Question from an Edexcel GCSE Maths Higher Paper (Nov 2018) - Solve the simultaneous equations: 5x + y = 21, x-3y = 9 (3 marks)

So since there are two unknowns in this question, its slightly more difficult than a normal equation where we only have 1 x or y that we're trying to work out. There are two ways to deal with this type of simultaneous equation. It's easiest to start by labelling each equation as 1 or 2
Equations: 5x+y=21 (1) x-3y=9 (2)
Method 1: We start with trying to see if we can get a simple equation for one unknown, and putting it into the other. This is known as the subsitution method. Looking at (2), we can see that by adding 3y to each side, we can get an equation that is equal to the unknown of x. This would be x=9+3y. So now we have an equation for what x is equal to. In (1), we can subsitute this because if we did that, we would have an equation that only has y in it, which we easily know how to solve. By subsituting (2) into (1), we get 5(9+3y)+y = 21. Expand out the brackets as you normally would, to get 45+15y+y=21. Add together all the like terms to get a total equation of 45+16y=21. You'd then solve this as a normal linear equation, starting by taking 45 away from both sides to get 16y= -24. You then divide both sides by 16 to get y=-3/2. Using this value of y, we can subsitute it into either 1 or 2 and solve to get x. x-3(-3/2)=9. x+9/2=9, x=9/2