Solve - x_{1} + 3x_{2} = 4; 3x_{1} - 2x_{2} + x_{3} = - 3; 2x_{1} + x_{2} - x_{3} = - 3 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\begin{cases}-x_{1}+3x_{2}=4\\ 3x_{1}-2x_{2}+x_{3}=-3\\ 2x_{1}+x_{2}-x_{3}=-3\end{cases}$$

Solve for x_1, x_2, x_3

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

Short Steps Using Substitution

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

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

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

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

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

$$x_{2}=\frac{9}{7}-\frac{1}{7}x_{3}$$ $$x_{3}=7x_{2}-5$$

Substitute $\frac{9}{7}-\frac{1}{7}x_{3}$ for $x_{2}$ in the equation $x_{3}=7x_{2}-5$.

$$x_{3}=7\left(\frac{9}{7}-\frac{1}{7}x_{3}\right)-5$$

Solve $x_{3}=7\left(\frac{9}{7}-\frac{1}{7}x_{3}\right)-5$ for $x_{3}$.

$$x_{3}=2$$

Substitute $2$ for $x_{3}$ in the equation $x_{2}=\frac{9}{7}-\frac{1}{7}x_{3}$.

$$x_{2}=\frac{9}{7}-\frac{1}{7}\times 2$$

Calculate $x_{2}$ from $x_{2}=\frac{9}{7}-\frac{1}{7}\times 2$.

$$x_{2}=1$$

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

$$x_{1}=3\times 1-4$$

Calculate $x_{1}$ from $x_{1}=3\times 1-4$.

$$x_{1}=-1$$

The system is now solved.

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