Math
posted by Anonymous on .
Given the following system is consistent
x+y=1
tx+y=t
(1+t)x+2y=3
Find the exact value of t.

subtract the first from the second
xt  x = t1
x(t1) = t1
x = 1, if tÂ±1
in 1st:
1+y = 1
y = 0
in the last:
(1+t)(1) + 0 = 3
1+t = 3
t = 2 
Subtracting the first equation from the second gives you:
t1 = (t1)x
Therefore x = 1.
The first equation tells you that y = 0.
The third equation requires that
1 + t = 3
t = 2