Solve the simultaneous equations: x^2 + y^2 = 5 and y = 3x + 1

x2 + y2 = 5 1y = 3x + 1 2Inserting 2 into 1: x2 + (3x +1)2 = 5 Expanding the brackets: x2 + 9x2 + 3x + 3x + 1 = 5Collecting like terms: 10x2 + 6x - 4 = 0Using the quadratic formula: x = (-6 ± √(62 - 4 * 10 * -4))/(2 * 10)Simplifying: x = ( -6 ± √(36 + 160))/20Simplifying further: x = (-6 ± √196)/20Solving for x: x = (-6 ± 14)/20 x1 = (-6 + 14)/20 or x2 = (-6 - 14)/20 x1 = 8/20 or x2 = -20/20 x1 = 0.4 or x2 = -1Substituting x1 and x2 into 2: y1 = 3x1 + 1 or y2 = 3x2 + 1 y1 = 3(0.4) + 1 or y2 = 3(-1) + 1 y1 = 1.2 + 1 or y2 = -3 + 1 y1 = 2.2 or y2 = -2Rewrite answers: x1 = 0.4 and y1= 2.2 x2 = -1 and y2 = -2

Answered by Sophie B. Maths tutor

4006 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Bhavin, Max and Imran share 6000 rupees in the ratios 2 : 3 : 7. Imran then gives 3/5 of his share of the money to Bhavin. What percentage of the 6000 rupees does Bhavin now have? Give your answer correct to the nearest whole number.


solve (y+1)^2 = 4


Fully expand (2x+4)(4x-3).


Solve ((3x-2)/4) - ((2x+5)/3) = ((1-x)/6)


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