Question

find the solutions of the system.
10. y = x² + 3x – 4
y = 2x + 2
(–3, 6) and (2, –4)
(–3, –4) and (2, 6)
(–3, –4) and (–2, –2)
no solution
11. y = x² – 2x – 2
y = 4x + 5
(–1, 1) and (–7, –23)
(–1, 1) and (7, 33)
(–1, 33) and (7, 1)
no solution

Answer 1

you have:
\[ \LARGE y = x^2 + 3x – 4\]
\[ \LARGE y =2x+2 \]
substitute the second equation into the first one...
\[ \LARGE 2x+2 = x^2 + 3x – 4\]
\[ \LARGE 2x-3x-x^2+4+2 = 0\]
\[ \LARGE-x^2-x+6=0\]
multiply both sides by -1
\[ \LARGE x^2+x-6=0\]
now you have a quadratic equation there...

use its formula:
\[\LARGE {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
substitute and there you go :)

Answer 2

so what the answer

Answer 3

0

Answer 4

so no solution

Answer 5

nope

Answer 6

(–3, 6) and (2, –4)
(–3, –4) and (2, 6)
(–3, –4) and (–2, –2)
no solution

Answer 7

hey wait...
there are Two solutions !!!
\[\LARGE x_1/2=\frac{-1\pm\sqrt{1^2-4\cdot (-6)}}{2 }\]
\[\LARGE x_1/2=\frac{-1\pm\sqrt{1+24 }}{2 }\]
\[\LARGE x_1/2=\frac{-1\pm5}{2 }\]

!!!

Answer 8

hold on... I think I just Got dizzy ! O_O

Answer 9

what a grind lol

Answer 10

(-3,-4) and (2,6)

this is the solution...

Answer 11

I haven't got dizzy yet :F

Answer 12

lol^

Answer 13

^^