How do you solve the system of equasions x y z-10 2x 4y 2z-32 and-x 8y-3z-42?

Answer 1

Assuming the equations are:

x + y. + z = 10 1

2x + 4y + 2z =32 2

-x +8y - 3z = 42 3

Multiply eq 3 by 2 and add to eq 2:

20y -4z = 116 4

Multiply eq 1 by 2 and subtract it from eq 2:

2y = 12; then y = 6

Substitute in eq 4:

Then -4z = 116 - 120 = -4; z = 1

Substitute in eq 1; then

X = 10 - 6 - 1 = 3

Answer:

X = 3

Y = 6

Z = 1