Question

x2 + y2 = 13
2x - y = 4

Answer 1

System of equations I assume?

Answer 2

yes UGH!! quadratic systems...I hate this stuff

Answer 3

1) x^2 + y^2 = 13
2) 2x - y =4
rearrange the second one to be y=ax + b
y = 2x + 4
sub y = 2x + 4 into the first equation
x^2 + (2x + 4)^2 = 13
x^2 + 4x^2 + 16x + 16 = 13
5x^2 + 16x + 16 - 13 =0
5x^2 + 16x + 3 = 0
(5x + 1)(x+3)

if u have to solve further.
5x + 1 = 0 x + 3 = 0
5x = -1 x = -3
x = -1/5

Answer 4

All right, it is easiest to solve for y in the second equation. This gives us\[y=2x-4\]You can substitute for y in the first equation and you get\[x^2+(2x-4)^2=13\]Expanding the (2x-4)^2 and collecting like terms gives us\[5x^2-16x+3=0\]We can use the quadratic formula with a=5 b=-16 and c=3\[\frac{16 \pm \sqrt{(-16)^2-4(5)(3)}}{2(5)} \]Simplifying we get, \[\frac{16 \pm 14}{10}\]Which is 3 or 1/5

Answer 5

oh okay I was almost there I just was lost at the final equation

Answer 6

follow @SBurchette, i mistakenly forgot the minus. sorry!

Answer 7

the negative answers I got should be positive.

Answer 8

okie dokie THANKS THE BOTH OF YOU!!!!!!!!

Answer 9

No problem =)

Answer 10

:)