Solve {l}{x+8y=9}{x+y=3}{x+y+z=99}

$$\left\{ \begin{array} { l } { x + 8 y = 9 } \\ { x + y = 3 } \\ { x + y + z = 99 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{15}{7} = 2\frac{1}{7} \approx 2.142857143$$

$$y=\frac{6}{7}\approx 0.857142857$$

$$z=96$$

Short Steps Using Substitution

Solve x+8y=9 for x.

$$x=-8y+9$$

Substitute -8y+9 for x in the second and third equation.

$$-8y+9+y=3 -8y+9+y+z=99$$

Solve these equations for y and z respectively.

$$y=\frac{6}{7} z=7y+90$$

Substitute \frac{6}{7} for y in the equation z=7y+90.

$$z=7\times \frac{6}{7}+90$$

Calculate z from z=7\times \frac{6}{7}+90.

$$z=96$$

Substitute \frac{6}{7} for y in the equation x=-8y+9.

$$x=-8\times \frac{6}{7}+9$$

Calculate x from x=-8\times \frac{6}{7}+9.

$$x=\frac{15}{7}$$

The system is now solved.

$$x=\frac{15}{7} y=\frac{6}{7} z=96$$