These are equations you need to solve at the same time…but there’s two types you could get..one of which is probably the toughest thing in GCSE maths.

Ones Without A Square (easy)

Example: Solve the simultaneous equations 5x = 26 + 3y and 4x + 2y = 34

Write both equations in the form ax + by = c and label them (1) and (2)

5x – 3y = 26 —> (equation 1)
4x + 2y = 34 —> (equation 2)

Multiply the equations so either the y or x values are the same (doesn’t matter if one is negative and other is positive)

(equation 1) x 2: 10x – 6y = 52
(equation 2) x 3: 12x + 6y = 102

Add or subtract the equations to get rid of the value you just matched (if have same sign then subtract, if different multiply)

(equation 1) + (equation 2):  22x + 0y = 154

So 22x = 154

Therefore x = 7

Now just plug that back into one of the original equations to get the y value

(equation 1): 10x – 6y = 52

70 – 6y = 52

-6y = -18

y = 3

Ones With A Square (hard)

Here you have to use a method called substitution.

Example: Solve the simultaneous equations y + x = 7 and x² + y² = 25

Make y or x the subject of the first equation

y = 7 – x

Now the know what y is equal to, we can substitute it into the second equation

x² + y²  = 25

Therefore  x² + (7-x)² = 25

Expand the brackets

x² + 49 – 14x + x² = 25

2x² – 14x + 49 = 25

2x² – 14x + 24 = 0

Thats right…you have a quadratic to solve…told you it was hard!

2x² – 14x + 24 = 0

Luckily everything divides by 2…so go ahead and do that. See the quadratic equations section if you’ve forgotten how to do this

x² – 7x + 12 = 0

(x – 3)(x – 4) = 0

Therefore x = 3 or x = 4

Plug these values back into the non-squared original equation to get the y values

y – x = 7

Therefore y = 7 – x

So when x = 3, y = 4
and when x = 4, y = 3

This is real A* stuff so well done if you have kept up!