Solve {l}{5x+y-2z=4}{2x+y=1}{5x+3y-2z=6}

$$\left. \begin{array} { l } { 5 x + y - 2 z = 4 } \\ { 2 x + y = 1 } \\ { 5 x + 3 y - 2 z = 6 } \end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=1$$

$$z = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Short Steps Using Substitution

Solve 5x+y-2z=4 for y.

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

Substitute -5x+2z+4 for y in the second and third equation.

$$2x-5x+2z+4=1 5x+3\left(-5x+2z+4\right)-2z=6$$

Solve these equations for x and z respectively.

$$x=\frac{2}{3}z+1 z=-\frac{3}{2}+\frac{5}{2}x$$

Substitute \frac{2}{3}z+1 for x in the equation z=-\frac{3}{2}+\frac{5}{2}x.

$$z=-\frac{3}{2}+\frac{5}{2}\left(\frac{2}{3}z+1\right)$$

Solve z=-\frac{3}{2}+\frac{5}{2}\left(\frac{2}{3}z+1\right) for z.

$$z=-\frac{3}{2}$$

Substitute -\frac{3}{2} for z in the equation x=\frac{2}{3}z+1.

$$x=\frac{2}{3}\left(-\frac{3}{2}\right)+1$$

Calculate x from x=\frac{2}{3}\left(-\frac{3}{2}\right)+1.

$$x=0$$

Substitute 0 for x and -\frac{3}{2} for z in the equation y=-5x+2z+4.

$$y=-5\times 0+2\left(-\frac{3}{2}\right)+4$$

Calculate y from y=-5\times 0+2\left(-\frac{3}{2}\right)+4.

$$y=1$$

The system is now solved.

$$x=0 y=1 z=-\frac{3}{2}$$