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

2z=10

z=10/2=5 so that is easy

substitute z=5 in x-y+3z=13

x-y+3*5=13

x-y=-2

subtract x-y=-2 from x+2y + (?) = 9

3y + (?) = 11

if (?) is nothing, then y=11/3 or 3.67 and x=5/3 or 1.67

but if (?) is 1z, then 3y+5=11

3y=6

y=2

x=-2+y=-2+2=0

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

Answer: x = 0; y = 2; z = 5