<PRE><B>Question 634643</B>
solve the following system of equations
X – 3z = -5
2x- y + 2z= 16
7x – 3y -5z = 19 
There is more than one way to solve this, but this is a pretty simple one.
A combination of substitution and elimination.
:
Rearrange the 1st equation to use for substitution:
x = (3z-5)
:
Multiply the 2nd equation by 3, subtract from the 3rd equation
7x – 3y - 5z = 19
6x - 3y + 6z = 48
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x + 0 - 11z = -29
Replace x with (3z-5) from the 1st equation
(3z-5) - 11z =-29
3z - 11z = -29 + 5
-8z = - 24
z = {{{(-24)/(-8)}}}
z = +3
:
Back to the 1st equation, replace z with 3
x = 3(3) - 5
x = 9 - 5
x = 4
:
Use the original 2nd equation to find y, Replace x and z
2x - y + 2z = 16
2(4) - y + 2(3) = 16 
8 - y + 6 = 16
-y = 16 - 8 - 6
-y = 2
Y has to be positive, multiply both sides by -1
y = -2
:
Our Solution: x=4; y=-2; z=3
:
:
Check these in the 3rd original equation
7x – 3y -5z = 19 
7(4) - 3(-2) - 5(3) = 
28 + 6 - 15 = 19; confirms our solutions</PRE>