Simultaneous Equations activity - Free Printable

Let's solve each pair of simultaneous equations step by step using either substitution or elimination methods. I'll show the working for each.

---

Q1:

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

Step: Add equations (1) and (2):
$$
(2x + y) + (2x - y) = 6 + 2 \Rightarrow 4x = 8 \Rightarrow x = 2
$$

Substitute $ x = 2 $ into (1):
$$
2(2) + y = 6 \Rightarrow 4 + y = 6 \Rightarrow y = 2
$$

✔ Answer:
- $ x = 2 $
- $ y = 2 $

---

Q2:

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

Solve (1) for $ x $:
$$
x = 5 - 2y
$$

Substitute into (2):
$$
3(5 - 2y) - y = -6 \Rightarrow 15 - 6y - y = -6 \Rightarrow 15 - 7y = -6
\Rightarrow -7y = -21 \Rightarrow y = 3
$$

Now $ x = 5 - 2(3) = 5 - 6 = -1 $

✔ Answer:
- $ x = -1 $
- $ y = 3 $

---

Q3:

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

From (1): $ x = y + 4 $

Substitute into (2):
$$
(y + 4) + 5y = -2 \Rightarrow 6y + 4 = -2 \Rightarrow 6y = -6 \Rightarrow y = -1
$$

Then $ x = -1 + 4 = 3 $

✔ Answer:
- $ x = 3 $
- $ y = -1 $

---

Q4:

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

Subtract (1) from (2):
$$
(2x - 5y) - (2x - y) = -6 - 2 \Rightarrow -4y = -8 \Rightarrow y = 2
$$

Substitute into (1):
$$
2x - 2 = 2 \Rightarrow 2x = 4 \Rightarrow x = 2
$$

✔ Answer:
- $ x = 2 $
- $ y = 2 $

---

Q5:

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

From (1): $ x = 2 - 3y $

Substitute into (2):
$$
2(2 - 3y) + y = 9 \Rightarrow 4 - 6y + y = 9 \Rightarrow 4 - 5y = 9 \Rightarrow -5y = 5 \Rightarrow y = -1
$$

Then $ x = 2 - 3(-1) = 2 + 3 = 5 $

✔ Answer:
- $ x = 5 $
- $ y = -1 $

---

Q6:

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

From (1): $ x = 4 + 3y $

Substitute into (2):
$$
5(4 + 3y) - 2y = 7 \Rightarrow 20 + 15y - 2y = 7 \Rightarrow 20 + 13y = 7 \Rightarrow 13y = -13 \Rightarrow y = -1
$$

Then $ x = 4 + 3(-1) = 4 - 3 = 1 $

✔ Answer:
- $ x = 1 $
- $ y = -1 $

---

Q7:

$$
\begin{align*}
(1)\quad 5x - 3y &= 1 \\
(2)\quad x - y &= 0
\end{align*}
$$

From (2): $ x = y $

Substitute into (1):
$$
5y - 3y = 1 \Rightarrow 2y = 1 \Rightarrow y = \frac{1}{2}
\Rightarrow x = \frac{1}{2}
$$

✔ Answer:
- $ x = \frac{1}{2} $
- $ y = \frac{1}{2} $

---

Q8:

$$
\begin{align*}
(1)\quad 2x - 3y - 2 &= 0 \Rightarrow 2x - 3y = 2 \\
(2)\quad 3x + 8y - 3 &= 0 \Rightarrow 3x + 8y = 3
\end{align*}
$$

Use elimination:

Multiply (1) by 3: $ 6x - 9y = 6 $

Multiply (2) by 2: $ 6x + 16y = 6 $

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

Substitute $ y = 0 $ into (1): $ 2x = 2 \Rightarrow x = 1 $

✔ Answer:
- $ x = 1 $
- $ y = 0 $

---

Q9:

$$
\begin{align*}
(1)\quad x + 9y &= 14 \\
(2)\quad -2x + y &= 10
\end{align*}
$$

From (2): $ y = 2x + 10 $

Substitute into (1):
$$
x + 9(2x + 10) = 14 \Rightarrow x + 18x + 90 = 14 \Rightarrow 19x = -76 \Rightarrow x = -4
$$

Then $ y = 2(-4) + 10 = -8 + 10 = 2 $

✔ Answer:
- $ x = -4 $
- $ y = 2 $

---

Q10:

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

Use elimination:

Multiply (1) by 3: $ 9x - 6y = 18 $

Multiply (2) by 2: $ 4x + 6y = 8 $

Add:
$$
(9x - 6y) + (4x + 6y) = 18 + 8 \Rightarrow 13x = 26 \Rightarrow x = 2
$$

Substitute $ x = 2 $ into (1):
$$
3(2) - 2y = 6 \Rightarrow 6 - 2y = 6 \Rightarrow -2y = 0 \Rightarrow y = 0
$$

✔ Answer:
- $ x = 2 $
- $ y = 0 $

---

✔ Final Answers:

| Question | $ x $ | $ y $ |
|--------|--------|--------|
| Q1 | 2 | 2 |
| Q2 | -1 | 3 |
| Q3 | 3 | -1 |
| Q4 | 2 | 2 |
| Q5 | 5 | -1 |
| Q6 | 1 | -1 |
| Q7 | ½ | ½ |
| Q8 | 1 | 0 |
| Q9 | -4 | 2 |
| Q10 | 2 | 0 |

Let me know if you'd like these formatted for a worksheet!