How do you solve these simultaneous equations? 4x + 5y = 8; 2x + 3y = 5

By elimination:(eq1) 4x + 5y = 8(eq2) 2x + 3y = 5By eye you can see that 2x multiplied by 2 is 4x.If we multiply the whole of (eq2) by 2 we get:(eq2.1) 4x + 6y = 10(eq1) 4x + 5y = 8If we subtract the whole of (eq1) from the whole of (eq2.1) component by component, we get:(6y-5y) = (10-8)y = 2.Then if we substitute this value of y back in to either (eq1) or (eq2) (we will use (eq1)):4x + 52 = 84x + 10 = 8 (then subtract 10 from either side)4x = -2x = -2/4 = -0.5By substitution:Rearrange either (eq1) or (eq2) with either x or y as the subject. We will use (eq1) and rearrange with x as the subject:4x + 5y = 8 (subtract 5y from both sides)4x = 8 - 5y (divide both sides by 4)x = (8 - 5y)/4Substitute this expression into the other simultaneous equation, in this case (eq2):2x + 3y = 5 (substitute above expression for x)2 (8 - 5y)/4 + 3y = 5 (multiply both sides by 4)2* (8 - 5y) + 12y = 20 (multiply out the brackets)16 - 10y + 12y = 20 (collect y terms)2y + 16 = 20 (subtract 16 from both sides)2y = 4 (divide both sides by 2)y = 2Substitute this value of y into either (eq1) or (eq2) as shown in the first technique, elimination, to obtain x = -0.5.