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

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

Solve for x, y, z

$$x=\beta +5$$

$$y=\beta$$

$$z=-\left(\beta +2\right)$$

Short Steps Using Substitution

Solve 2x-y+3z=4-2\beta for y.

$$y=2x+3z-4+2\beta$$

Substitute 2x+3z-4+2\beta for y in the second and third equation.

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

Solve these equations for x and z respectively.

$$x=-\frac{4}{3}z+\frac{7}{3}-\frac{1}{3}\beta z=-x+3$$

Substitute -\frac{4}{3}z+\frac{7}{3}-\frac{1}{3}\beta for x in the equation z=-x+3.

$$z=-\left(-\frac{4}{3}z+\frac{7}{3}-\frac{1}{3}\beta \right)+3$$

Solve z=-\left(-\frac{4}{3}z+\frac{7}{3}-\frac{1}{3}\beta \right)+3 for z.

$$z=-2-\beta$$

Substitute -2-\beta for z in the equation x=-\frac{4}{3}z+\frac{7}{3}-\frac{1}{3}\beta .

$$x=-\frac{4}{3}\left(-2-\beta \right)+\frac{7}{3}-\frac{1}{3}\beta$$

Calculate x from x=-\frac{4}{3}\left(-2-\beta \right)+\frac{7}{3}-\frac{1}{3}\beta .

$$x=5+\beta$$

Substitute 5+\beta for x and -2-\beta for z in the equation y=2x+3z-4+2\beta .

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

Calculate y from y=2\left(5+\beta \right)+3\left(-2-\beta \right)-4+2\beta .

$$y=\beta$$

The system is now solved.

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