Question

Solve the following system.

x^2 + y^2 = 13

2x - y = 4

Answer 1

@Chlorophyll

Answer 2

(3, 2) and (1/5, -3.6) ??

Answer 3

y=2x-4
x^2+(2x-4)^2=13
x^2+4x^2-16x+16=13
5x^2-16x+3=0
5x^2-15x-x+3=0
5x(x-3)-(x-3)=0
(5x-1)(x-3)=0
x=1/5 or x=3

Answer 4

y = 2x - 4
=> x^2 + ( 2x -4 ) ^2 = 13

Answer 5

so what are the points? (?,?)(?,?)

Answer 6

From @ajprincess's correct solution
x = 3 => y = 2x - 4 = 2
=> ( 3,2)

Answer 7

when x=3
y=2*3-4
=6-4
=2
when x=1/5
y=2*1/5-4
=2/5-4
=-18/5
=-3.6

Answer 8

Similarly for x = 1/5 --> y = ...

Answer 9

but there's two sets

Answer 10

That's why you should look at airprincess's solution, will you!