Solve algebraically the simultaneous equations x^2 + y^2 = 25 and y - 3x = 13

Start to solve by substitution: eqn 1 x^2 + y^2 = 25eqn 2 y - 3x = 13 => y = 3x + 13Substitute eqn 2 into 1: x^2 + (3x +13)^2 = 25expand and simplify the equation ...5x^2 + 39x + 72 = 0Factorise the equation: (5x+24)(x+3) = 05x = -24 => x = -24/5x = -3Substitute back into equation 2 to find equivalent y values: x = -3 and y = 4, x = -24/5 and y = -7/5

Answered by Rosita R. Maths tutor

2418 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Work out the value of 81^(-1/4) (Non-Calculator)


Work out 90% of 130


The graph of y = x^2 + 4x - 3 (Graph A) is translated by the vector (3 | 2), find the equation of the new graph (Graph B)


Solve the simultaneous equations; 5x + 2y = 11, 4x – 3y = 18


We're here to help

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