Math Problem #15

Question

x^2+y^2=8 & 3x –y= 4 . Solve the system algebraically and determine if they intersect

Answer 1

Solves the second equation for y. It is 3x - 4 = y

Substitutes into the first:

x^2 + (3x-4)^2 = 8

x^2 + 9x^2 -24x + 16 = 8

10x^2 - 24x + 8 = 0

5x^2 - 12x + 4 = 0

( 5x - 2)(x - 2 ) = 0

5x-2 = 0 ---> x = 2/5 ----> y=3(2/5) - 4 = 6/5 - 4 = 6/5 - 20/5 = -14/5

x-2=0 --> x=2 ---> y = 3(2)-4 = 6-4 = 2

check:

(2/5)^2 + (-14/5)^2 = 4/25 + 196/25 = 200/25 = 8

2^2 + 2^2 = 4+4=8

3(2/5) - -14/5 = 6/5 + 14/5 = 20/5 = 4

3(2) - 2 = 6-2 = 4

so the circle and the line meet at (2/5, -14/5) and (2,2)