Solve {l}{a=1}{3a+5b+2c=0}{a+5b=0}

$$\left. \begin{array} { l } { a = 1 } \\ { 3 a + 5 b + 2 c = 0 } \\ { a + 5 b = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=-\frac{1}{5}=-0.2$$

$$c=-1$$

Solution Steps

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

$$1+5b=0$$

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

$$5b=-1$$

Divide both sides by 5.

$$b=-\frac{1}{5}$$

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

$$3\times 1+5\left(-\frac{1}{5}\right)+2c=0$$

Multiply 3 and 1 to get 3.

$$3+5\left(-\frac{1}{5}\right)+2c=0$$

Multiply 5 and -\frac{1}{5} to get -1.

$$3-1+2c=0$$

Subtract 1 from 3 to get 2.

$$2+2c=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2c=-2$$

Divide both sides by 2.

$$c=\frac{-2}{2}$$

Divide -2 by 2 to get -1.

$$c=-1$$

The system is now solved.

$$a=1 b=-\frac{1}{5} c=-1$$