Solve {l}{x_{1}+x_{2}=x_{3}}{x_{1}-0.6x_{2}=13}{x_{1}+24x_{3}=130}

$$\left. \begin{array} { l } { x _ { 1 } + x _ { 2 } = x _ { 3 } } \\ { x _ { 1 } - 0.6 x _ { 2 } = 13 } \\ { x _ { 1 } + 24 x _ { 3 } = 130 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=10$$

$$x_{2}=-5$$

$$x_{3}=5$$

Short Steps Using Substitution

Solve x_{1}+x_{2}=x_{3} for x_{3}.

$$x_{3}=x_{1}+x_{2}$$

Substitute x_{1}+x_{2} for x_{3} in the equation x_{1}+24x_{3}=130.

$$x_{1}+24\left(x_{1}+x_{2}\right)=130$$

Solve the second equation for x_{2} and the third equation for x_{1}.

$$x_{2}=\frac{5}{3}x_{1}-\frac{65}{3} x_{1}=\frac{26}{5}-\frac{24}{25}x_{2}$$

Substitute \frac{5}{3}x_{1}-\frac{65}{3} for x_{2} in the equation x_{1}=\frac{26}{5}-\frac{24}{25}x_{2}.

$$x_{1}=\frac{26}{5}-\frac{24}{25}\left(\frac{5}{3}x_{1}-\frac{65}{3}\right)$$

Solve x_{1}=\frac{26}{5}-\frac{24}{25}\left(\frac{5}{3}x_{1}-\frac{65}{3}\right) for x_{1}.

$$x_{1}=10$$

Substitute 10 for x_{1} in the equation x_{2}=\frac{5}{3}x_{1}-\frac{65}{3}.

$$x_{2}=\frac{5}{3}\times 10-\frac{65}{3}$$

Calculate x_{2} from x_{2}=\frac{5}{3}\times 10-\frac{65}{3}.

$$x_{2}=-5$$

Substitute -5 for x_{2} and 10 for x_{1} in the equation x_{3}=x_{1}+x_{2}.

$$x_{3}=10-5$$

Calculate x_{3} from x_{3}=10-5.

$$x_{3}=5$$

The system is now solved.

$$x_{1}=10 x_{2}=-5 x_{3}=5$$