same method as the last one distribute first
@satellite73 ohh okay thanks
i can check your answer if you like
Okay sounds good, Give me a minute
I got 2x+y=2
lets go slow
what do you get when you remove the partentheses using the distributive law for \[2(x-3)\]?
so now we are at \[y+4=2x-6\]
Do we move all the variables to one side?
yes and no
we don't "move" in math, just add, subtract, multiply, divide
you want the variables on the left or the right?
ok so don't "move" the \(2x\) to the left, "subtract" \(2x\) from both sides
yes, got that. so 2x-y+4=6
then we subtract -4 to both sides? right?
the minus sign goes in front of the \(2x\) not behind it \[-2x+y+4=-6\]
Ohh okay! I see now.
now you would subtract the \(4x\) from both sides also don't drop the minus sign in front of the \(-6\) on the right
oops i mean subtract \(4\) from both sides, sorry b
Right no worries so, -2x+y=-10
right maybe one more step, maybe not
some people like the coefficient of the x term to be positive if so, just change the sign of everything, go form \[-2x+y=-10\] to \[2x-y=10\] same equation
Okay. I understand this more now