Solving Systems of Equations Worksheets with Two Variables - Free Printable

Let’s solve each system of equations using the substitution method.
We’ll go one by one, step by step.

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Problem 1:
> 5x + 2y = 16
> x + 8y = 26

Step 1: Solve one equation for one variable. Let’s solve the second equation for x:

x + 8y = 26 → x = 26 - 8y

Step 2: Substitute this into the first equation:

5(26 - 8y) + 2y = 16
→ 130 - 40y + 2y = 16
→ 130 - 38y = 16
→ -38y = 16 - 130 = -114
→ y = (-114)/(-38) = 3

Step 3: Plug y = 3 back into x = 26 - 8y:

x = 26 - 8(3) = 26 - 24 = 2

✔ Solution: (2, 3)

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Problem 2:
> 4x + y = 7
> -x - 6y = -2

Step 1: Solve first equation for y:

4x + y = 7 → y = 7 - 4x

Step 2: Substitute into second equation:

-x - 6(7 - 4x) = -2
→ -x - 42 + 24x = -2
→ 23x - 42 = -2
→ 23x = 40
→ x = 40/23

Wait — that’s a fraction. Let me double-check.

Original equations:
4x + y = 7 → y = 7 - 4x
Second: -x -6y = -2

Substitute:
-x -6*(7 - 4x) = -2
→ -x -42 + 24x = -2
→ 23x = 40 → x = 40/23

Then y = 7 - 4*(40/23) = 7 - 160/23 = (161 - 160)/23 = 1/23

Hmm… fractions are okay, but let me check if I copied right? The problem says:

“4x + y = 7” and “-x -6y = -2”

Yes. So solution is (40/23, 1/23). But maybe we can leave as improper fractions or mixed numbers? Actually, better to keep as fractions.

But wait — let me verify by plugging back in.

Check first equation: 4*(40/23) + 1/23 = 160/23 + 1/23 = 161/23 = 7 ✔️

Second: -40/23 -6*(1/23) = -40/23 -6/23 = -46/23 = -2 ✔️

So it’s correct. But perhaps the problem expects integer answers? Maybe I misread?

Looking again at original image description — user didn’t provide text, but from context, Problem 2 is:

Actually, looking back — in the initial prompt, the user pasted:

"2) 4x + y = 7
-x - 6y = -2"

Yes. So answer is fractional. We’ll go with it.

✔ Solution: (40/23, 1/23)

But let me write as ordered pair: x = 40/23, y = 1/23

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Problem 3:
> 8p + 7q = 45
> 2p - 7 = q

Step 1: Second equation already gives q in terms of p: q = 2p - 7

Step 2: Substitute into first equation:

8p + 7(2p - 7) = 45
→ 8p + 14p - 49 = 45
→ 22p = 45 + 49 = 94
→ p = 94 / 22 = 47 / 11

Then q = 2*(47/11) - 7 = 94/11 - 77/11 = 17/11

Check:

First equation: 8*(47/11) + 7*(17/11) = (376 + 119)/11 = 495/11 = 45 ✔️

Second: 2*(47/11) - 7 = 94/11 - 77/11 = 17/11 = q ✔️

✔ Solution: (47/11, 17/11)

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Problem 4:
> -5a + b = 8
> 7a + 9b = -32

Step 1: Solve first equation for b:

b = 8 + 5a

Step 2: Substitute into second:

7a + 9(8 + 5a) = -32
→ 7a + 72 + 45a = -32
→ 52a + 72 = -32
→ 52a = -104
→ a = -2

Then b = 8 + 5*(-2) = 8 - 10 = -2

Check:

First: -5*(-2) + (-2) = 10 - 2 = 8 ✔️
Second: 7*(-2) + 9*(-2) = -14 -18 = -32 ✔️

✔ Solution: (-2, -2)

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Problem 5:
> -5 + 2m = 6n
> 4m + 5n - 18 = 0

Rewrite both equations clearly:

Eq1: 2m - 6n = 5 (added 5 to both sides)
Eq2: 4m + 5n = 18

Step 1: Solve Eq1 for m:

2m = 5 + 6n → m = (5 + 6n)/2

Step 2: Substitute into Eq2:

4*((5 + 6n)/2) + 5n = 18
→ 2*(5 + 6n) + 5n = 18
→ 10 + 12n + 5n = 18
→ 10 + 17n = 18
→ 17n = 8
→ n = 8/17

Then m = (5 + 6*(8/17)) / 2 = (5 + 48/17) / 2 = (85/17 + 48/17) / 2 = (133/17) / 2 = 133/34

Check:

Eq1: 2*(133/34) - 6*(8/17) = 266/34 - 48/17 = 133/17 - 48/17 = 85/17 = 5 ✔️
Eq2: 4*(133/34) + 5*(8/17) = 532/34 + 40/17 = 266/17 + 40/17 = 306/17 = 18 ✔️

✔ Solution: (133/34, 8/17)

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Problem 6:
> v = 2 - 4u
> 9u + 2v = 3

Step 1: First equation already gives v in terms of u: v = 2 - 4u

Step 2: Substitute into second:

9u + 2(2 - 4u) = 3
→ 9u + 4 - 8u = 3
→ u + 4 = 3
→ u = -1

Then v = 2 - 4*(-1) = 2 + 4 = 6

Check:

First: v = 2 - 4*(-1) = 6 ✔️
Second: 9*(-1) + 2*6 = -9 + 12 = 3 ✔️

✔ Solution: (-1, 6)

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Problem 7:
> r + 2s = 4
> 3r + s = 1

Step 1: Solve second equation for s:

s = 1 - 3r

Step 2: Substitute into first:

r + 2(1 - 3r) = 4
→ r + 2 - 6r = 4
→ -5r + 2 = 4
→ -5r = 2
→ r = -2/5

Then s = 1 - 3*(-2/5) = 1 + 6/5 = 11/5

Check:

First: -2/5 + 2*(11/5) = -2/5 + 22/5 = 20/5 = 4 ✔️
Second: 3*(-2/5) + 11/5 = -6/5 + 11/5 = 5/5 = 1 ✔️

✔ Solution: (-2/5, 11/5)

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Problem 8:
> 4y + 5z = 0
> 3z + 2y = 53

Let me rewrite them neatly:

Eq1: 4y + 5z = 0
Eq2: 2y + 3z = 53 (reordered)

Step 1: Solve Eq1 for y:

4y = -5z → y = -5z/4

Step 2: Substitute into Eq2:

2*(-5z/4) + 3z = 53
→ -10z/4 + 3z = 53
→ -5z/2 + 3z = 53
Convert 3z to 6z/2:
→ (-5z + 6z)/2 = 53
→ z/2 = 53
→ z = 106

Then y = -5*(106)/4 = -530/4 = -265/2

Check:

Eq1: 4*(-265/2) + 5*106 = -530 + 530 = 0 ✔️
Eq2: 2*(-265/2) + 3*106 = -265 + 318 = 53 ✔️

✔ Solution: (-265/2, 106)

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Now, compiling all final answers:

Final Answer:
1) (2, 3)
2) (40/23, 1/23)
3) (47/11, 17/11)
4) (-2, -2)
5) (133/34, 8/17)
6) (-1, 6)
7) (-2/5, 11/5)
8) (-265/2, 106)