{c}{x_{1}+2x_{2}-x_{3}+3x_{4}=0}{2x_{1}+3x_{2}-x_{3}+2x_{4}=0}{x_{1}quad+3x_{3}+3x_{4}=0}-এর সমাধান করুন

$$\left. \begin{array} { c } { x _ { 1 } + 2 x _ { 2 } - x _ { 3 } + 3 x _ { 4 } = 0 } \\ { 2 x _ { 1 } + 3 x _ { 2 } - x _ { 3 } + 2 x _ { 4 } = 0 } \\ { x _ { 1 } \quad + 3 x _ { 3 } + 3 x _ { 4 } = 0 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=9x_{4}$$

$$x_{2}=-8x_{4}$$

$$x_{3}=-4x_{4}$$

Short Steps Using Substitution

Solve x_{1}+2x_{2}-x_{3}+3x_{4}=0 for x_{1}.

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

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

$$2\left(-2x_{2}+x_{3}-3x_{4}\right)+3x_{2}-x_{3}+2x_{4}=0 -2x_{2}+x_{3}-3x_{4}+3x_{3}+3x_{4}=0$$

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

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

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

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

Solve x_{3}=\frac{1}{2}\left(x_{3}-4x_{4}\right) for x_{3}.

$$x_{3}=-4x_{4}$$

Substitute -4x_{4} for x_{3} in the equation x_{2}=x_{3}-4x_{4}.

$$x_{2}=-4x_{4}-4x_{4}$$

Calculate x_{2} from x_{2}=-4x_{4}-4x_{4}.

$$x_{2}=-8x_{4}$$

Substitute -8x_{4} for x_{2} and -4x_{4} for x_{3} in the equation x_{1}=-2x_{2}+x_{3}-3x_{4}.

$$x_{1}=-2\left(-8\right)x_{4}-4x_{4}-3x_{4}$$

Calculate x_{1} from x_{1}=-2\left(-8\right)x_{4}-4x_{4}-3x_{4}.

$$x_{1}=9x_{4}$$

The system is now solved.

$$x_{1}=9x_{4} x_{2}=-8x_{4} x_{3}=-4x_{4}$$