Solving Systems of Equations by Elimination Worksheets - Math Monks - Free Printable

Let's solve each system of equations using the elimination method. The goal is to eliminate one variable by adding or subtracting the equations (after possibly multiplying them by constants), then solve for the remaining variable.

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Problem ①

$$
\begin{align*}
(1)\quad & 2x + 2y = 6 \\
(2)\quad & 3x - 5 = y \quad \text{(Rewrite as: } y = 3x - 5)
\end{align*}
$$

We can substitute $ y = 3x - 5 $ into the first equation:

$$
2x + 2(3x - 5) = 6 \\
2x + 6x - 10 = 6 \\
8x - 10 = 6 \\
8x = 16 \\
x = 2
$$

Now plug back into $ y = 3x - 5 $:
$$
y = 3(2) - 5 = 6 - 5 = 1
$$

✔ Solution: $ x = 2, y = 1 $

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Problem ②

$$
\begin{align*}
(1)\quad & -20x + 6y = -6 \\
(2)\quad & -10x - 4y = 4
\end{align*}
$$

We want to eliminate one variable. Let's eliminate $ x $. Multiply equation (2) by 2 so that $ -10x \times 2 = -20x $:

$$
(2) \times 2: \quad -20x - 8y = 8
$$

Now subtract from equation (1):

$$
(-20x + 6y) - (-20x - 8y) = -6 - 8 \\
-20x + 6y + 20x + 8y = -14 \\
14y = -14 \Rightarrow y = -1
$$

Now plug $ y = -1 $ into one of the original equations. Use (2):
$$
-10x - 4(-1) = 4 \\
-10x + 4 = 4 \\
-10x = 0 \Rightarrow x = 0
$$

✔ Solution: $ x = 0, y = -1 $

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Problem ③

$$
\begin{align*}
(1)\quad & 3x + 5y = -9 \\
(2)\quad & -6x - 5y = -18
\end{align*}
$$

Notice: coefficients of $ y $ are $ +5 $ and $ -5 $. Add the two equations:

$$
(3x + 5y) + (-6x - 5y) = -9 + (-18) \\
-3x = -27 \Rightarrow x = 9
$$

Plug $ x = 9 $ into (1):
$$
3(9) + 5y = -9 \\
27 + 5y = -9 \\
5y = -36 \Rightarrow y = -\frac{36}{5} = -7.2
$$

Wait! But let’s check if this is correct.

Wait — double-check addition:

Equation (1): $ 3x + 5y = -9 $
Equation (2): $ -6x - 5y = -18 $

Add:
$$
(3x - 6x) + (5y - 5y) = -9 - 18 \Rightarrow -3x = -27 \Rightarrow x = 9
$$

Now plug into (1):
$$
3(9) + 5y = -9 \Rightarrow 27 + 5y = -9 \Rightarrow 5y = -36 \Rightarrow y = -\frac{36}{5}
$$

But let's verify in equation (2):
$$
-6(9) - 5(-36/5) = -54 + 36 = -18 \quad \text{✓ Correct}
$$

So it's valid.

✔ Solution: $ x = 9, y = -\frac{36}{5} $ or $ x = 9, y = -7.2 $

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Problem ④

$$
\begin{align*}
(1)\quad & -2x + 3y = -1 \\
(2)\quad & 2x + 5y = 25
\end{align*}
$$

Add the two equations (the $ x $ terms will cancel):

$$
(-2x + 3y) + (2x + 5y) = -1 + 25 \\
8y = 24 \Rightarrow y = 3
$$

Now plug into (1):
$$
-2x + 3(3) = -1 \Rightarrow -2x + 9 = -1 \Rightarrow -2x = -10 \Rightarrow x = 5
$$

✔ Solution: $ x = 5, y = 3 $

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Problem ⑤

$$
\begin{align*}
(1)\quad & x + 4y = 7 \\
(2)\quad & 4x - 3y = 9
\end{align*}
$$

Eliminate one variable. Let's eliminate $ x $. Multiply equation (1) by 4:

$$
(1) \times 4: \quad 4x + 16y = 28
$$

Now subtract equation (2):

$$
(4x + 16y) - (4x - 3y) = 28 - 9 \\
4x + 16y - 4x + 3y = 19 \\
19y = 19 \Rightarrow y = 1
$$

Now plug into (1):
$$
x + 4(1) = 7 \Rightarrow x = 3
$$

✔ Solution: $ x = 3, y = 1 $

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Problem ⑥

$$
\begin{align*}
(1)\quad & 2x - 5y = 30 \\
(2)\quad & x + 5y = -45
\end{align*}
$$

Notice: $ -5y $ and $ +5y $. Add the equations:

$$
(2x - 5y) + (x + 5y) = 30 + (-45) \\
3x = -15 \Rightarrow x = -5
$$

Plug into (2):
$$
-5 + 5y = -45 \Rightarrow 5y = -40 \Rightarrow y = -8
$$

✔ Solution: $ x = -5, y = -8 $

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Problem ⑦

$$
\begin{align*}
(1)\quad & -7x - 6y = 11 \\
(2)\quad & -8x - 12y = 28
\end{align*}
$$

We can simplify both equations.

First, divide equation (2) by -4:
$$
(-8x - 12y)/(-4) = 28 / (-4) \Rightarrow 2x + 3y = -7
$$

Now equation (1): $ -7x - 6y = 11 $

Let’s make the coefficients of $ y $ match. Multiply simplified (2) by 2:
$$
2(2x + 3y) = 2(-7) \Rightarrow 4x + 6y = -14
$$

Now add to equation (1):

$$
(-7x - 6y) + (4x + 6y) = 11 + (-14) \\
-3x = -3 \Rightarrow x = 1
$$

Now plug into $ 2x + 3y = -7 $:
$$
2(1) + 3y = -7 \Rightarrow 2 + 3y = -7 \Rightarrow 3y = -9 \Rightarrow y = -3
$$

✔ Solution: $ x = 1, y = -3 $

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Problem ⑧

$$
\begin{align*}
(1)\quad & 3x + 4y = -1 \\
(2)\quad & 4x - 3y = 7
\end{align*}
$$

We need to eliminate one variable. Let’s eliminate $ y $. Find LCM of 4 and 3 → 12.

Multiply (1) by 3: $ 9x + 12y = -3 $
Multiply (2) by 4: $ 16x - 12y = 28 $

Now add:
$$
(9x + 12y) + (16x - 12y) = -3 + 28 \\
25x = 25 \Rightarrow x = 1
$$

Now plug into (1):
$$
3(1) + 4y = -1 \Rightarrow 3 + 4y = -1 \Rightarrow 4y = -4 \Rightarrow y = -1
$$

✔ Solution: $ x = 1, y = -1 $

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✔ Final Answers Summary:

| Problem | Solution |
|--------|----------|
| ① | $ x = 2, y = 1 $ |
| ② | $ x = 0, y = -1 $ |
| ③ | $ x = 9, y = -\frac{36}{5} $ |
| ④ | $ x = 5, y = 3 $ |
| ⑤ | $ x = 3, y = 1 $ |
| ⑥ | $ x = -5, y = -8 $ |
| ⑦ | $ x = 1, y = -3 $ |
| ⑧ | $ x = 1, y = -1 $ |

Let me know if you'd like these checked with substitution or graphed!