Simultaneous Equations Textbook Exercise - Corbettmaths - Free Printable

Let's solve Question 1 from the image step by step, using the elimination method for solving simultaneous equations. We'll go through each part (a) to (o), but since you asked to "solve the problem", I’ll provide a detailed explanation and solution for one example, then summarize how to approach the rest.

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🔍 Understanding the Elimination Method

The elimination method involves:
1. Making the coefficients of one variable (either $x$ or $y$) the same in both equations.
2. Adding or subtracting the equations to eliminate that variable.
3. Solving for the remaining variable.
4. Substituting back to find the other variable.

We’ll use this process on several parts.

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## ✔ Example: Part (a)

(a)
$$
\begin{aligned}
(1)\quad & 6x + y = 18 \\
(2)\quad & 4x + y = 14
\end{aligned}
$$

Step 1: Eliminate one variable

Both equations have $+y$, so we can subtract equation (2) from equation (1):

$$
(6x + y) - (4x + y) = 18 - 14
$$
$$
6x + y - 4x - y = 4
$$
$$
2x = 4 \Rightarrow x = 2
$$

Step 2: Substitute $x = 2$ into one of the original equations

Use equation (2):
$$
4x + y = 14 \Rightarrow 4(2) + y = 14 \Rightarrow 8 + y = 14 \Rightarrow y = 6
$$

✔ Solution: $x = 2$, $y = 6$

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## ✔ Example: Part (b)

(b)
$$
\begin{aligned}
(1)\quad & 4x + 2y = 10 \\
(2)\quad & x + 2y = 7
\end{aligned}
$$

Step 1: Eliminate $y$

Both equations have $+2y$, so subtract (2) from (1):

$$
(4x + 2y) - (x + 2y) = 10 - 7
\Rightarrow 3x = 3 \Rightarrow x = 1
$$

Step 2: Substitute $x = 1$ into (2):

$$
1 + 2y = 7 \Rightarrow 2y = 6 \Rightarrow y = 3
$$

✔ Solution: $x = 1$, $y = 3$

---

## ✔ Example: Part (c)

(c)
$$
\begin{aligned}
(1)\quad & 9x - 4y = 19 \\
(2)\quad & 4x + 4y = 20
\end{aligned}
$$

Step 1: Eliminate $y$

Add equations (1) and (2) — because $-4y + 4y = 0$:

$$
(9x - 4y) + (4x + 4y) = 19 + 20
\Rightarrow 13x = 39 \Rightarrow x = 3
$$

Step 2: Substitute $x = 3$ into (2):

$$
4(3) + 4y = 20 \Rightarrow 12 + 4y = 20 \Rightarrow 4y = 8 \Rightarrow y = 2
$$

✔ Solution: $x = 3$, $y = 2$

---

## ✔ Example: Part (d)

(d)
$$
\begin{aligned}
(1)\quad & 2x + y = 36 \\
(2)\quad & x - y = 9
\end{aligned}
$$

Step 1: Eliminate $y$

Add equations:
$$
(2x + y) + (x - y) = 36 + 9 \Rightarrow 3x = 45 \Rightarrow x = 15
$$

Step 2: Substitute into (2):

$$
15 - y = 9 \Rightarrow y = 6
$$

✔ Solution: $x = 15$, $y = 6$

---

## ✔ Example: Part (e)

(e)
$$
\begin{aligned}
(1)\quad & 6x - 3y = 12 \\
(2)\quad & 4x - 3y = 2
\end{aligned}
$$

Step 1: Eliminate $y$

Subtract (2) from (1):
$$
(6x - 3y) - (4x - 3y) = 12 - 2 \Rightarrow 2x = 10 \Rightarrow x = 5
$$

Step 2: Substitute $x = 5$ into (2):

$$
4(5) - 3y = 2 \Rightarrow 20 - 3y = 2 \Rightarrow -3y = -18 \Rightarrow y = 6
$$

✔ Solution: $x = 5$, $y = 6$

---

## ✔ Example: Part (f)

(f)
$$
\begin{aligned}
(1)\quad & 3x - 6y = 6 \\
(2)\quad & 2x - 6y = 3
\end{aligned}
$$

Step 1: Eliminate $y$

Subtract (2) from (1):
$$
(3x - 6y) - (2x - 6y) = 6 - 3 \Rightarrow x = 3
$$

Step 2: Substitute $x = 3$ into (2):

$$
2(3) - 6y = 3 \Rightarrow 6 - 6y = 3 \Rightarrow -6y = -3 \Rightarrow y = 0.5
$$

✔ Solution: $x = 3$, $y = 0.5$

---

## ✔ Example: Part (g)

(g)
$$
\begin{aligned}
(1)\quad & 8x + 7y = 39 \\
(2)\quad & 8x + 2y = 34
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (2) from (1):
$$
(8x + 7y) - (8x + 2y) = 39 - 34 \Rightarrow 5y = 5 \Rightarrow y = 1
$$

Step 2: Substitute $y = 1$ into (2):

$$
8x + 2(1) = 34 \Rightarrow 8x = 32 \Rightarrow x = 4
$$

✔ Solution: $x = 4$, $y = 1$

---

## ✔ Example: Part (h)

(h)
$$
\begin{aligned}
(1)\quad & x + 3y = 38 \\
(2)\quad & x + 6y = 53
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (1) from (2):
$$
(x + 6y) - (x + 3y) = 53 - 38 \Rightarrow 3y = 15 \Rightarrow y = 5
$$

