How do you solve the simultaneous equations $y=3x - 5$ and $x^2 + y^2 = 25$ ?

Question

7594 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T23:42:48

Solution pairs of $(x, y)$ are $(0, -5)$ and $(3, 4)$

After plugging first equation to second one,

$x^2+(3x-5)^2=25$

$x^2+9x^2-30x+25=25$

$10x^2-30x+25=25$

$10x^2-30x=0$

$x^2-3x=0$

$x*(x-3)=0$

Hence $x_1=0$ and $x_2=3$ . So $y_1=-5$ and $y_2=4$ . Thus, solution pairs of $(x, y)$ are $(0, -5)$ and $(3, 4)$

How do you solve the simultaneous equations y=3x - 5y=3x - 5 and x^2 + y^2 = 25x^2 + y^2 = 25?