Solve {l}{3x_{1}+x+2+x_{3}-x_{4}=0}{-2x_{3}-x_{4}=0}{x_{1}+x_{4}=0}{x_{1}=1}

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

Solve for x_1, x, x_3, x_4

$$x = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

$$x_{1}=1$$

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

$$x_{4}=-1$$

Solution Steps

Consider the third equation. Insert the known values of variables into the equation.

$$1+x_{4}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$x_{4}=-1$$

Consider the second equation. Insert the known values of variables into the equation.

$$-2x_{3}-\left(-1\right)=0$$

Multiply -1 and -1 to get 1.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Fraction \frac{-1}{-2} can be simplified to \frac{1}{2} by removing the negative sign from both the numerator and the denominator.

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

Consider the first equation. Insert the known values of variables into the equation.

$$3\times 1+x+2+\frac{1}{2}-\left(-1\right)=0$$

Do the multiplications.

$$3+x+2+\frac{1}{2}+1=0$$

Add 3 and 2 to get 5.

$$5+x+\frac{1}{2}+1=0$$

Add 5 and \frac{1}{2} to get \frac{11}{2}.

$$\frac{11}{2}+x+1=0$$

Add \frac{11}{2} and 1 to get \frac{13}{2}.

$$\frac{13}{2}+x=0$$

Subtract \frac{13}{2} from both sides. Anything subtracted from zero gives its negation.

$$x=-\frac{13}{2}$$

The system is now solved.

$$x_{1}=1 x=-\frac{13}{2} x_{3}=\frac{1}{2} x_{4}=-1$$