Step 2: Substitute into (1):

$$
x + 3(5) = 38 \Rightarrow x + 15 = 38 \Rightarrow x = 23
$$

✔ Solution: $x = 23$, $y = 5$

---

## ✔ Example: Part (i)

(i)
$$
\begin{aligned}
(1)\quad & 6x + 3y = 48 \\
(2)\quad & 6x + y = 26
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (2) from (1):
$$
(6x + 3y) - (6x + y) = 48 - 26 \Rightarrow 2y = 22 \Rightarrow y = 11
$$

Step 2: Substitute into (2):

$$
6x + 11 = 26 \Rightarrow 6x = 15 \Rightarrow x = 2.5
$$

✔ Solution: $x = 2.5$, $y = 11$

---

## ✔ Example: Part (j)

(j)
$$
\begin{aligned}
(1)\quad & 2x - 4y = 10 \\
(2)\quad & 2x + 3y = 24
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (1) from (2):
$$
(2x + 3y) - (2x - 4y) = 24 - 10 \Rightarrow 7y = 14 \Rightarrow y = 2
$$

Step 2: Substitute into (1):

$$
2x - 4(2) = 10 \Rightarrow 2x - 8 = 10 \Rightarrow 2x = 18 \Rightarrow x = 9
$$

✔ Solution: $x = 9$, $y = 2$

---

## ✔ Example: Part (k)

(k)
$$
\begin{aligned}
(1)\quad & 5x - 2y = 120 \\
(2)\quad & 5x + y = 165
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (1) from (2):
$$
(5x + y) - (5x - 2y) = 165 - 120 \Rightarrow 3y = 45 \Rightarrow y = 15
$$

Step 2: Substitute into (2):

$$
5x + 15 = 165 \Rightarrow 5x = 150 \Rightarrow x = 30
$$

✔ Solution: $x = 30$, $y = 15$

---

## ✔ Example: Part (l)

(l)
$$
\begin{aligned}
(1)\quad & x - 2y = 8 \\
(2)\quad & x - 3y = 3
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (2) from (1):
$$
(x - 2y) - (x - 3y) = 8 - 3 \Rightarrow y = 5
$$

Step 2: Substitute into (1):

$$
x - 2(5) = 8 \Rightarrow x - 10 = 8 \Rightarrow x = 18
$$

✔ Solution: $x = 18$, $y = 5$

---

## ✔ Example: Part (m)

(m)
$$
\begin{aligned}
(1)\quad & 3x + 2y = 54 \\
(2)\quad & 2x - 2y = 16
\end{aligned}
$$

Step 1: Eliminate $y$

Add equations:
$$
(3x + 2y) + (2x - 2y) = 54 + 16 \Rightarrow 5x = 70 \Rightarrow x = 14
$$

Step 2: Substitute into (2):

$$
2(14) - 2y = 16 \Rightarrow 28 - 2y = 16 \Rightarrow -2y = -12 \Rightarrow y = 6
$$

✔ Solution: $x = 14$, $y = 6$

---

## ✔ Example: Part (n)

(n)
$$
\begin{aligned}
(1)\quad & 7x - 4y = 80 \\
(2)\quad & 3x - 4y = -80
\end{aligned}
$$

Step 1: Eliminate $y$

Subtract (2) from (1):
$$
(7x - 4y) - (3x - 4y) = 80 - (-80) \Rightarrow 4x = 160 \Rightarrow x = 40
$$

Step 2: Substitute into (2):

$$
3(40) - 4y = -80 \Rightarrow 120 - 4y = -80 \Rightarrow -4y = -200 \Rightarrow y = 50
$$

✔ Solution: $x = 40$, $y = 50$

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## ✔ Example: Part (o)

(o)
$$
\begin{aligned}
(1)\quad & 5x - 2y = -23 \\
(2)\quad & 5x - 6y = -39
\end{aligned}
$$

Step 1: Eliminate $x$

Subtract (1) from (2):
$$
(5x - 6y) - (5x - 2y) = -39 - (-23) \Rightarrow -4y = -16 \Rightarrow y = 4
$$

Step 2: Substitute into (1):

$$
5x - 2(4) = -23 \Rightarrow 5x - 8 = -23 \Rightarrow 5x = -15 \Rightarrow x = -3
$$

✔ Solution: $x = -3$, $y = 4$

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## 📌 Summary of All Answers

| Part | $x$ | $y$ |
|------|-----|-----|
| (a) | 2 | 6 |
| (b) | 1 | 3 |
| (c) | 3 | 2 |
| (d) | 15 | 6 |
| (e) | 5 | 6 |
| (f) | 3 | 0.5 |
| (g) | 4 | 1 |
| (h) | 23 | 5 |
| (i) | 2.5 | 11 |
| (j) | 9 | 2 |
| (k) | 30 | 15 |
| (l) | 18 | 5 |
| (m) | 14 | 6 |
| (n) | 40 | 50 |
| (o) | -3 | 4 |

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## ✔ Final Notes

All problems were solved using the elimination method:
- Look for common variables with same or opposite signs.
- Add or subtract to eliminate one variable.
- Solve for the remaining variable.
- Back-substitute to find the second.

This method is efficient when coefficients are manageable.

Let me know if you'd like a video explanation or practice worksheet!