9
→ x = 3

Then y = 4 - 2(3) = 4 - 6 = -2

✔ Answer: x = 3, y = -2

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2)

> 14x + 2y = 8
> x + y = 1

Method: Substitution

From second equation: y = 1 - x

Plug into first:
14x + 2(1 - x) = 8
→ 14x + 2 - 2x = 8
→ 12x = 6
→ x = 0.5

Then y = 1 - 0.5 = 0.5

✔ Answer: x = 0.5, y = 0.5

*(Or as fractions: x = 1/2, y = 1/2)*

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3)

> 7x + 8y = 3
> 3x + 4y = 7

Method: Elimination

Multiply second equation by 2 to match coefficients of y:

→ 6x + 8y = 14

Now subtract from first equation:

(7x + 8y) - (6x + 8y) = 3 - 14
→ x = -11

Plug into second equation:
3(-11) + 4y = 7 → -33 + 4y = 7 → 4y = 40 → y = 10

✔ Answer: x = -11, y = 10

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4)

> 3x + 4y = 29
> 4x - 2y = 2

Method: Elimination

Multiply second equation by 2 to eliminate y:

→ 8x - 4y = 4

Now add to first equation:

(3x + 4y) + (8x - 4y) = 29 + 4
→ 11x = 33
→ x = 3

Plug into second original equation:
4(3) - 2y = 2 → 12 - 2y = 2 → -2y = -10 → y = 5

✔ Answer: x = 3, y = 5

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5)

> 6x - 5y = -1
> 3x - y = -2

Method: Substitution

From second equation: y = 3x + 2

Plug into first:
6x - 5(3x + 2) = -1
→ 6x - 15x - 10 = -1
→ -9x = 9
→ x = -1

Then y = 3(-1) + 2 = -3 + 2 = -1

✔ Answer: x = -1, y = -1

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6)

> 3x - 2y = 6
> 5x + 6y = 38

Method: Elimination

Multiply first equation by 3 to match coefficients of y:

→ 9x - 6y = 18

Now add to second equation:

(9x - 6y) + (5x + 6y) = 18 + 38
→ 14x = 56
→ x = 4

Plug into first equation:
3(4) - 2y = 6 → 12 - 2y = 6 → -2y = -6 → y = 3

✔ Answer: x = 4, y = 3

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## 📝 Final Answers Summary

Section A

1. x = 4, y = 1
2. x = 3, y = 1
3. x = 5, y = 2
4. x = 7, y = 1
5. x = 2, y = 4
6. x = 4, y = 3

Section B

1. x = 3, y = -2
2. x = 0.5, y = 0.5
3. x = -11, y = 10
4. x = 3, y = 5
5. x = -1, y = -1
6. x = 4, y = 3

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Let me know if you’d like these explained with graphs or in another method!