GCSE Maths Revision Worksheets & Exam Practice - Teachwire - Free Printable

Let's solve each problem step by step.

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1. Solve:

$$
2x - 3y = 11 \quad \text{(1)} \\
5x + 2y = 18 \quad \text{(2)}
$$

We'll use the elimination method.

#### Step 1: Eliminate one variable.
Let’s eliminate $ y $. To do this, make the coefficients of $ y $ the same.

Multiply equation (1) by 2 and equation (2) by 3:

- $ 2(2x - 3y) = 2(11) \Rightarrow 4x - 6y = 22 $
- $ 3(5x + 2y) = 3(18) \Rightarrow 15x + 6y = 54 $

Now add the two equations:
$$
(4x - 6y) + (15x + 6y) = 22 + 54 \\
19x = 76 \\
x = \frac{76}{19} = 4
$$

#### Step 2: Substitute $ x = 4 $ into one of the original equations.

Use equation (1):
$ 2(4) - 3y = 11 $
$ 8 - 3y = 11 $
$ -3y = 3 $
$ y = -1 $

✔ Answer: $ x = 4, y = -1 $

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2. Solve the simultaneous equations:

$$
2x + 3y = -3 \quad \text{(1)} \\
3x - 2y = 28 \quad \text{(2)}
$$

Again, use elimination.

We want to eliminate one variable. Let’s eliminate $ y $. Multiply (1) by 2 and (2) by 3:

- $ 2(2x + 3y) = 2(-3) \Rightarrow 4x + 6y = -6 $
- $ 3(3x - 2y) = 3(28) \Rightarrow 9x - 6y = 84 $

Add them:
$$
(4x + 6y) + (9x - 6y) = -6 + 84 \\
13x = 78 \\
x = \frac{78}{13} = 6
$$

Substitute $ x = 6 $ into equation (1):
$ 2(6) + 3y = -3 $
$ 12 + 3y = -3 $
$ 3y = -15 $
$ y = -5 $

✔ Answer: $ x = 6, y = -5 $

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3. Solve the simultaneous equations:

$$
6x - 2y = 35 \quad \text{(1)} \\
4x - 3y = 9 \quad \text{(2)}
$$

Let’s use elimination again.

We’ll eliminate $ y $. Multiply (1) by 3 and (2) by 2:

- $ 3(6x - 2y) = 3(35) \Rightarrow 18x - 6y = 105 $
- $ 2(4x - 3y) = 2(9) \Rightarrow 8x - 6y = 18 $

Now subtract the second from the first:
$$
(18x - 6y) - (8x - 6y) = 105 - 18 \\
10x = 87 \\
x = \frac{87}{10} = 8.7
$$

Now substitute $ x = 8.7 $ into equation (1):

$ 6(8.7) - 2y = 35 $
$ 52.2 - 2y = 35 $
$ -2y = 35 - 52.2 = -17.2 $
$ y = \frac{-17.2}{-2} = 8.6 $

Alternatively, write as fractions:

$ x = \frac{87}{10}, y = \frac{86}{10} = \frac{43}{5} $

But let’s check if it's better to keep as decimals or fractions.

Wait — let's double-check with exact fractions.

From earlier:

$ 10x = 87 \Rightarrow x = \frac{87}{10} $

Now plug into (1):

$ 6(\frac{87}{10}) - 2y = 35 $
$ \frac{522}{10} - 2y = 35 $
$ 52.2 - 2y = 35 $
$ -2y = -17.2 $
$ y = 8.6 = \frac{86}{10} = \frac{43}{5} $

✔ Answer: $ x = \frac{87}{10}, y = \frac{43}{5} $ or $ x = 8.7, y = 8.6 $

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4. Triangle ABC is isosceles. AB = AC, angle A = x°

Diagram shows triangle ABC with AB = AC, so it's isosceles with AB = AC, meaning base angles at B and C are equal.

Let’s denote:

- $ \angle A = x^\circ $
- Since AB = AC, then $ \angle B = \angle C $

Sum of angles in a triangle = $ 180^\circ $

So:
$$
\angle A + \angle B + \angle C = 180^\circ \\
x + \angle B + \angle B = 180^\circ \\
x + 2\angle B = 180^\circ \\
2\angle B = 180 - x \\
\angle B = \frac{180 - x}{2}
$$

#### (a) Find an expression, in terms of $ x $, for the size of angle B

✔ Answer: $ \angle B = \frac{180 - x}{2} $ degrees

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#### (b) Solve the simultaneous equations:
$$
3p + q = 11 \quad \text{(1)} \\
\text{No second equation given?}
$$

Wait — only one equation is shown: $ 3p + q = 11 $

But it says "solve the simultaneous equations" — but only one equation is provided.

Possibly there's a typo or missing information. But perhaps this was meant to be part of the previous problem?

Wait — looking back at your image description, you said:

> "Solve the simultaneous equations.
> $ 3p + q = 11 $"

But only one equation is listed. That’s not enough to solve for two variables.

Unless this is part (b) of question 4, and maybe it's linked to the triangle?

But no connection is apparent.

Alternatively, maybe the second equation was omitted.

Wait — could it be that this is a typo and the second equation was supposed to be included?

Alternatively, perhaps the second equation is implied from the triangle?

But no — unless there's more context.

Given that, we must assume either:

- There's a mistake in the problem.
- Or the second equation is missing.

But since only one equation is given, we cannot find unique values for $ p $ and $ q $.

We can express $ q $ in terms of $ p $:

From $ 3p + q = 11 $, we get:
$$
q = 11 - 3p
$$

So the solution is all pairs $ (p, q) $ such that $ q = 11 - 3p $

But without a second equation, we can't solve uniquely.

✔ Conclusion: The system has infinitely many solutions unless a second equation is provided.

If this was meant to be a separate problem, please provide the second equation.

But based on what’s given:

✔ Answer: $ q = 11 - 3p $ (infinite solutions)

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✔ Final Answers Summary:

1. $ x = 4, y = -1 $
2. $ x = 6, y = -5 $
3. $ x = \frac{87}{10} = 8.7, y = \frac{43}{5} = 8.6 $
4. (a) $ \angle B = \frac{180 - x}{2} $
(b) $ q = 11 - 3p $ (infinitely many solutions; needs second equation)

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Let me know if you have the second equation for part (b), or if there was a formatting issue!