Question

Solve for x, y, and z

(system of equations with 3 equations)

2x−3y+4z=8, 3x+4y−5z=−4, 4x−5y+6z=12

Answer:
x = __, y = __, z = __

Answer 1

The solution is x=1,y=2,z=3

Explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

`x=(8+3y-4z)/(2)...(4)`

Put equation (4) into equation (2) and (3)

`3((8+3y-4z)/(2))+4y-5z=-4`

Multiply through by;

`3(8+3y-4z)+8y-10z=-8`

Expand;

`24+9y-12z+8y-10z=-8`

Simplify;

`17y-22z=-32...(5)`

Equation (4) in (3)

`4((8+3y-4z)/(2))-5y+6z=12`

`2(8+3y-4z)-5y+6z=12`

`16+6y-8z-5y+6z=12`

`y-2z=-4`

`y=2z-4...(6)`

Put equation (6) into equation (5)

`17(2z-4)-22z=-32`

`34z-68-22z=-32`

`34z-22z=-32+68`

`12z=36`

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

`x=(8+3(2)-4(3))/(2)=1`

The solution is x=1,y=2,z=3