Solve a - 4b - 2c = 21; 2a + b - 2c = 3; 3a + 2b - c = - 2 | 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

$$a-4b-2c=21; 2a+b-2c=3; 3a+2b-c=-2$$

Solve for a, b, c

$a = \frac{19}{17} = 1\frac{2}{17} \approx 1.117647059$
$b = -\frac{65}{17} = -3\frac{14}{17} \approx -3.823529412$
$c = -\frac{39}{17} = -2\frac{5}{17} \approx -2.294117647$

Short Steps Using Substitution

Solve $a-4b-2c=21$ for $a$.

$$a=4b+2c+21$$

Substitute $4b+2c+21$ for $a$ in the second and third equation.

$$2\left(4b+2c+21\right)+b-2c=3$$ $$3\left(4b+2c+21\right)+2b-c=-2$$

Solve these equations for $b$ and $c$ respectively.

$$b=-\frac{2}{9}c-\frac{13}{3}$$ $$c=-\frac{14}{5}b-13$$

Substitute $-\frac{2}{9}c-\frac{13}{3}$ for $b$ in the equation $c=-\frac{14}{5}b-13$.

$$c=-\frac{14}{5}\left(-\frac{2}{9}c-\frac{13}{3}\right)-13$$

Solve $c=-\frac{14}{5}\left(-\frac{2}{9}c-\frac{13}{3}\right)-13$ for $c$.

$$c=-\frac{39}{17}$$

Substitute $-\frac{39}{17}$ for $c$ in the equation $b=-\frac{2}{9}c-\frac{13}{3}$.

$$b=-\frac{2}{9}\left(-\frac{39}{17}\right)-\frac{13}{3}$$

Calculate $b$ from $b=-\frac{2}{9}\left(-\frac{39}{17}\right)-\frac{13}{3}$.

$$b=-\frac{65}{17}$$

Substitute $-\frac{65}{17}$ for $b$ and $-\frac{39}{17}$ for $c$ in the equation $a=4b+2c+21$.

$$a=4\left(-\frac{65}{17}\right)+2\left(-\frac{39}{17}\right)+21$$

Calculate $a$ from $a=4\left(-\frac{65}{17}\right)+2\left(-\frac{39}{17}\right)+21$.

$$a=\frac{19}{17}$$

The system is now solved.

$$a=\frac{19}{17}$$ $$b=-\frac{65}{17}$$ $$c=-\frac{39}{17}$$