Anyone give me two equations and two variables left for solving 3 variable system

Question

5x+3y-z=5

3x-2y+4z=13

4x+3y-5z=22

find x,y,z

give me two simple equations with 2 variables left to help me start

Answer 1

Step 1 is to look over the equations and decide which of the 3 variables would be easiest to eliminate.

When I look at the equations:

(1) 5x + 3y - z = 5

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

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

I see that the "y" coefficients seem to require the least amount of change to eliminate so I'll choose 1st to eliminate "y".

Notice that if I multiply equation (1) by 2, the result would be that the "3y" would become "6y" and if I multiply equation (2) by 3, the result would be that the "-2y" would become "-6y" and this would allow me to add the equations together and eliminate the "y" because 6y + -6y = 0

So:

Modified (1): 10x + 6y - 2z = 10

Modified (2): 9x - 6y + 12z = 39

And adding these together results in:

10x + 6y - 2z = 10

9x - 6y + 12z = 39

--------------------------

19x +10z = 49

So, we can make that Equation 4:

(4) 19x + 10z = 49

Now, repeat with Equations (1) and (3) except the only thing I need to do to eliminate "y" is multiply one of the equations by "-1". I'll choose to multiply (3) by -1:

(1) 5x + 3y - z = 5

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

And adding these together gives us:

5x + 3y - z = 5

-4x - 3y + 5z = -22

------------------------

x + 4z = -17

So we can make this equation 5:

(5) x + 4z = -17

Now, you have two equations in two unknowns:

(4) 19x + 10z = 49

(5) x + 4z = -17

From here, you can eliminate "x" by multiplying (5) by "-19" and adding the result to (4).