Question

Ok let me say this first i need to find two variables for both x and y via substitution method and only substitution method. If possible may someone answer this for me:

6x - y = 14
y = x^2 - 2

Answer 1

what have you tried ?

Answer 2

ok si will show you one sec please

Answer 3

6x - x^2 - 2 = 14
-x^2 + 6x - 16 = 0
?

Answer 4

this is after substituting the equation y = x^2 -2 into 6x - y = 14

Answer 5

y = x^2-2
-y = -x^2 +2
!

Answer 6

so
6x - x^2 + 2 = 14
did you get how

Answer 7

yeah im subbing the y in 6x - y = 14 with x^2 - 2 but after that is as far as i get

Answer 8

i meant you wrote -2
it should be +2

Answer 9

huh why?

Answer 10

y = x^2-2
-y = ... ?

Answer 11

6x - (x^2-2)
6x -x^2- (-2)
6x-x^2+2

Answer 12

oh so 6x - x^2 + 2 = 14

Answer 13

yep, proceed

Answer 14

take x^2 term on right side
infact every term on right side...

Answer 15

ok so here is what i have so far:

6x -(x^2 - 2) = 14
6x - x^2 + 2 = 14
-x^2 + 6x - 12 = 0
(-x-3) (x -2) = 0 - this is the problem i have no clue what im supposed to do in order to get that exact variable

Answer 16

if ab= 0
a=0 or b=0

Answer 17

also (-x-3) (x -2) is incorrect!

Answer 18

ok so what is the correct on?

Answer 19

cause all i know is 2 * 3 or 4 *3 and the latter dosen't seem like the one to help me

Answer 20

and that is to get -12 and the 6x with two of the same numbers

Answer 21

are you sure the question is correct ?
-x^2 + 6x - 12 = 0
does not have any real solutions!

Answer 22

ok here is the variable sry i must have messed up somewhere:

6x - y = 14
y = x^2 - 2

Answer 23

im ust have doen something wrong cause this is an assingment given by a teacher

Answer 24

still same!

Answer 25

or you can show that -x^2 + 6x - 12 = 0
does not have any real solutions
and hence no real solution exist for the system

Answer 26

oh ok thank you well that is all for today thank you very much

Answer 27

welcome ^_^