Solve the simultaneous equations: y=3x+2, x^2+y^2=20

y=3x+2 x^2+y^2=20 x^2+(3x+2)^2=20 x^2+(3x+2)(3x+2)=20 x^2+(9x^2+6x+6x+4)=20 x^2+(9x^2+12x+4)=20 10x^2+12x+4=20 10x^2+12x-16=0 5x^2+6x-8=0 (5x-4)(x+2)=0 5x-4=0, x+2=0 x=4/5=0.8, x=-2 y=3x+2, y=3x+2 y=3(0.8)+2, y=3(-2)+2 y=4.4, y=-4 When x=0.8, y = 4.4 When x=-2, y=-4

Answered by Daniela C. Maths tutor

13513 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve 3(x-2)=21


how do i calcualte the length of an unknown side of a right angled triangle


Solve the simultaneous equations 5x + 2y = 4 and x - y - 5 =0


Expand the brackets and simplify: 7(2x+3y)-x(14-y)


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 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences