Solve {l}{4x_{1}+5x_{2}+6x_{3}=1}{3x_{1}+4x_{2}+5x_{3}=2}{3x_{1}+6x_{2}+7x_{3}=0}

$$\left\{ \begin{array} { l } { 4 x _ { 1 } + 5 x _ { 2 } + 6 x _ { 3 } = 1 } \\ { 3 x _ { 1 } + 4 x _ { 2 } + 5 x _ { 3 } = 2 } \\ { 3 x _ { 1 } + 6 x _ { 2 } + 7 x _ { 3 } = 0 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=0$$

$$x_{2}=-7$$

$$x_{3}=6$$

Short Steps Using Substitution

Solve 4x_{1}+5x_{2}+6x_{3}=1 for x_{1}.

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

Substitute -\frac{5}{4}x_{2}-\frac{3}{2}x_{3}+\frac{1}{4} for x_{1} in the second and third equation.

$$3\left(-\frac{5}{4}x_{2}-\frac{3}{2}x_{3}+\frac{1}{4}\right)+4x_{2}+5x_{3}=2 3\left(-\frac{5}{4}x_{2}-\frac{3}{2}x_{3}+\frac{1}{4}\right)+6x_{2}+7x_{3}=0$$

Solve these equations for x_{2} and x_{3} respectively.

$$x_{2}=5-2x_{3} x_{3}=-\frac{3}{10}-\frac{9}{10}x_{2}$$

Substitute 5-2x_{3} for x_{2} in the equation x_{3}=-\frac{3}{10}-\frac{9}{10}x_{2}.

$$x_{3}=-\frac{3}{10}-\frac{9}{10}\left(5-2x_{3}\right)$$

Solve x_{3}=-\frac{3}{10}-\frac{9}{10}\left(5-2x_{3}\right) for x_{3}.

$$x_{3}=6$$

Substitute 6 for x_{3} in the equation x_{2}=5-2x_{3}.

$$x_{2}=5-2\times 6$$

Calculate x_{2} from x_{2}=5-2\times 6.

$$x_{2}=-7$$

Substitute -7 for x_{2} and 6 for x_{3} in the equation x_{1}=-\frac{5}{4}x_{2}-\frac{3}{2}x_{3}+\frac{1}{4}.

$$x_{1}=-\frac{5}{4}\left(-7\right)-\frac{3}{2}\times 6+\frac{1}{4}$$

Calculate x_{1} from x_{1}=-\frac{5}{4}\left(-7\right)-\frac{3}{2}\times 6+\frac{1}{4}.

$$x_{1}=0$$

The system is now solved.

$$x_{1}=0 x_{2}=-7 x_{3}=6$$