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ok, hold on reaally quick,.
do u know this stuff @LegendarySadist
Yes, I do know it. Do you have any method you want to solve it with? Substitution?
Well then let's just solve it with substitution. First, get x alone in the bottom equation.
so do i add 3 to y or y to 3
Y to 3.
The order doesn't matter. But it should look like \[\large \sf x=3+y\]
so the first one is 4x=2+y?
So now that you know what x is, substitute "3+y" into the first equation. And no, we are only getting x alone for the second equation.
the second equation is done?
No, we'll get back to it.
so with those substitutions how should the first equation look?
So like I said, for the equation \[\large \sf 4x+y=2\] you will substitute\[\large \sf y+3\] in for every x.
Sooo, the first equation should look like \[\large \sf 4(y+3)+y=2\] and then solve for y
k im confused do i minus y from 2
Well we are solving for y. So we want all of the y values on the left side and all the "regular" numbers, on the right. So first we would distribute the 4 onto the y and the 3. Then we would move "regular" numbers to the right side and combine like terms on both sides.
4y+12+y=2 5y=-10 y=-2 right?
Correct. Now we will substitute "y=-2" into either equation and solve for x.
No problem :)