Question

Consider the equations y = |x − 1| and y = 3x + 2.



The approximate solution of this system of equations is
a.(-0.2, 1.2)
b.(0.2, -1.2)
c.(0.2, 1.2)
d.(-0.2, -1.2)

Answer 1

From the first equation: y^2 = x^2 - 2x + 1
From the second equation: y^2 = 9x^2 + 12x + 4
Subtracting one from the other:
8x^2 + 14x - 3 = 0
Find x, then substitute x into one of the equations to obtain y.

Answer 2

Idk how to do that

Answer 3

@LollyLau

Answer 4

Absolute value means the result must be non-negative.
By squaring it we remove the absolute value symbol and replace it with a square instead, which is easier to work with.
The equation of order two is solved with the quadratic equation.

Answer 5

I forget how to do quadratic equation :( math used to be my #1 subject and I had a seizure so I forgot everything in life

Answer 6

\[x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}\]

Answer 7

._. I don't understand any of that

Answer 8

Alternatively, you can graph the two equations and read the values of the intersections.

Answer 9

HI!!

Answer 10

ugh :/ I still don't understand

Answer 11

ok forget the quadratic equation

Answer 12

alright that's forgotten

Answer 13

if \(x>1\) then \(|x-1|=x-1\) solve \[x-1=3x+2\] in three steps

Answer 14

I think I solved it backwards cause I got -1.5

Answer 15

\[x-1=3x+2\\
-1=2x+2\\
-3=2x\\
-\frac{3}{2}=x\] but that is not a solution, because we assumed that \(x>1\)and it isn't so now one more step

Answer 16

if \(x<1\) then \(|x-1|=1-x\) solve \[1-x=3x+2\]

Answer 17

let me know what you get

Answer 18

pretty positive I did it totally wrong I got -3.3

Answer 19

add \(x\) to both sides
then subtract 2 from both sides, then divide both sides by 4

Answer 20

-.25?

Answer 21

yes

Answer 22

awesome :DD

Answer 23

which for some reason they rounded to \(-.2\)

Answer 24

Someone please help with this last part?