What do want done ?
I can "solve" #2 :
sub the first into the 2nd ...
3(y-2) - y = 6
3y - 6 - y = 6
2y = 12
y = 6
then x = y-2 = 6-2 = 4
for the 1st and 3rd, I have no clue what the
xt1 and 2xt3 are supposed to be.
Is the t a new variable?
is it t^3 ?
1.
y=xt1
2x+y=7
2.
x=y-2
3x-y=6
3.
y=2xt3
5x-y=-3
1 answer
1 answer