Systems of Linear Equations -- Two Variables (A) - Free Printable

Let's solve each system of linear equations step by step. We'll use either substitution or elimination methods depending on which is more efficient.

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1.

$$
\begin{align*}
3u + z &= 15 \quad \text{(1)}\\
u + 2z &= 10 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve equation (2) for $ u $:
$$
u = 10 - 2z
$$

Step 2: Substitute into equation (1):
$$
3(10 - 2z) + z = 15 \\
30 - 6z + z = 15 \\
30 - 5z = 15 \\
-5z = -15 \\
z = 3
$$

Step 3: Plug back into $ u = 10 - 2z $:
$$
u = 10 - 2(3) = 4
$$

✔ Solution: $ u = 4, z = 3 $

---

2.

$$
\begin{align*}
u + 6y &= 32 \quad \text{(1)}\\
u + 3y &= 17 \quad \text{(2)}
\end{align*}
$$

Step 1: Subtract (2) from (1):
$$
(u + 6y) - (u + 3y) = 32 - 17 \\
3y = 15 \Rightarrow y = 5
$$

Step 2: Plug into (2):
$$
u + 3(5) = 17 \Rightarrow u + 15 = 17 \Rightarrow u = 2
$$

✔ Solution: $ u = 2, y = 5 $

---

3.

$$
\begin{align*}
3c + 4u &= 33 \quad \text{(1)}\\
6c + 3u &= 36 \quad \text{(2)}
\end{align*}
$$

Step 1: Simplify equation (2) by dividing by 3:
$$
2c + u = 12 \Rightarrow u = 12 - 2c
$$

Step 2: Substitute into (1):
$$
3c + 4(12 - 2c) = 33 \\
3c + 48 - 8c = 33 \\
-5c = -15 \Rightarrow c = 3
$$

Step 3: Plug back:
$$
u = 12 - 2(3) = 6
$$

✔ Solution: $ c = 3, u = 6 $

---

4.

$$
\begin{align*}
6u + v &= 18 \quad \text{(1)}\\
5u + 2v &= 22 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1) for $ v $:
$$
v = 18 - 6u
$$

Step 2: Substitute into (2):
$$
5u + 2(18 - 6u) = 22 \\
5u + 36 - 12u = 22 \\
-7u = -14 \Rightarrow u = 2
$$

Step 3: Plug back:
$$
v = 18 - 6(2) = 6
$$

✔ Solution: $ u = 2, v = 6 $

---

5.

$$
\begin{align*}
2a + 2x &= 18 \quad \text{(1)}\\
a + 3x &= 17 \quad \text{(2)}
\end{align*}
$$

Step 1: Divide (1) by 2:
$$
a + x = 9 \Rightarrow a = 9 - x
$$

Step 2: Substitute into (2):
$$
(9 - x) + 3x = 17 \\
9 + 2x = 17 \Rightarrow 2x = 8 \Rightarrow x = 4
$$

Step 3: Plug back:
$$
a = 9 - 4 = 5
$$

✔ Solution: $ a = 5, x = 4 $

---

6.

$$
\begin{align*}
5a + 2v &= 32 \quad \text{(1)}\\
6a + 6v &= 42 \quad \text{(2)}
\end{align*}
$$

Step 1: Simplify (2) by dividing by 6:
$$
a + v = 7 \Rightarrow v = 7 - a
$$

Step 2: Substitute into (1):
$$
5a + 2(7 - a) = 32 \\
5a + 14 - 2a = 32 \\
3a = 18 \Rightarrow a = 6
$$

Step 3: Plug back:
$$
v = 7 - 6 = 1
$$

✔ Solution: $ a = 6, v = 1 $

---

7.

$$
\begin{align*}
2b + v &= 13 \quad \text{(1)}\\
b + v &= 8 \quad \text{(2)}
\end{align*}
$$

Step 1: Subtract (2) from (1):
$$
(2b + v) - (b + v) = 13 - 8 \\
b = 5
$$

Step 2: Plug into (2):
$$
5 + v = 8 \Rightarrow v = 3
$$

✔ Solution: $ b = 5, v = 3 $

---

8.

$$
\begin{align*}
3a + 5u &= 17 \quad \text{(1)}\\
2a + u &= 9 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (2) for $ u $:
$$
u = 9 - 2a
$$

Step 2: Substitute into (1):
$$
3a + 5(9 - 2a) = 17 \\
3a + 45 - 10a = 17 \\
-7a = -28 \Rightarrow a = 4
$$

Step 3: Plug back:
$$
u = 9 - 2(4) = 1
$$

✔ Solution: $ a = 4, u = 1 $

---

✔ Final Answers:

| Problem | Solution |
|--------|---------|
| 1 | $ u = 4, z = 3 $ |
| 2 | $ u = 2, y = 5 $ |
| 3 | $ c = 3, u = 6 $ |
| 4 | $ u = 2, v = 6 $ |
| 5 | $ a = 5, x = 4 $ |
| 6 | $ a = 6, v = 1 $ |
| 7 | $ b = 5, v = 3 $ |
| 8 | $ a = 4, u = 1 $ |

Let me know if you'd like these solutions presented in a table or formatted differently!