Question

find the solutions of the systems
y = x² + 3x – 4
y = 2x + 2


y = x² – 2x – 2
y = 4x + 5

Answer 1

i'm not completely sure, but i'll try to help anyway...

first one:
\[x² + 3x - 4 = 2x + 2\]
\[x²+x-6=0\]

use the quadratic formula:
\[x= (-b \pm \sqrt{b²-4ac})/(2a)\]
\[x=(-1\pm \sqrt{1²-4.(1)(-6)})/(2.(1)\]
\[x=(-1\pm \sqrt{25}) /(2)\]
\[x= (-1\pm5)/2\]

x= 2 or x= -3

maybe you can try to second one yourself now