Solve X_{1} + x_{2} - x_{3} = 1; 8x_{1} + 3x_{2} - 6x_{3} = 2; - 4x_{1} - x_{2} + 3x_{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

$$\begin{cases}X_{1}+x_{2}-x_{3}=1\\ 8x_{1}+3x_{2}-6x_{3}=2\\ -4x_{1}-x_{2}+3x_{3}=-3\end{cases}$$

Solve for x_1, x_2, x_3

$x_{1}=\frac{3X_{1}}{4}-2$
$x_{2}=-4$
$x_{3}=X_{1}-5$

Short Steps Using Substitution

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

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

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

$$8x_{1}+3\left(-X_{1}+x_{3}+1\right)-6x_{3}=2$$ $$-4x_{1}-\left(-X_{1}+x_{3}+1\right)+3x_{3}=-3$$

Solve these equations for $x_{1}$ and $x_{3}$ respectively.

$$x_{1}=\frac{3}{8}X_{1}+\frac{3}{8}x_{3}-\frac{1}{8}$$ $$x_{3}=-\frac{1}{2}X_{1}+2x_{1}-1$$

Substitute $\frac{3}{8}X_{1}+\frac{3}{8}x_{3}-\frac{1}{8}$ for $x_{1}$ in the equation $x_{3}=-\frac{1}{2}X_{1}+2x_{1}-1$.

$$x_{3}=-\frac{1}{2}X_{1}+2\left(\frac{3}{8}X_{1}+\frac{3}{8}x_{3}-\frac{1}{8}\right)-1$$

Solve $x_{3}=-\frac{1}{2}X_{1}+2\left(\frac{3}{8}X_{1}+\frac{3}{8}x_{3}-\frac{1}{8}\right)-1$ for $x_{3}$.

$$x_{3}=-5+X_{1}$$

Substitute $-5+X_{1}$ for $x_{3}$ in the equation $x_{1}=\frac{3}{8}X_{1}+\frac{3}{8}x_{3}-\frac{1}{8}$.

$$x_{1}=\frac{3}{8}X_{1}+\frac{3}{8}\left(-5+X_{1}\right)-\frac{1}{8}$$

Calculate $x_{1}$ from $x_{1}=\frac{3}{8}X_{1}+\frac{3}{8}\left(-5+X_{1}\right)-\frac{1}{8}$.

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

Substitute $-5+X_{1}$ for $x_{3}$ in the equation $x_{2}=-X_{1}+x_{3}+1$.

$$x_{2}=-X_{1}-5+X_{1}+1$$

Calculate $x_{2}$ from $x_{2}=-X_{1}-5+X_{1}+1$.

$$x_{2}=-4$$

The system is now solved.

$$x_{1}=-2+\frac{3}{4}X_{1}$$ $$x_{2}=-4$$ $$x_{3}=-5+X_{1}$$