Solve the following simultanious equations: zy=28 and 2z-3y=13

zy=28 so y=28/z13=2z-3y13= 2z - (28 x 3)/z13=2z-84/zmultiply each side by z to give 13z = 2z2-84rearange for a quadratic2z2-13z-84=0solve by factorising(2z+8)(z-21/2)z= -4 0r 21/2substitute -4 into both initial equations-4y=28 and -8-3y=13 both give y=-7substitute 21/2 into both inital equations21y/2=28 and 21-3y=13 gives y=8/3(-4,-7) and (21/2,8/3)

OB
Answered by Oliver B. Further Mathematics tutor

2550 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Find the tangent to the equation y=x^2 -2x +4 when x=2


Can you explain rationalising surds?


How can a system of two linear equations be solved?


Let Curve C be f(x)=(1/3)(x^2)+8 and line L be y=3x+k where k is a positive constant. Given that L is tangent to C, find the value of k. (6 marks approx)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning