Solve {l}{8x_{1}-2x_{2}=15}{x_{2}-5x_{3}=1}{2x_{1}-3x_{3}=4}

$$\left. \begin{array} { l } { 8 x _ { 1 } - 2 x _ { 2 } = 15 } \\ { x _ { 2 } - 5 x _ { 3 } = 1 } \\ { 2 x _ { 1 } - 3 x _ { 3 } = 4 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$x_{2} = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x_{3}=\frac{1}{2}=0.5$$

Short Steps Using Substitution

Reorder the equations.

$$x_{2}-5x_{3}=1 8x_{1}-2x_{2}=15 2x_{1}-3x_{3}=4$$

Solve x_{2}-5x_{3}=1 for x_{2}.

$$x_{2}=5x_{3}+1$$

Substitute 5x_{3}+1 for x_{2} in the equation 8x_{1}-2x_{2}=15.

$$8x_{1}-2\left(5x_{3}+1\right)=15$$

Solve the second equation for x_{1} and the third equation for x_{3}.

$$x_{1}=\frac{5}{4}x_{3}+\frac{17}{8} x_{3}=-\frac{4}{3}+\frac{2}{3}x_{1}$$

Substitute \frac{5}{4}x_{3}+\frac{17}{8} for x_{1} in the equation x_{3}=-\frac{4}{3}+\frac{2}{3}x_{1}.

$$x_{3}=-\frac{4}{3}+\frac{2}{3}\left(\frac{5}{4}x_{3}+\frac{17}{8}\right)$$

Solve x_{3}=-\frac{4}{3}+\frac{2}{3}\left(\frac{5}{4}x_{3}+\frac{17}{8}\right) for x_{3}.

$$x_{3}=\frac{1}{2}$$

Substitute \frac{1}{2} for x_{3} in the equation x_{1}=\frac{5}{4}x_{3}+\frac{17}{8}.

$$x_{1}=\frac{5}{4}\times \frac{1}{2}+\frac{17}{8}$$

Calculate x_{1} from x_{1}=\frac{5}{4}\times \frac{1}{2}+\frac{17}{8}.

$$x_{1}=\frac{11}{4}$$

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

$$x_{2}=5\times \frac{1}{2}+1$$

Calculate x_{2} from x_{2}=5\times \frac{1}{2}+1.

$$x_{2}=\frac{7}{2}$$

The system is now solved.

$$x_{1}=\frac{11}{4} x_{2}=\frac{7}{2} x_{3}=\frac{1}{2}$$