I did the earlier ones with elimination
here is substitution
y = (-x+4)
so
(-x+4) = -2x - 4
x = -8
then go back and get y
y = -x + 4 = 8+4 = 12
Solve the following system of equations using both substitutio and elimination
y=-x+4
y=-2x-4
Do I do what I do with the previous problems
10 answers
for elimination subtract as is
y=-x+4
y=-2x-4
-------------
0 = 1 x + 8
so
x = -8
then get y = 8+4 = 12
y=-x+4
y=-2x-4
-------------
0 = 1 x + 8
so
x = -8
then get y = 8+4 = 12
I know where you get 1x but where did you get the +8 from
On the Elimination part
y = -x + 4
y = -(-8) + 4
y = -(-8) + 4
So you first do 2x-x and get 1x. What do you do after that? Because I see it says 0=1 x + 8 but I still don't know how to get it
y=-x+4
y=-2x-4
------------- SUBTRACT these two equations
0 = -x -(-2x) + 4 - (-4)
0 = -x + 2x + 4 + 4
0 = x + 8
x = -8
y=-2x-4
------------- SUBTRACT these two equations
0 = -x -(-2x) + 4 - (-4)
0 = -x + 2x + 4 + 4
0 = x + 8
x = -8
I think I'm confused because this one is in Slope Intercept Form
Get it now. When you subtract you reverse the sign of the bottom one and add.
So when you plug in the x's as -8 would it be y=-x+4 and y=2(-8)-4