Solve {l}{2x+4z-5y=-3}{2z-2y+x=5}{-4y+x+5z=10}

$$\left. \begin{array} { l } { 2 x + 4 z - 5 y = - 3 } \\ { 2 z - 2 y + x = 5 } \\ { - 4 y + x + 5 z = 10 } \end{array} \right.$$

Solve for x, z, y

$$x = \frac{31}{3} = 10\frac{1}{3} \approx 10.333333333$$

$$y=13$$

$$z = \frac{31}{3} = 10\frac{1}{3} \approx 10.333333333$$

Short Steps Using Substitution

Reorder the equations.

$$2z-2y+x=5 2x+4z-5y=-3 -4y+x+5z=10$$

Solve 2z-2y+x=5 for x.

$$x=-2z+2y+5$$

Substitute -2z+2y+5 for x in the second and third equation.

$$2\left(-2z+2y+5\right)+4z-5y=-3 -4y-2z+2y+5+5z=10$$

Solve these equations for y and z respectively.

$$y=13 z=\frac{2}{3}y+\frac{5}{3}$$

Substitute 13 for y in the equation z=\frac{2}{3}y+\frac{5}{3}.

$$z=\frac{2}{3}\times 13+\frac{5}{3}$$

Calculate z from z=\frac{2}{3}\times 13+\frac{5}{3}.

$$z=\frac{31}{3}$$

Substitute 13 for y and \frac{31}{3} for z in the equation x=-2z+2y+5.

$$x=-2\times \frac{31}{3}+2\times 13+5$$

Calculate x from x=-2\times \frac{31}{3}+2\times 13+5.

$$x=\frac{31}{3}$$

The system is now solved.

$$x=\frac{31}{3} z=\frac{31}{3} y=13$$