Solve {c}{11x_{1}+13x_{2}+4x_{3}=37}{12x_{1}+14x_{2}+5x_{3}=40}{9x_{1}+3x_{2}+3x_{3}=15}

$$\left. \begin{array} { c } { 11 x _ { 1 } + 13 x _ { 2 } + 4 x _ { 3 } = 37 } \\ { 12 x _ { 1 } + 14 x _ { 2 } + 5 x _ { 3 } = 40 } \\ { 9 x _ { 1 } + 3 x _ { 2 } + 3 x _ { 3 } = 15 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=1$$

$$x_{2}=2$$

$$x_{3}=0$$

Short Steps Using Substitution

Solve 11x_{1}+13x_{2}+4x_{3}=37 for x_{1}.

$$x_{1}=-\frac{13}{11}x_{2}-\frac{4}{11}x_{3}+\frac{37}{11}$$

Substitute -\frac{13}{11}x_{2}-\frac{4}{11}x_{3}+\frac{37}{11} for x_{1} in the second and third equation.

$$12\left(-\frac{13}{11}x_{2}-\frac{4}{11}x_{3}+\frac{37}{11}\right)+14x_{2}+5x_{3}=40 9\left(-\frac{13}{11}x_{2}-\frac{4}{11}x_{3}+\frac{37}{11}\right)+3x_{2}+3x_{3}=15$$

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

$$x_{2}=2+\frac{7}{2}x_{3} x_{3}=56-28x_{2}$$

Substitute 2+\frac{7}{2}x_{3} for x_{2} in the equation x_{3}=56-28x_{2}.

$$x_{3}=56-28\left(2+\frac{7}{2}x_{3}\right)$$

Solve x_{3}=56-28\left(2+\frac{7}{2}x_{3}\right) for x_{3}.

$$x_{3}=0$$

Substitute 0 for x_{3} in the equation x_{2}=2+\frac{7}{2}x_{3}.

$$x_{2}=2+\frac{7}{2}\times 0$$

Calculate x_{2} from x_{2}=2+\frac{7}{2}\times 0.

$$x_{2}=2$$

Substitute 2 for x_{2} and 0 for x_{3} in the equation x_{1}=-\frac{13}{11}x_{2}-\frac{4}{11}x_{3}+\frac{37}{11}.

$$x_{1}=-\frac{13}{11}\times 2-\frac{4}{11}\times 0+\frac{37}{11}$$

Calculate x_{1} from x_{1}=-\frac{13}{11}\times 2-\frac{4}{11}\times 0+\frac{37}{11}.

$$x_{1}=1$$

The system is now solved.

$$x_{1}=1 x_{2}=2 x_{3}=0$$