IM 3 solving systems of 3 equations part 2

Answer 1

x - y + z = -3

x - y - z = -3

5x - 5y + z = -15

Subtract the second equation from the first and add -5 times the first equation to the third:

x - y + z = -3

0 = 0

-4z = 0 So, z = 0.

Since x - y + z = -3 and z = 0, we have x - y = -3. So, x = y - 3.

If we let y = c, then x = c - 3, y = c, z = 0, where c is any real number.

Solution set: (x, y, z) = (c-3, c, 0) = (c, c, 0) + (-3,0,0) = c(1, 1, 0) + (-3, 0, 0), where c is any real number.

There are infinitely many solutions, one for each choice of c:

For example, if c = 0, then (-3, 0, 0) is a solution. If c = 2, then (-1, 2, 0) is another solution.

Answer 2