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

2410 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Use the factor theorem to show that (x-1) is a factor of x^3 - 3x^2 -13x + 15


Find the coordinates of any stationary points of the curve y(x)=x^3-3x^2+3x+2


In the expansion of (x-7)(3x**2+kx-3) the coefficient of x**2 is 0. i) Find the value of k ii) Find the coefficient of x. iii) write the fully expanded equation in terms of x


What is the distance between two points with x-coordinates 4 and 8 on the straight line with the equation y=(3/4)x-2


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