Solve {l}{2x+y+2z=3}{x+y-z=b}{3x+2y-2z=c}

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

Solve for x, y, z

$$x=c-2b$$

$$y=\frac{10b}{3}-\frac{4c}{3}+1$$

$$z=\frac{b}{3}-\frac{c}{3}+1$$

Short Steps Using Substitution

Solve 2x+y+2z=3 for y.

$$y=-2x-2z+3$$

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

$$x-2x-2z+3-z=b 3x+2\left(-2x-2z+3\right)-2z=c$$

Solve these equations for x and z respectively.

$$x=-3z+3-b z=1-\frac{1}{6}x-\frac{1}{6}c$$

Substitute -3z+3-b for x in the equation z=1-\frac{1}{6}x-\frac{1}{6}c.

$$z=1-\frac{1}{6}\left(-3z+3-b\right)-\frac{1}{6}c$$

Solve z=1-\frac{1}{6}\left(-3z+3-b\right)-\frac{1}{6}c for z.

$$z=1+\frac{1}{3}b-\frac{1}{3}c$$

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

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

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

$$x=-2b+c$$

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

$$y=-2\left(-2b+c\right)-2\left(1+\frac{1}{3}b-\frac{1}{3}c\right)+3$$

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

$$y=1+\frac{10}{3}b-\frac{4}{3}c$$

The system is now solved.

$$x=-2b+c y=1+\frac{10}{3}b-\frac{4}{3}c z=1+\frac{1}{3}b-\frac{1}{3}c$$