Question

For questions 10–11, find the solutions of the system.

10. y = x² + 3x – 4
y = 2x + 2

(a) (–3, 6) and (2, –4)
(b) (–3, –4) and (2, 6)
(c) (–3, –4) and (–2, –2)
(d) no solution

11. y = x² – 2x – 2
y = 4x + 5

(a) (–1, 1) and (–7, –23)
(b) (–1, 1) and (7, 33)
(c) (–1, 33) and (7, 1)
(d) no solution

thankyou!!!

Answer 1

For 10
y = x² + 3x – 4 -(1)
y = 2x + 2 -(2)
(1) - (2)
0 = x² + 3x – 4 -(2x+2)
0 = x² +x -6
Can you solve for x and y from here?

Answer 2

not really

Answer 3

Alright , do you understand so far?

Answer 4

i think

Answer 5

then i'll continue..
0 = x² +x -6
0 =(x+3)(x-2)
x = -3 or x=2

Put x=-3 into y = 2x + 2
y = 2(-3) + 2 = -2

Put x=2 into y = 2x + 2
y = 2(2) + 2 =6
Can you choose the answer for 10 now?

Answer 6

Sorry.. for the middle part, it should be like this

Put x=-3 into y = 2x + 2
y = 2(-3) + 2 = -4

Answer 7

For 11, it's similar to 10
y = x² – 2x – 2 -(1)
y = 4x + 5 -(2)
(1) -(2)
0 = x² – 2x – 2 -(4x + 5)
0 = x² - 6x -7

Can you do it from here then?

Answer 8

A plot is attached.