linear equations

6x-4y+5z=31
5x+2y+2z=13
x+y+z=2

solve for x, y ,z

to solve for x wouldn't you use the 2nd and 3rd equation?

having problem finding y and z

1--6x-4y+5z=31
2--5x+2y+2z=13
3--x+y+z=2

solve for x, y ,z

to solve for x wouldn't you use the 2nd and 3rd equation?

having problem finding y and z.

1--Multiply (3) by 5 and subtract (2) from the result.
2--Multiply (3) by 6 and subtract 3 from the result
3--You now have two equations in y and z.
4--Multiply either the y or z coefficient by whatever makes their coefficients equal and subtract again.
5--You now have one equation in either y or z to solve for either y or z.
6--Substitute back into the earlier equations and you will have x, y and z.

2--5x+2y+2z=13
3--x+y+z=2

Easier: multiply 3 by -2, then add 'em:
2: 5x + 2y + 2z= 13
3: -2x - 2y - 2z = -4

3x = 9 ; x = 3
Now go back and substitute x = 3; capish?