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)

Related Further Mathematics GCSE answers

All answers ▸

The equation 3x^2 – 5x + 4 = 0 has roots P and Q, find a quadratic equation with the roots (P + 1/2Q) and (Q + 1/2P)


Why does the discriminant b^2-4ac determine the number of roots of the quadratic equation ax^2+bx+c=0?


f'(x) = 3x^2 - 5cos(3x) + 90. Find f(x) and f''(x).


Why is it that when 'transformation A' is followed by 'transformation B', that the combined transformation is BA and not AB?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences