Expert: Scott A Wilson - 11/29/2011

Question
Solve this system by the addition methrod
(2/5)x - (3/2)y=1
(1/3)x +(1/2)y=2

Answer
If we say equation 1 is [1] 2x/5 - 3y/2 = 1 and equation 2 is [2] x/3 + y/2 = 2,
we can compute [1]+3[2] to eliminate y.

That is 2x/5 - 3y/2 + 3(x/3 + y/2) = 2x/5 + x - 3y/2 + 3y/2 = 2x/5 + 5x/5 = 7x/5 on the left and 1 + 3(2) = 1 + 6 = 7 on the right.

Since that gives 7x/5 = 7, multiply both side by 5/7.  That gives x = (5/7)7 = 5, so x is 5.
Putting that in [1] gives 2 - 3y/2 = 1; add -2 to both sides gives -3y/2 = -1; this means
y = 2/3.

Checking out [2] x/3 + y/2 = 2 with x = 5 and y = 2/3 giveds 5/3 + 1/3 = 6/3,
and 6/3 is the same as 2, so the answer is correct.