Solve {l}{x+y+5z=10}{2x+5y+2z=20}{100x+2y+z=30}

$$\left\{ \begin{array} { l } { x + y + 5 z = 10 } \\ { 2 x + 5 y + 2 z = 20 } \\ { 100 x + 2 y + z = 30 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{500}{2281}\approx 0.219202104$$

$$y = \frac{7760}{2281} = 3\frac{917}{2281} \approx 3.402016659$$

$$z = \frac{2910}{2281} = 1\frac{629}{2281} \approx 1.275756247$$

Short Steps Using Substitution

Solve x+y+5z=10 for x.

$$x=-y-5z+10$$

Substitute -y-5z+10 for x in the second and third equation.

$$2\left(-y-5z+10\right)+5y+2z=20 100\left(-y-5z+10\right)+2y+z=30$$

Solve these equations for y and z respectively.

$$y=\frac{8}{3}z z=\frac{970}{499}-\frac{98}{499}y$$

Substitute \frac{8}{3}z for y in the equation z=\frac{970}{499}-\frac{98}{499}y.

$$z=\frac{970}{499}-\frac{98}{499}\times \frac{8}{3}z$$

Solve z=\frac{970}{499}-\frac{98}{499}\times \frac{8}{3}z for z.

$$z=\frac{2910}{2281}$$

Substitute \frac{2910}{2281} for z in the equation y=\frac{8}{3}z.

$$y=\frac{8}{3}\times \frac{2910}{2281}$$

Calculate y from y=\frac{8}{3}\times \frac{2910}{2281}.

$$y=\frac{7760}{2281}$$

Substitute \frac{7760}{2281} for y and \frac{2910}{2281} for z in the equation x=-y-5z+10.

$$x=-\frac{7760}{2281}-5\times \frac{2910}{2281}+10$$

Calculate x from x=-\frac{7760}{2281}-5\times \frac{2910}{2281}+10.

$$x=\frac{500}{2281}$$

The system is now solved.

$$x=\frac{500}{2281} y=\frac{7760}{2281} z=\frac{2910}{2281}$$