Solve {l}{text{I}5x+2y+z=19}{piquad2x-3y-2z=10}{text{IIT}10x+y-2z=42}

$$\left. \begin{array} { l } { \text { I } 5 x + 2 y + z = 19 } \\ { \pi \quad 2 x - 3 y - 2 z = 10 } \\ { \text { IIT } 10 x + y - 2 z = 42 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{16}{\pi + 3} \approx 2.605187433$$

$$y = \frac{16 {(\pi - 1)}}{\pi + 3} \approx 5.579250268$$

$$z=\frac{9-13\pi }{\pi +3}\approx -5.184437701$$

Short Steps Using Substitution

Solve 5x+2y+z=19 for z.

$$z=-5x-2y+19$$

Substitute -5x-2y+19 for z in the second and third equation.

$$\pi \times 2x-3y-2\left(-5x-2y+19\right)=10 10x+y-2\left(-5x-2y+19\right)=42$$

Solve these equations for y and x respectively.

$$y=-10x-2\pi x+48 x=4-\frac{1}{4}y$$

Substitute -10x-2\pi x+48 for y in the equation x=4-\frac{1}{4}y.

$$x=4-\frac{1}{4}\left(-10x-2\pi x+48\right)$$

Solve x=4-\frac{1}{4}\left(-10x-2\pi x+48\right) for x.

$$x=16\left(\pi +3\right)^{-1}$$

Substitute 16\left(\pi +3\right)^{-1} for x in the equation y=-10x-2\pi x+48.

$$y=-10\times 16\left(\pi +3\right)^{-1}-2\pi \times 16\left(\pi +3\right)^{-1}+48$$

Calculate y from y=-10\times 16\left(\pi +3\right)^{-1}-2\pi \times 16\left(\pi +3\right)^{-1}+48.

$$y=-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi$$

Substitute -16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi for y and 16\left(\pi +3\right)^{-1} for x in the equation z=-5x-2y+19.

$$z=-5\times 16\left(\pi +3\right)^{-1}-2\left(-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi \right)+19$$

Calculate z from z=-5\times 16\left(\pi +3\right)^{-1}-2\left(-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi \right)+19.

$$z=9\left(3+\pi \right)^{-1}-13\left(3+\pi \right)^{-1}\pi$$

The system is now solved.

$$x=16\left(\pi +3\right)^{-1} y=-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi z=9\left(3+\pi \right)^{-1}-13\left(3+\pi \right)^{-1}\pi$$