Consider the 3 by 3 system of equations: x-3y-5z=-5, 4x-2y+3z=13, 5x+3y+4z=22

Question

A.) show how to create a 2 by 2 system of equations by eliminating the variable “x” and then state your 2 by 2 system

b.) now solve the 2 by 2 system for “y” and “z”

c.) now state the final solution to the original 3 by 3 system as an ordered triple.

Answer 1

x-3y-5z=-5

4x-2y+3z=13

5x+3y+4z=22

equation 1 becomes x=3y+5z-5

substitute the right side into each x in equation 2 and 3

4(3y+5z-5)-2y+3z=13

5(3y+5z-5)+3y+4z=22

12y+20z-20-2y+3z=13

15y+25z-25+3y+4z=22

combine similar terms

10y+23z-20=13

18y+29z-25=22

10y+23z=33

18y+29z=47

multiply top equation by 9 and multiply bottom equation by 5 to get...

90y+207z=297

90y+145z=235

subtract to get...

62z=62

z=1

10y+23z=33

18y+29z=47

z=1 so substitute 1 for each z to get...

10y+23=33

18y+29=47

10y=10

18y=18

It's obvious that y=1 also.

Go up to the three original equations and choose one of them and replace z by 1 and replace y by 1 also.

I chose x-3y-5z=-5

x-3-5=-5

x-8=-5

x=8-5

x=3

Now check the other two equations with x=3, y=1 and z=1

12-2+3=13, 13=13

15+3+4=22, 22=22

Everything checks out so the solution is (x,y,z)=(3,1,1)