Solve these simultaneous equations. 2x + y = 10 and 3x + 4y = 25.

2x + y = 10 (1)

3x + 4y = 25 (2)

1. Eliminate y.

Method 1:

(1) x 4: 8x + 4y = 40 (1')

(1') - (2): 5x = 15

Method 2:

Rearrange (1): y = 10 - 2x (1")

Substitute (1") into (2): 3x + 4(10 - 2x) = 25 (3)

Expand (3): 3x + 40 - 8x = 25 (3")

Simplify (3"): 5x = 15

2. Find x.

5x = 15

=> x = 3

3. Substitute x into either (1) or (2).

e.g. Sub. into (1): 2(3) + y = 10

=> y = 4

4. Check your answer by substituting into the other equation (2) or (1).

e.g. Sub. into (2): LHS = 3x + 4y = 3(3) + 4(4) = 9 + 16 = 25 = RHS (note: if it does not equal the RHS, go back and find the mistake).

Solution: x = 3 and y = 4

DD

Related Maths GCSE answers

All answers ▸

How to solve the following for x: (2x+3)/(x-4) - (2x-8)(2x+1) = 1


How do you solve an algebraic equation?


How do I expand the following equation (x+4)(x+2)


What are the differences between arithmetic and geometric sequences?