Algebra 1, 5.1: Solving Systems of Linear Equations by Graphing - Free Printable

Let’s solve each system of linear equations by graphing. That means we’ll find where the two lines cross — that point (x, y) is the solution.

We’ll go one problem at a time. For each, I’ll:

1. Find two points for each line (usually using x = 0 and y = 0 or easy values).
2. Plot them mentally (or on paper if you’re following along).
3. See where they intersect.
4. Write the solution as an ordered pair (x, y).

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Problem 10:
y = -x + 3
y = x + 5

First equation: y = -x + 3
- When x = 0 → y = 3 → point (0, 3)
- When x = 3 → y = 0 → point (3, 0)

Second equation: y = x + 5
- When x = 0 → y = 5 → point (0, 5)
- When x = -5 → y = 0 → point (-5, 0)

But let’s pick easier points to see intersection. Try x = -1:

First line: y = -(-1) + 3 = 1 + 3 = 4 → (-1, 4)
Second line: y = -1 + 5 = 4 → (-1, 4) ← same point!

So they meet at (-1, 4). ✔

Check: Plug x = -1 into both:
- First: y = -(-1)+3 = 4 ✔️
- Second: y = -1+5 = 4 ✔️

Solution: (-1, 4)

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Problem 11:
y = (1/2)x + 2
y = -(1/2)x + 4

First line: y = (1/2)x + 2
- x = 0 → y = 2 → (0, 2)
- x = 2 → y = 1 + 2 = 3 → (2, 3)
- x = -2 → y = -1 + 2 = 1 → (-2, 1)

Second line: y = -(1/2)x + 4
- x = 0 → y = 4 → (0, 4)
- x = 2 → y = -1 + 4 = 3 → (2, 3) ← same as above!
- x = -2 → y = 1 + 4 = 5 → (-2, 5)

They both pass through (2, 3). So that’s the intersection.

Check:
First: y = (1/2)(2) + 2 = 1 + 2 = 3 ✔️
Second: y = -(1/2)(2) + 4 = -1 + 4 = 3 ✔️

Solution: (2, 3)

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Problem 12:
3x - 2y = 6
y = -3

This one is easy because the second equation is already solved: y = -3.

Plug y = -3 into first equation:

3x - 2(-3) = 6
→ 3x + 6 = 6
→ 3x = 0
→ x = 0

So solution is (0, -3)

Graphically: y = -3 is a horizontal line. The other line crosses it at x=0.

Check:
First equation: 3(0) - 2(-3) = 0 + 6 = 6 ✔️
Second: y = -3 ✔️

Solution: (0, -3)

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Problem 13:
y = 4x
y = -4x + 8

Set them equal since both equal y:

4x = -4x + 8
Add 4x to both sides:
8x = 8
x = 1

Then y = 4x = 4(1) = 4

So solution: (1, 4)

Check second equation: y = -4(1) + 8 = 4 ✔️

Solution: (1, 4)

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Problem 14:
y = (1/4)x + 3
y = (3/4)x + 5

Set equal:

(1/4)x + 3 = (3/4)x + 5

Subtract (1/4)x from both sides:

3 = (2/4)x + 5
→ 3 = (1/2)x + 5

Subtract 5:

-2 = (1/2)x

Multiply both sides by 2:

x = -4

Now plug into first equation: y = (1/4)(-4) + 3 = -1 + 3 = 2

Check second: y = (3/4)(-4) + 5 = -3 + 5 = 2 ✔️

Solution: (-4, 2)

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Problem 15:
3x - 4y = 7
5x + 2y = 3

We can solve this algebraically to find the intersection point (since graphing might be messy without grid precision).

Let’s use elimination.

Multiply second equation by 2 so the y terms cancel with first:

Original:
Eq1: 3x - 4y = 7
Eq2: 5x + 2y = 3 → multiply by 2: 10x + 4y = 6

Now add Eq1 and new Eq2:

3x - 4y = 7
10x + 4y = 6
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13x + 0 = 13 → x = 1

Plug x = 1 into Eq2: 5(1) + 2y = 3 → 5 + 2y = 3 → 2y = -2 → y = -1

Check in Eq1: 3(1) - 4(-1) = 3 + 4 = 7 ✔️

Solution: (1, -1)

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Final Answer:
10. (-1, 4)
11. (2, 3)
12. (0, -3)
13. (1, 4)
14. (-4, 2)
15. (1, -1)