MathSystem of Equations
posted by Lilly on .
1. Solve for x, y, z in the following system:
2x3y2=z
yx+z=1
y2z=7
The answers are: x=7 y=5 and z=1 however I don't know how to go about this.

From the 3rd: y = 2z7
in the 1st:
2x  3(2z7)  z = 2
2x +6z +21  z = 2
2x + 5z = 19 (#4)
in the 2nd:
2z  7  x + z = 1
x + z = 8
2x + 2z = 16 (#5)
#4  #5 :
3z = 3
z = 1
sub into y = 2z7 > 27 = 5
sub into 2nd:
5  x  1 = 1
x = 7
x = 7
so , x = 7, y = 5 and z = 1