Solve {l}{x+y-z=3}{x-y+2z=15}{6x-2y=3}

$$\left. \begin{array} { l } { x + y - z = 3 } \\ { x - y + 2 z = 15 } \\ { 6 x - 2 y = 3 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{15}{4} = 3\frac{3}{4} = 3.75$$

$$y = \frac{39}{4} = 9\frac{3}{4} = 9.75$$

$$z = \frac{21}{2} = 10\frac{1}{2} = 10.5$$

Short Steps Using Substitution

Solve x+y-z=3 for x.

$$x=-y+z+3$$

Substitute -y+z+3 for x in the second and third equation.

$$-y+z+3-y+2z=15 6\left(-y+z+3\right)-2y=3$$

Solve these equations for y and z respectively.

$$y=-6+\frac{3}{2}z z=-\frac{5}{2}+\frac{4}{3}y$$

Substitute -6+\frac{3}{2}z for y in the equation z=-\frac{5}{2}+\frac{4}{3}y.

$$z=-\frac{5}{2}+\frac{4}{3}\left(-6+\frac{3}{2}z\right)$$

Solve z=-\frac{5}{2}+\frac{4}{3}\left(-6+\frac{3}{2}z\right) for z.

$$z=\frac{21}{2}$$

Substitute \frac{21}{2} for z in the equation y=-6+\frac{3}{2}z.

$$y=-6+\frac{3}{2}\times \frac{21}{2}$$

Calculate y from y=-6+\frac{3}{2}\times \frac{21}{2}.

$$y=\frac{39}{4}$$

Substitute \frac{39}{4} for y and \frac{21}{2} for z in the equation x=-y+z+3.

$$x=-\frac{39}{4}+\frac{21}{2}+3$$

Calculate x from x=-\frac{39}{4}+\frac{21}{2}+3.

$$x=\frac{15}{4}$$

The system is now solved.

$$x=\frac{15}{4} y=\frac{39}{4} z=\frac{21}{2}$$