Solving Systems of Equations - Graphing Worksheet

Let’s solve each system of equations by graphing. Since we’re doing this without actual graph paper, I’ll find the solution algebraically (by solving the systems) — which is what you’d check your graph against anyway. The “solution” is the point (x, y) where the two lines cross.

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Problem 1:
y = x + 2
y = -2x + 2

Set them equal:
x + 2 = -2x + 2
Add 2x to both sides:
3x + 2 = 2
Subtract 2:
3x = 0 → x = 0
Plug into first equation: y = 0 + 2 = 2
→ Solution: (0, 2)

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Problem 2:
y = (2/3)x
y = (-4/3)x + 6

Set equal:
(2/3)x = (-4/3)x + 6
Add (4/3)x to both sides:
(6/3)x = 6 → 2x = 6 → x = 3
Plug into first: y = (2/3)(3) = 2
→ Solution: (3, 2)

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Problem 3:
y = (1/5)x + 1
y = (-2/5)x - 4

Set equal:
(1/5)x + 1 = (-2/5)x - 4
Add (2/5)x to both sides:
(3/5)x + 1 = -4
Subtract 1:
(3/5)x = -5
Multiply both sides by 5/3:
x = -5 * (5/3)? Wait — no:

Wait: (3/5)x = -5 → multiply both sides by 5: 3x = -25 → x = -25/3? That seems messy. Let me double-check.

Actually:
(1/5)x + 1 = (-2/5)x - 4
Add (2/5)x to both sides:
(3/5)x + 1 = -4
Subtract 1:
(3/5)x = -5
Now multiply both sides by 5: 3x = -25 → x = -25/3 ≈ -8.333...

Then y = (1/5)(-25/3) + 1 = -5/3 + 1 = -5/3 + 3/3 = -2/3

So solution: (-25/3, -2/3) — but that’s unusual for a graphing worksheet. Maybe I made a mistake?

Wait — let me re-solve carefully:

Equation 1: y = (1/5)x + 1
Equation 2: y = (-2/5)x - 4

Set equal:
(1/5)x + 1 = (-2/5)x - 4
Add (2/5)x to both sides:
(3/5)x + 1 = -4
Subtract 1:
(3/5)x = -5
Multiply both sides by 5: 3x = -25 → x = -25/3
Yes, correct.

But maybe the problem expects integer solutions? Let me check if I copied right.

Looking back at original:
3. y = (1/5)x + 1
y = (-2/5)x - 4

Hmm. Maybe it's correct. We'll go with it.

→ Solution: (-25/3, -2/3) or approximately (-8.33, -0.67)

But wait — perhaps there’s a typo in my reading? Let me assume it’s correct and move on.

Actually, let me try plugging x = -5 into both:

Eq1: y = (1/5)(-5) + 1 = -1 + 1 = 0
Eq2: y = (-2/5)(-5) - 4 = 2 - 4 = -2 → not equal.

x = -10:
Eq1: (1/5)(-10)+1 = -2+1=-1
Eq2: (-2/5)(-10)-4=4-4=0 → no.

x = -25/3 ≈ -8.333:
Eq1: (1/5)(-25/3) +1 = -5/3 + 3/3 = -2/3
Eq2: (-2/5)(-25/3) -4 = (50/15) -4 = 10/3 - 12/3 = -2/3 → yes, matches.

So it’s correct. But maybe the worksheet has a typo? Or perhaps I should leave as fraction.

We’ll keep it as (-25/3, -2/3)

But let me check problem 4 — maybe similar issue.

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Problem 4:
y = -2x + 2
y = -4x - 2

Set equal:
-2x + 2 = -4x - 2
Add 4x to both sides:
2x + 2 = -2
Subtract 2:
2x = -4 → x = -2
Plug into first: y = -2(-2) + 2 = 4 + 2 = 6
→ Solution: (-2, 6)

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Problem 5:
y = (2/7)x - 4
y = -x + 5

Set equal:
(2/7)x - 4 = -x + 5
Add x to both sides:
(2/7)x + x - 4 = 5 → (9/7)x - 4 = 5
Add 4:
(9/7)x = 9
Multiply both sides by 7/9:
x = 9 * (7/9) = 7
Then y = -7 + 5 = -2
→ Solution: (7, -2)

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Problem 6:
5x - 4y = 16
-3x + 2y = -12

Let’s solve using substitution or elimination. I’ll use elimination.

First, make coefficients of y match. Multiply second equation by 2:

Eq2 * 2: -6x + 4y = -24

Now add to Eq1:

5x - 4y = 16
-6x + 4y = -24
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-x = -8 → x = 8

Plug into Eq2: -3(8) + 2y = -12 → -24 + 2y = -12 → 2y = 12 → y = 6

→ Solution: (8, 6)

Check in Eq1: 5(8) - 4(6) = 40 - 24 = 16 ✔️

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Problem 7:
y = (1/3)x - 1
5x + 3y = 15

Substitute y from first into second:

5x + 3[(1/3)x - 1] = 15
5x + (3*(1/3)x) - 3 = 15
5x + x - 3 = 15
6x - 3 = 15
6x = 18 → x = 3

Then y = (1/3)(3) - 1 = 1 - 1 = 0
→ Solution: (3, 0)

Check in second equation: 5(3) + 3(0) = 15 ✔️

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Problem 8:
-x + 2y = 6
-3x + 2y = 14

Subtract first equation from second to eliminate y:

(-3x + 2y) - (-x + 2y) = 14 - 6
-3x + 2y + x - 2y = 8
-2x = 8 → x = -4

Plug into first equation: -(-4) + 2y = 6 → 4 + 2y = 6 → 2y = 2 → y = 1

→ Solution: (-4, 1)

Check in second: -3(-4) + 2(1) = 12 + 2 = 14 ✔️

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Now, going back to Problem 3 — I think I did it right, but let me see if maybe the equations were meant to be different? For example, if it was y = (1/5)x + 1 and y = (-2/5)x + 4, then:

(1/5)x + 1 = (-2/5)x + 4 → (3/5)x = 3 → x=5, y=2 — nice numbers.

But as written, it’s -4, so we have to go with fractions.

Alternatively, maybe it’s y = (1/5)x + 1 and y = (-2/5)x - 4 — and the solution is indeed (-25/3, -2/3). But that’s very unusual for a graphing worksheet. Perhaps I misread the sign?

Looking again at user input:

"3. y = \frac{1}{5}x + 1
y = -\frac{2}{5}x - 4"

Yes, that’s what it says. So we’ll keep it.

But let me write it as mixed number or decimal? No, better as fraction.

So final answers:

1. (0, 2)
2. (3, 2)
3. (-25/3, -2/3)
4. (-2, 6)
5. (7, -2)
6. (8, 6)
7. (3, 0)
8. (-4, 1)

Wait — for problem 3, let me double-check once more.

y = (1/5)x + 1
y = (-2/5)x - 4

Set equal:
(1/5)x + 1 = (-2/5)x - 4
Bring all terms to left:
(1/5)x + 1 + (2/5)x + 4 = 0
(3/5)x + 5 = 0
(3/5)x = -5
x = -5 * 5/3 = -25/3 — same as before.

y = (1/5)(-25/3) + 1 = -5/3 + 3/3 = -2/3 — correct.

So it’s right. Maybe the worksheet has non-integer solutions sometimes.

Okay, compiling all:

Final Answer:
1. (0, 2)
2. (3, 2)
3. (-25/3, -2/3)
4. (-2, 6)
5. (7, -2)
6. (8, 6)
7. (3, 0)
8. (-4, 1)