What is the solution to the system? 1. x-y + 2 z = -7 2. y + z =1 3. x-2 y - 3 z = 0

You'd find this problem easier to understand and do if you'd please list the defining equations vertically and line up variables:

1. x - 1y + 2 z = -7

2. y + 1z = 1

3. x - 2 y - 3 z = 0 Now eliminate the line numbers:

x - 1y + 2 z = -7

1y + 1 z = 1

x - 2 y - 3 z = 0

Let's use the elimination method to eliminate variable z: Seeing that z = 1 - y, we transform the first equation into 1x - 1y + 2(1-y) = -7

and the third into x - 2y - 3(1-y) = 0.

Simplifying 1x - 1y + 2(1-y) = -7

and x - 2y - 3(1-y) = 0,

we get

1x - 2y - 3 + 3y) = 0 and 1x - 1y + 2 - 2y = -7

which in turn simplify to

1x + y = 3 and 1x - 3y = -9

Having eliminated the variable z, we now focus on eliminating x. Mult. the 1st equation by -1, obtaining -1x - 1y = -3. Add this result to 1x - 3y = -9:

0 - 4y = -12, which tells us that y = 3. Subbing 3 for y in 1x + 1y = 3 tells us that x = 0.

All we have left to determine is the vaue of z.

Borrowing Equation 3, from above, we get x - 2 y - 3 z = 0, and into this equation we substitute x = 0 and y = 3: 0 -2(3) - 3z = 0.

Thus, -3z = 6, and z = -2.

The solution set is (0, 3, -2). You should check this by substitution.

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