I am helping my daughter and this one has me stumped - it is a three-variable equation and their asking her to use "elimination" to solve the problem. The text also states that "If you obtain a false equation, such as 0=1, in any of the steps, then the system has no solution. If you do not obtain a false equation, but obtain an identity such as 0=0, then the system has infinitely many solutions."

#13 in Mcougal Little Algebra 2, pp 182:

Eq#1: 5x + y - z = 6

Eq#2: x + y + z = 2

Eq#3: 3x + y = 4

My steps:

"bounce" EQ#1 of EQ#2 and eliminate the "z" variable:

5x + y - z = 6

x + y + z = 2

____________

6x + 2y = 8 (New EQ#1')

Bounce EQ#1' against EQ#3 and eliminate the x variable:

6x + 2y = 8 (EQ#1')

-2*(3x + y = 4)

yields...

6x + 2 y = 8

-6x - 2y = -8

_____________

0 + 0 = 0 (or 0=0, infinitely many solutions).

Did it several times, comes out the same.

Yet the answer in the back of the book gives (0, 4, -2) as the ordered triplet - I plug this into all three equations, and it works!

What gives? and where am I going wrong?

Thanks.

- Jeff (frustrated dad...)