Solve x_{1} + 0 + 2x_{3} = 6; - 3x_{1} + 4x_{2} + 6x_{3} = 30; - x_{1} + 2x_{2} + 3x_{3} = 8 | 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

$$x_{1}+0+2x_{3}=6; -3x_{1}+4x_{2}+6x_{3}=30; -x_{1}+2x_{2}+3x_{3}=8$$

Solve for x_1, x_3, x_2

$x_{1}=-14$
$x_{3}=10$
$x_{2}=-18$

Short Steps Using Substitution

Solve $x_{1}+0+2x_{3}=6$ for $x_{1}$.

$$x_{1}=6-2x_{3}$$

Substitute $6-2x_{3}$ for $x_{1}$ in the second and third equation.

$$-3\left(6-2x_{3}\right)+4x_{2}+6x_{3}=30$$ $$-\left(6-2x_{3}\right)+2x_{2}+3x_{3}=8$$

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

$$x_{3}=4-\frac{1}{3}x_{2}$$ $$x_{2}=7-\frac{5}{2}x_{3}$$

Substitute $4-\frac{1}{3}x_{2}$ for $x_{3}$ in the equation $x_{2}=7-\frac{5}{2}x_{3}$.

$$x_{2}=7-\frac{5}{2}\left(4-\frac{1}{3}x_{2}\right)$$

Solve $x_{2}=7-\frac{5}{2}\left(4-\frac{1}{3}x_{2}\right)$ for $x_{2}$.

$$x_{2}=-18$$

Substitute $-18$ for $x_{2}$ in the equation $x_{3}=4-\frac{1}{3}x_{2}$.

$$x_{3}=4-\frac{1}{3}\left(-18\right)$$

Calculate $x_{3}$ from $x_{3}=4-\frac{1}{3}\left(-18\right)$.

$$x_{3}=10$$

Substitute $10$ for $x_{3}$ in the equation $x_{1}=6-2x_{3}$.

$$x_{1}=6-2\times 10$$

Calculate $x_{1}$ from $x_{1}=6-2\times 10$.

$$x_{1}=-14$$

The system is now solved.

$$x_{1}=-14$$ $$x_{3}=10$$ $$x_{2}=-18$$