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

Let's solve each system of equations by substitution, step by step. The goal is to solve one equation for one variable and substitute that expression into the other equation.

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Problem 1:

$$
\begin{align*}
(1)\quad & -7x - 2y = -13 \\
(2)\quad & x - 2y = 11
\end{align*}
$$

Step 1: Solve equation (2) for $ x $:
$$
x = 11 + 2y
$$

Step 2: Substitute into equation (1):
$$
-7(11 + 2y) - 2y = -13 \\
-77 - 14y - 2y = -13 \\
-77 - 16y = -13 \\
-16y = 64 \\
y = -4
$$

Step 3: Plug back into $ x = 11 + 2y $:
$$
x = 11 + 2(-4) = 11 - 8 = 3
$$

✔ Solution: $ \boxed{(3, -4)} $

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Problem 2:

$$
\begin{align*}
(1)\quad & x - 2y = -13 \\
(2)\quad & 4x + 2y = 18
\end{align*}
$$

Step 1: Solve equation (1) for $ x $:
$$
x = -13 + 2y
$$

Step 2: Substitute into equation (2):
$$
4(-13 + 2y) + 2y = 18 \\
-52 + 8y + 2y = 18 \\
-52 + 10y = 18 \\
10y = 70 \\
y = 7
$$

Step 3: Plug back into $ x = -13 + 2y $:
$$
x = -13 + 2(7) = -13 + 14 = 1
$$

✔ Solution: $ \boxed{(1, 7)} $

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Problem 3:

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

Step 1: Since $ y = 7 $, substitute into equation (1):
$$
7x - 2(7) = -7 \\
7x - 14 = -7 \\
7x = 7 \\
x = 1
$$

✔ Solution: $ \boxed{(1, 7)} $

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Problem 4:

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

Step 1: Set equations equal since both equal $ y $:
$$
9x - 9 = 9 \\
9x = 18 \\
x = 2
$$

✔ Solution: $ \boxed{(2, 9)} $

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Problem 5:

$$
\begin{align*}
(1)\quad & -4y - 11x = 36 \\
(2)\quad & 20 = -10x - 10y
\end{align*}
$$

First, simplify equation (2):
$$
-10x - 10y = 20 \Rightarrow x + y = -2 \quad \text{(Divide both sides by -10)}
$$

So now:
$$
(1)\quad -4y - 11x = 36 \\
(2)\quad x + y = -2
$$

Step 1: Solve (2) for $ x $:
$$
x = -2 - y
$$

Step 2: Substitute into (1):
$$
-4y - 11(-2 - y) = 36 \\
-4y + 22 + 11y = 36 \\
7y + 22 = 36 \\
7y = 14 \\
y = 2
$$

Step 3: Plug into $ x = -2 - y $:
$$
x = -2 - 2 = -4
$$

✔ Solution: $ \boxed{(-4, 2)} $

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Problem 6:

$$
\begin{align*}
(1)\quad & 3x + 6y = -9 \\
(2)\quad & 2x + 9y = -26
\end{align*}
$$

Step 1: Simplify equation (1): divide by 3
$$
x + 2y = -3 \Rightarrow x = -3 - 2y
$$

Step 2: Substitute into (2):
$$
2(-3 - 2y) + 9y = -26 \\
-6 - 4y + 9y = -26 \\
-6 + 5y = -26 \\
5y = -20 \\
y = -4
$$

Step 3: Plug into $ x = -3 - 2y $:
$$
x = -3 - 2(-4) = -3 + 8 = 5
$$

✔ Solution: $ \boxed{(5, -4)} $

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Problem 7:

$$
\begin{align*}
(1)\quad & x + 3y = -1 \\
(2)\quad & -3x - 3y = -15
\end{align*}
$$

Step 1: Solve (1) for $ x $:
$$
x = -1 - 3y
$$

Step 2: Substitute into (2):
$$
-3(-1 - 3y) - 3y = -15 \\
3 + 9y - 3y = -15 \\
3 + 6y = -15 \\
6y = -18 \\
y = -3
$$

Step 3: Plug into $ x = -1 - 3y $:
$$
x = -1 - 3(-3) = -1 + 9 = 8
$$

✔ Solution: $ \boxed{(8, -3)} $

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Problem 8:

$$
\begin{align*}
(1)\quad & y = 6x - 11 \\
(2)\quad & 2x + 3y = 7
\end{align*}
$$

Step 1: Substitute (1) into (2):
$$
2x + 3(6x - 11) = 7 \\
2x + 18x - 33 = 7 \\
20x - 33 = 7 \\
20x = 40 \\
x = 2
$$

Step 2: Plug into $ y = 6x - 11 $:
$$
y = 6(2) - 11 = 12 - 11 = 1
$$

✔ Solution: $ \boxed{(2, 1)} $

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Problem 9:

$$
\begin{align*}
(1)\quad & x = y - 4 \\
(2)\quad & 2x + 3y = 6
\end{align*}
$$

Step 1: Substitute (1) into (2):
$$
2(y - 4) + 3y = 6 \\
2y - 8 + 3y = 6 \\
5y - 8 = 6 \\
5y = 14 \\
y = \frac{14}{5}
$$

Step 2: Plug into $ x = y - 4 $:
$$
x = \frac{14}{5} - 4 = \frac{14}{5} - \frac{20}{5} = -\frac{6}{5}
$$

✔ Solution: $ \boxed{\left(-\frac{6}{5}, \frac{14}{5}\right)} $

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Problem 10:

$$
\begin{align*}
(1)\quad & x = 5 \\
(2)\quad & x + y = 12
\end{align*}
$$

Step 1: Substitute $ x = 5 $ into (2):
$$
5 + y = 12 \Rightarrow y = 7
$$

✔ Solution: $ \boxed{(5, 7)} $

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✔ Final Answers:

| Problem | Solution |
|--------|----------|
| 1 | $ (3, -4) $ |
| 2 | $ (1, 7) $ |
| 3 | $ (1, 7) $ |
| 4 | $ (2, 9) $ |
| 5 | $ (-4, 2) $ |
| 6 | $ (5, -4) $ |
| 7 | $ (8, -3) $ |
| 8 | $ (2, 1) $ |
| 9 | $ \left(-\frac{6}{5}, \frac{14}{5}\right) $ |
| 10 | $ (5, 7) $ |

Let me know if you'd like a visual explanation or help checking any steps!