Past papers › JAMB 1983 · UME › Question 24 Question JAMB General Maths 1983 Objective Quadratics & their graphs Linear & simultaneous equations Quadratics & their graphs, Linear & simultaneous equations
Solve the simultaneous equations x 2 + y − 8 = 0 x^2 + y - 8 = 0 x 2 + y − 8 = 0 y + 5 x − 2 = 0 y + 5x - 2 = 0 y + 5 x − 2 = 0 x x x
A − 28 , 7 -28, 7 − 28 , 7 B 6 , − 28 6, -28 6 , − 28 C 6 , − 1 6, -1 6 , − 1 D − 1 , 7 -1, 7 − 1 , 7 E 3 , 2 3, 2 3 , 2
Worked solution (try it first) Make
y y y the subject of the second equation:
y = 2 − 5 x y = 2 - 5x y = 2 − 5 x .
Substitute into the first:
x 2 + ( 2 − 5 x ) − 8 = 0 x^2 + (2 - 5x) - 8 = 0 x 2 + ( 2 − 5 x ) − 8 = 0 , so
x 2 − 5 x − 6 = 0 x^2 - 5x - 6 = 0 x 2 − 5 x − 6 = 0 .
Factorise:
( x − 6 ) ( x + 1 ) = 0 (x - 6)(x + 1) = 0 ( x − 6 ) ( x + 1 ) = 0 .
So
x = 6 x = 6 x = 6 or
x = − 1 x = -1 x = − 1 , option C.
Watch out
Collect the numbers carefully: 2 − 8 = − 6 2 - 8 = -6 2 − 8 = − 6 + 6 +6 + 6 x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) x^2 - 5x + 6 = (x - 2)(x - 3) x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) Report a problem with this question