Solving Systems of Equations Worksheets with Two Variables - Free Printable

Let's solve each system of equations using the substitution method as requested. I'll go step by step for each problem.

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1.

$$
\begin{cases}
3x - 2y = 16 \\
x + y = -8
\end{cases}
$$

Step 1: Solve one equation for one variable.
From the second equation:
$$
x + y = -8 \Rightarrow x = -8 - y
$$

Step 2: Substitute into the first equation:
$$
3(-8 - y) - 2y = 16 \\
-24 - 3y - 2y = 16 \\
-24 - 5y = 16 \\
-5y = 40 \\
y = -8
$$

Step 3: Plug back to find $x$:
$$
x = -8 - (-8) = -8 + 8 = 0
$$

✔ Solution: $ (x, y) = (0, -8) $

---

2.

$$
\begin{cases}
2x - y = 7 \\
x + 2y = 1
\end{cases}
$$

Step 1: Solve the first equation for $y$:
$$
2x - y = 7 \Rightarrow y = 2x - 7
$$

Step 2: Substitute into the second equation:
$$
x + 2(2x - 7) = 1 \\
x + 4x - 14 = 1 \\
5x = 15 \\
x = 3
$$

Step 3: Find $y$:
$$
y = 2(3) - 7 = 6 - 7 = -1
$$

✔ Solution: $ (x, y) = (3, -1) $

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3.

$$
\begin{cases}
3p + 2q = 19 \\
2p - r = -6
\end{cases}
$$

Wait — this system has three variables ($p, q, r$), but only two equations. That means we cannot find a unique solution unless more info is given. But since it's labeled as a "system of equations" with two equations, perhaps there's a typo or missing equation?

But let’s check: The problem says “solve systems of equations” and gives two equations with three variables → underdetermined.

We can express in terms of one variable, but no unique solution exists.

However, if this was meant to be two equations with two variables, maybe the third variable is a typo?

Looking at the next problems, they are all two-variable systems.

Possibility: Maybe it's a typo and should be:

$$
\begin{cases}
3p + 2q = 19 \\
2p + q = -6
\end{cases}
$$

Let me assume that, because otherwise it's unsolvable uniquely.

So assume:
$$
\begin{cases}
3p + 2q = 19 \quad \text{(1)}\\
2p + q = -6 \quad \text{(2)}
\end{cases}
$$

Step 1: Solve equation (2) for $q$:
$$
q = -6 - 2p
$$

Step 2: Substitute into (1):
$$
3p + 2(-6 - 2p) = 19 \\
3p -12 -4p = 19 \\
-p = 31 \\
p = -31
$$

Then:
$$
q = -6 - 2(-31) = -6 + 62 = 56
$$

✔ Solution: $ (p, q) = (-31, 56) $

> ⚠️ Note: This assumes the second equation was meant to be $2p + q = -6$, not involving $r$. If $r$ is intended, then the system is incomplete.

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4.

$$
\begin{cases}
3a + b = 14 \\
3a + 8b = -32
\end{cases}
$$

Step 1: Solve first equation for $b$:
$$
b = 14 - 3a
$$

Step 2: Substitute into second equation:
$$
3a + 8(14 - 3a) = -32 \\
3a + 112 - 24a = -32 \\
-21a = -144 \\
a = \frac{144}{21} = \frac{48}{7}
$$

Now find $b$:
$$
b = 14 - 3\left(\frac{48}{7}\right) = 14 - \frac{144}{7} = \frac{98 - 144}{7} = \frac{-46}{7}
$$

✔ Solution: $ \left(a, b\right) = \left(\frac{48}{7}, -\frac{46}{7}\right) $

---

5.

$$
\begin{cases}
-2x + 3m = 46 \\
4m + 3x = 28
\end{cases}
$$

Step 1: Solve first equation for $m$:
$$
-2x + 3m = 46 \Rightarrow 3m = 46 + 2x \Rightarrow m = \frac{46 + 2x}{3}
$$

Step 2: Substitute into second equation:
$$
4\left(\frac{46 + 2x}{3}\right) + 3x = 28 \\
\frac{184 + 8x}{3} + 3x = 28
$$

Multiply both sides by 3:
$$
184 + 8x + 9x = 84 \\
184 + 17x = 84 \\
17x = -100 \\
x = -\frac{100}{17}
$$

