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

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

Solve for x, y, z

$$x=\frac{20}{99}\approx 0.202020202$$

$$y=0$$

$$z = \frac{970}{99} = 9\frac{79}{99} \approx 9.797979798$$

Short Steps Using Substitution

Solve x+y+z=10 for x.

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

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

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

Solve these equations for y and z respectively.

$$y=0 z=\frac{970}{99}-\frac{98}{99}y$$

Substitute 0 for y in the equation z=\frac{970}{99}-\frac{98}{99}y.

$$z=\frac{970}{99}-\frac{98}{99}\times 0$$

Calculate z from z=\frac{970}{99}-\frac{98}{99}\times 0.

$$z=\frac{970}{99}$$

Substitute 0 for y and \frac{970}{99} for z in the equation x=-y-z+10.

$$x=-0-\frac{970}{99}+10$$

Calculate x from x=-0-\frac{970}{99}+10.

$$x=\frac{20}{99}$$

The system is now solved.

$$x=\frac{20}{99} y=0 z=\frac{970}{99}$$