What is the solution to the following system? x+2y+=9; x-y+3z=13; 2z=10 x = 10, y = 2, z = 5 x = –4, y = 2, z = 5 x = 0, y = 2, z = 5 x = 4, y = 6, z = 5

x + 2y + z = 9
x - y + 3z = 13
2z = 10

Solve the 3rd equation for z.
z = 5

Now substitute z = 5 into both the first and seconds equations.


x + 2y + 5 = 9
x - y + 3(5) = 13

x + 2y = 4
x - y = -2

Subtract the first equation from the second equation above.

-3y = -6

y = 2

Now substitute z = 5 and y = -2 into the first original equation, and solve for x.

x + 2(2) + 5 = 9

x + 9 = 9

x = 0

x = 0; y = 2; z = 5

answere is (c)perpendicular