Now find $m$:
$$
m = \frac{46 + 2(-100/17)}{3} = \frac{46 - 200/17}{3} = \frac{(782 - 200)/17}{3} = \frac{582}{17 \cdot 3} = \frac{582}{51} = \frac{194}{17}
$$

✔ Solution: $ \left(x, m\right) = \left(-\frac{100}{17}, \frac{194}{17}\right) $

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6.

$$
\begin{cases}
x + 2y = 4n \\
3n + y = 1
\end{cases}
$$

This has three variables: $x, y, n$. Only two equations → again, underdetermined.

But likely, this is a typo and meant to be two variables.

Possibility: Maybe $n$ is a constant? Or maybe it's supposed to be:

$$
\begin{cases}
x + 2y = 4 \\
3x + y = 1
\end{cases}
$$

Assume that typo and proceed.

So:
$$
\begin{cases}
x + 2y = 4 \quad \text{(1)}\\
3x + y = 1 \quad \text{(2)}
\end{cases}
$$

Step 1: Solve (1) for $x$:
$$
x = 4 - 2y
$$

Step 2: Substitute into (2):
$$
3(4 - 2y) + y = 1 \\
12 - 6y + y = 1 \\
12 - 5y = 1 \\
-5y = -11 \\
y = \frac{11}{5}
$$

Then:
$$
x = 4 - 2\left(\frac{11}{5}\right) = 4 - \frac{22}{5} = \frac{20 - 22}{5} = -\frac{2}{5}
$$

✔ Solution: $ \left(x, y\right) = \left(-\frac{2}{5}, \frac{11}{5}\right) $

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7.

$$
\begin{cases}
x + 2z = 4 \\
x - z = 1
\end{cases}
$$

Two equations, two variables: $x$ and $z$

Step 1: Solve second equation for $x$:
$$
x = 1 + z
$$

Step 2: Substitute into first:
$$
(1 + z) + 2z = 4 \\
1 + 3z = 4 \\
3z = 3 \\
z = 1
$$

Then:
$$
x = 1 + 1 = 2
$$

✔ Solution: $ (x, z) = (2, 1) $

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8.

$$
\begin{cases}
4x + 3y = 16 \\
2x - 5y = 10
\end{cases}
$$

Step 1: Solve first equation for $x$:
$$
4x = 16 - 3y \Rightarrow x = \frac{16 - 3y}{4}
$$

Step 2: Substitute into second:
$$
2\left(\frac{16 - 3y}{4}\right) - 5y = 10 \\
\frac{32 - 6y}{4} - 5y = 10 \\
8 - 1.5y - 5y = 10 \\
8 - 6.5y = 10 \\
-6.5y = 2 \\
y = -\frac{2}{6.5} = -\frac{2}{13/2} = -\frac{4}{13}
$$

Now find $x$:
$$
x = \frac{16 - 3(-4/13)}{4} = \frac{16 + 12/13}{4} = \frac{(208 + 12)/13}{4} = \frac{220}{13 \cdot 4} = \frac{55}{13}
$$

✔ Solution: $ \left(x, y\right) = \left(\frac{55}{13}, -\frac{4}{13}\right) $

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✔ Final Answers:

| Problem | Solution |
|--------|----------|
| 1 | $ (0, -8) $ |
| 2 | $ (3, -1) $ |
| 3 | $ (-31, 56) $ *(assuming typo: $2p + q = -6$)* |
| 4 | $ \left(\frac{48}{7}, -\frac{46}{7}\right) $ |
| 5 | $ \left(-\frac{100}{17}, \frac{194}{17}\right) $ |
| 6 | $ \left(-\frac{2}{5}, \frac{11}{5}\right) $ *(assuming typo: $3x + y = 1$)* |
| 7 | $ (2, 1) $ |
| 8 | $ \left(\frac{55}{13}, -\frac{4}{13}\right) $ |

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⚠️ Note on Problems 3 and 6: They contain three variables but only two equations, which makes them unsolvable for unique solutions unless assumptions are made. Based on context, I assumed typos to make them solvable as two-variable systems.

If you have the original image and want to confirm the actual equations, please clarify!

Let me know if you'd like these checked with graphing or elimination method!