3x + 2y = 6, 5x+3y=11, solve for x and y.

There are two main methods to solve simultaneous equations. Elimination and Substitution. Usually one is easier and quicker than the other, but this is dependent upon the question you face in the exam so I will show you both ways. ELIMINATION  3x + 2y = 6 (Equation 1) 5x + 3y = 11 (Equation 2) Aim: to get a coefficient of a variable to be equal(Equation 1) x 3 9x + 6y = 18(Equation 2) x 2 10x + 6y = 22Aim is now achieved as we have 6y in both equations. I use the acronym STOP to remember what to do next. Same signs Take away Opposite signs Plus.So using this, the coefficient of the variable we made the same in both equations, +6y, has same signs, therefore we take one equation away from the other.I'm going to do (Equation 2) - (Equation 1).10x - 9x = x . 6y- 6y = 0 . 22-18 = 4 Therefore x=4 Don't forget to complete the question by finding y. Substitute your value of x into one of the first equations x --> (Equation 1) . 3(4) + 2y = 6 . 12 + 2y = 6 . 2y = 6-12 . 2y = -6 . y = -3 SUBSTITUTION rearrange equation 1: x = (6-2y)/3 substitute into equation 2: 5((6-2y)/3) + 3y = 11expand out the brackets: 5x6/3 - 10y/3 + 9y/3 = 11 . 10 - y/3 = 11 . 10-11 = y/3 -1 = y/3 y = -3 sub back into rearrange equation for x: x = (6-2(-3))/3 x = 12/3 = 4 Here is a prime example of where one method is easier than the other. I was much more likely to make a mistake in the substitution method compared to the elimination method. Therefore being confident with both methods in the exam will be very valuable to you.