How do you solve these equations by substitution 2x-5y plus 2z equals 16 3x plus 2y-3z equals -19 4x-3y plus 4z equals 18?

Answer 1

If you mean: 2x-5y+2z = 16 and 3x+2y-3z = -19 and 4x-3y+4z = 18 then the solutions are found as follows:-

3(2x-5y+2z = 16) => 6x-15y+6z = 48

2(3x+2y-3z = -19) => 6x+4y-6z = -38

Adding the above: 12x-11y = 10 thus eliminating z

4(3x+2y-3z = -19) => 12x+8y-12z = -76

3(4x-3y+4z = 18) => 12x-9y+12z = 54

Adding the above: 24x-y = -22 thus eliminating z

2(12x-11y = 10) => 24x-22y = 20

1(24x-y = -22) => 24x-y = -22

Subtracting the above: -21y = 42 thus eliminating x

If: -21y = 42 then y = -2

So by substitution: x = -1, y = -2 and z = 4

Check: (2*-1)-(5*-2)+(2*4) = 16

Check: (3*-1)+(2*-2)-(3*4) = -19

Check: (4*-1)-(3*-2)+(4*4) = 18