Question

(These questions are the ones i got wrong on my test I need someone to help me answer them please.) What are the solutions of the system?
y=x^2-3x+2

y=4x-4


B. (1,20) and (6,0)
C. (1,0) and (6,20)
D. No solution

Answer 1

Question?

Answer 2

I just posted it

Answer 3

use substitution method
as the terms in x both equal y, we have:-

4x - 4 = x^2 - 3x + 2

solve this for x

Answer 4

can you help me work it out step by step

Answer 5

ok
first subtract 4x from both sides
-4 = x^2 -3x - 4x + 2
add 4 to both sides:
0 = x^2 -7x + 6
now we have a quadratic equation to solve

can you do this?

Answer 6

no I guess thats why i got most of them wrong

Answer 7

this will factor to a binomial of the form

(x + a)(x + b) = 0

now
x^2 - 7x + 6 = 0

to work out a and b we need 2 numbers whose product = last number 6 and whose sum =
-7

look at possible factors of + 6:

+ 2 * +3
-2 * -3
+1 * +6
-1 * -6

which pair of factors will give us -7 when added?

Answer 8

hold on one minute let me write all this down in my nots on how to do it

Answer 9

*notes

Answer 10

would -1*-7, +1*+7

Answer 11

it would be -1 and -6 ( because -1*-6 = +6 and -1 + -6 = -7)

do you get that?

Answer 12

yes i justed figured since it was on the one for 6 it wouldnt be here

Answer 13

perhaps the * confused you maybe i should have listed the factors as +2 and +3 etc

see what i mean?

Answer 14

yes so whats next

Answer 15

we 've found -1 and -6 so
(x - 1)(x - 6) = 0
when two entities are multiplied to give 0 then either of them must = 0 , so
either x-1 = 0 or x - 6 = 0

giving x = 1 or x = 6

Answer 16

the y values are obtained by plugging in x = 1 and x = 6
into y = 4x - 4

so one of them is y = 4(1) - 4 = 0 and solution is (1.0)
the other one is 4(6) - 4 = ?

Answer 17

would it be (6,20)?

Answer 18

yea

Answer 19

Thanks so much

Answer 20

yw