Systems Of Equations Solving Two Variable Systems Of Equations By ... - Free Printable

Let's solve each system of equations by elimination, as requested. We'll go through each problem step by step and verify the solutions provided.

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🔷 Problem 1:

$$
\begin{align*}
3x + 2y &= 4 \quad \text{(1)} \\
8x - 3y &= -6 \quad \text{(2)}
\end{align*}
$$

Step 1: Eliminate one variable. Let's eliminate $ y $.
We need to make the coefficients of $ y $ equal in magnitude.

Multiply equation (1) by 3:
$ 9x + 6y = 12 $

Multiply equation (2) by 2:
$ 16x - 6y = -12 $

Now add:
$$
(9x + 6y) + (16x - 6y) = 12 + (-12) \\
25x = 0 \Rightarrow x = 0
$$

Substitute $ x = 0 $ into equation (1):
$$
3(0) + 2y = 4 \Rightarrow 2y = 4 \Rightarrow y = 2
$$

✔ Solution: $ (0, 2) $ — Correct

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🔷 Problem 2:

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

Substitute $ y = -6 $ into equation (1):
$$
-6 = \frac{5}{3}x - 1 \\
-5 = \frac{5}{3}x \\
x = -5 \cdot \frac{3}{5} = -3
$$

✔ Solution: $ (-3, -6) $ — Correct

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🔷 Problem 3:

$$
\begin{align*}
y &= 4x - 10 \quad \text{(1)} \\
y &= \frac{1}{3}x + 1 \quad \text{(2)}
\end{align*}
$$

Set equations equal:
$$
4x - 10 = \frac{1}{3}x + 1 \\
4x - \frac{1}{3}x = 1 + 10 \\
\frac{12}{3}x - \frac{1}{3}x = 11 \\
\frac{11}{3}x = 11 \Rightarrow x = 3
$$

Plug into (1): $ y = 4(3) - 10 = 12 - 10 = 2 $

✔ Solution: $ (3, 2) $ — Correct

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🔷 Problem 4:

$$
\begin{align*}
y &= -\frac{4}{9}x - 3 \quad \text{(1)} \\
y &= -\frac{7}{5}x - 3 \quad \text{(2)}
\end{align*}
$$

Set equal:
$$
-\frac{4}{9}x - 3 = -\frac{7}{5}x - 3 \\
-\frac{4}{9}x = -\frac{7}{5}x \\
\text{Add } \frac{7}{5}x \text{ to both sides: } \left(\frac{7}{5} - \frac{4}{9}\right)x = 0
$$

Find common denominator: $ \frac{63}{45} - \frac{20}{45} = \frac{43}{45}x = 0 \Rightarrow x = 0 $

Then $ y = -\frac{4}{9}(0) - 3 = -3 $

✔ Solution: $ (0, -3) $ — Correct

---

🔷 Problem 5:

$$
\begin{align*}
y &= \frac{5}{2}x - 4 \quad \text{(1)} \\
y &= -x + 3 \quad \text{(2)}
\end{align*}
$$

Set equal:
$$
\frac{5}{2}x - 4 = -x + 3 \\
\frac{5}{2}x + x = 3 + 4 \\
\frac{7}{2}x = 7 \Rightarrow x = 2
$$

Then $ y = -2 + 3 = 1 $

✔ Solution: $ (2, 1) $ — Correct

---

🔷 Problem 6:

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

Substitute $ y = 4 $ into (1):
$$
4 = \frac{7}{4}x - 3 \\
7 = \frac{7}{4}x \Rightarrow x = 4
$$

✔ Solution: $ (4, 4) $ — Correct

---

🔷 Problem 7:

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

Substitute $ y = -3 $:
$$
-3 = \frac{3}{2}x + 3 \\
-6 = \frac{3}{2}x \Rightarrow x = -4
$$

✔ Solution: $ (-4, -3) $ — Correct

---

🔷 Problem 8:

$$
\begin{align*}
y &= 8x - 9 \quad \text{(1)} \\
y &= 7 \quad \text{(2)}
\end{align*}
$$

Substitute $ y = 7 $:
$$
7 = 8x - 9 \Rightarrow 16 = 8x \Rightarrow x = 2
$$

✔ Solution: $ (2, 7) $ — Correct

---

🔷 Problem 9:

$$
\begin{align*}
-5x + 9y &= -12 \quad \text{(1)} \\
3x + 2y &= 22 \quad \text{(2)}
\end{align*}
$$

Use elimination. Make coefficients of $ x $ match.

Multiply (1) by 3: $ -15x + 27y = -36 $
Multiply (2) by 5: $ 15x + 10y = 110 $

Add:
$$
(-15x + 27y) + (15x + 10y) = -36 + 110 \\
37y = 74 \Rightarrow y = 2
$$

Substitute into (2): $ 3x + 2(2) = 22 \Rightarrow 3x + 4 = 22 \Rightarrow 3x = 18 \Rightarrow x = 6 $

✔ Solution: $ (6, 2) $ — Correct

---

🔷 Problem 10:

$$
\begin{align*}
-3x + 2y &= -5 \quad \text{(1)} \\
3x &= 21 \quad \text{(2)}
\end{align*}
$$

From (2): $ x = 7 $

Substitute into (1):
$ -3(7) + 2y = -5 \Rightarrow -21 + 2y = -5 \Rightarrow 2y = 16 \Rightarrow y = 8 $

✔ Solution: $ (7, 8) $ — Correct

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✔ Final Answer:

All the given solutions are correct.

| Problem | Solution |
|--------|----------|
| 1 | (0, 2) |
| 2 | (-3, -6) |
| 3 | (3, 2) |
| 4 | (0, -3) |
| 5 | (2, 1) |
| 6 | (4, 4) |
| 7 | (-4, -3) |
| 8 | (2, 7) |
| 9 | (6, 2) |
| 10 | (7, 8) |

✔️ All verified using elimination or substitution methods.

Let me know if you'd like a visual explanation or graphing help!