Lutasin ang {l}{x_{1}+x_{2}-3x_{3}-2x_{4}=2}{x_{1}+x_{3}-x_{4}=1}{2x_{1}+6x_{2}-13x_{3}-5x_{4}=6}

$$\left. \begin{array}{l}{ x _ { 1 } + x _ { 2 } - 3 x _ { 3 } - 2 x _ { 4 } = 2 }\\{ x _ { 1 } + x _ { 3 } - x _ { 4 } = 1 }\\{ 2 x _ { 1 } + 6 x _ { 2 } - 13 x _ { 3 } - 5 x _ { 4 } = 6 }\end{array} \right.$$

I-solve ang x_1, x_2, x_3

$$x_{1}=\frac{4x_{4}}{3}+\frac{11}{9}$$

$$x_{2}=-\frac{x_{4}}{3}+\frac{1}{9}$$

$$x_{3}=-\frac{x_{4}}{3}-\frac{2}{9}$$

Maiikling Hakbang sa Paggamit ng Substitution

Lutasin ang x_{1}+x_{2}-3x_{3}-2x_{4}=2 para sa x_{1}.

$$x_{1}=-x_{2}+3x_{3}+2x_{4}+2$$

I-substitute ang -x_{2}+3x_{3}+2x_{4}+2 para sa x_{1} sa pangalawa at pangatlong equation.

$$-x_{2}+3x_{3}+2x_{4}+2+x_{3}-x_{4}=1 2\left(-x_{2}+3x_{3}+2x_{4}+2\right)+6x_{2}-13x_{3}-5x_{4}=6$$

Lutasin ang mga equation na ito para sa x_{2} at x_{3} nang naaayon.

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

I-substitute ang 4x_{3}+x_{4}+1 para sa x_{2} sa equation na x_{3}=\frac{4}{7}x_{2}-\frac{1}{7}x_{4}-\frac{2}{7}.

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

Lutasin ang x_{3}=\frac{4}{7}\left(4x_{3}+x_{4}+1\right)-\frac{1}{7}x_{4}-\frac{2}{7} para sa x_{3}.

$$x_{3}=-\frac{2}{9}-\frac{1}{3}x_{4}$$

I-substitute ang -\frac{2}{9}-\frac{1}{3}x_{4} para sa x_{3} sa equation na x_{2}=4x_{3}+x_{4}+1.

$$x_{2}=4\left(-\frac{2}{9}-\frac{1}{3}x_{4}\right)+x_{4}+1$$

Kalkulahin ang x_{2} mula sa x_{2}=4\left(-\frac{2}{9}-\frac{1}{3}x_{4}\right)+x_{4}+1.

$$x_{2}=\frac{1}{9}-\frac{1}{3}x_{4}$$

I-substitute ang \frac{1}{9}-\frac{1}{3}x_{4} para sa x_{2} at ang -\frac{2}{9}-\frac{1}{3}x_{4} para sa x_{3} sa equation na x_{1}=-x_{2}+3x_{3}+2x_{4}+2.

$$x_{1}=-\left(\frac{1}{9}-\frac{1}{3}x_{4}\right)+3\left(-\frac{2}{9}-\frac{1}{3}x_{4}\right)+2x_{4}+2$$

Kalkulahin ang x_{1} mula sa x_{1}=-\left(\frac{1}{9}-\frac{1}{3}x_{4}\right)+3\left(-\frac{2}{9}-\frac{1}{3}x_{4}\right)+2x_{4}+2.

$$x_{1}=\frac{11}{9}+\frac{4}{3}x_{4}$$

Nalutas na ang system.

$$x_{1}=\frac{11}{9}+\frac{4}{3}x_{4} x_{2}=\frac{1}{9}-\frac{1}{3}x_{4} x_{3}=-\frac{2}{9}-\frac{1}{3}x_{4}$$