Solve the simultaneous equations. x^2 + y^2 = 29. y-x = 3

The 2 equations are: 1) x2 + y2 = 29 2) y-x=3 In this kind of simultaneous equation, you cannot take it away from each other. Instead, you need to substitue. To do so, rearrange the 2nd equation to make x or y the subject. To make y the subject, we add x on each side: y= 3+ x We can now substitute y into the first equation and solve it. x2 + y2 = 29 x2 + (3+ x)(3+ x) = 29 x2 + x2 + 6x + 9 =29 2x2 + 6x -20 = 0 x2 + 3x -10 = 0 (x-2)(x+5) = 0 So x can be: x-2 = 0   or   x+5 = 0 x = 2             x = -5 This means y can be: y-x = 3 so y = 3 + x When x = 2, y = 3+2 = 5. When x = -5, y = 3+(-5) = -2.

PS
Answered by Pallave S. Maths tutor

13350 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the quadratic equation (x^2)-x-12=0 (easy), (x^2)-9=0 (special case), (x^2)+5x-13=0 (quadratic formula)


The length of a plank of wood is 80cm to the nearest 1cm. What is the largest and smallest possible value for the actual length of the plank?


find the gradient of the line y=2x^2-12x+16 at the coordinates (5,6)


How do I factorise a quadratic equation?


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:

© 2025 by IXL Learning