SOLUTION: How do you solve this equation by substitution {{{system(2x + y-z=5,x+3y+z=14,3x-y+2z=1)}}}

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Question 212741This question is from textbook
: How do you solve this equation by substitution
system%282x+%2B+y-z=5%2Cx%2B3y%2Bz=14%2C3x-y%2B2z=1%29
This question is from textbook

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
system%282x+%2B+y-z=5%2Cx%2B3y%2Bz=14%2C3x-y%2B2z=1%29

Solve for any letter in any equation.

I'll arbitrarily choose to solve the middle equation for z.

x%2B3y%2Bz=14
z=14-x-3y

Now substitute %2814-x-3y%29 for z in both
the other two equations:

system%282x+%2B+y-z=5%2C3x-y%2B2z=1%29

so we get

system%282x+%2B+y-%2814-x-3y%29=5%2C3x-y%2B2%2814-x-3y%29=1%29

Then we simplify:

system%282x+%2B+y-14%2Bx%2B3y=5%2C3x-y%2B28-2x-6y%29=1%29

system%283x+%2B4y=19%2Cx-7y=-27%29

Now we have it down to 2 equations and 2 unknowns

We solve the 2nd one here for x:

x-7y=-27%29
x=7y-27%29

Then we substitute %287y-27%29 for x in the
other equation:

3x+%2B4y=19

becomes:

3%287y-27%29%2B4y=19

Then we solve for y:

21y-81%2B4y=19%29

25y=100%29

y=4

Now we substitute %284%29 for y in
the equation above solved for x:
 
x-7y=-27%29
x-7%284%29=-27%29
x-28=-27%29
x=1

Finally we substitute %284%29 for y and
%281%29 for x in the equation above solved for z:

z=14-x-3y

z=14-%281%29-3%284%29

z=14-1-12

z=1

So the final solution is %22%28x%2Cy%2Cz%29%22=%22%281%2C4%2C1%29%22

Edwin