Junior Cert Maths Samples

_________________________________________ juniorcertsolutions.com ________________________________________ 2016 Question 5 – Higher Paper 2 Solution (a) If (3,–5) is on 5 3 6 0 x y + + = then if we substitute the point into the line we will get 0 = 0 ( ) ( ) 5 3 3 5 6 6 0 + − + = ≠ ∴ (3,–5) is not on 5 3 6 0 x y + + = ( 5 marks ) (b) Rearrange the lines into a pair of simultaneous equations 3 2 7 3 2 7 2 5 2 4 2 10 3 3 x y x y x y x y x x + = + =  + + = × − − − = −  ⇒ = − = − Now find y : ( ) 2 5 2 3 5 6 5 5 6 1 x y y y y + = ⇒ + = ⇒ + = ⇒ = − = − The point of intersection is (3,–1) ( 15 marks ) Comment: Basic algebra.