Solve {l}{300k+12=3}{l=100k+1}{text{Solvefor}mtext{where}}{m=l}

$$\left. \begin{array} { l } { 300 k + 12 = 3 }\\ { l = 100 k + 1 }\\ { \text{Solve for } m \text{ where} } \\ { m = l } \end{array} \right.$$

Solve for k, l, m

$$m=-2$$

Solution Steps

Consider the first equation. Subtract 12 from both sides.

$$300k=3-12$$

Subtract 12 from 3 to get -9.

$$300k=-9$$

Divide both sides by 300.

$$k=\frac{-9}{300}$$

Reduce the fraction \frac{-9}{300} to lowest terms by extracting and canceling out 3.

$$k=-\frac{3}{100}$$

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

$$l=100\left(-\frac{3}{100}\right)+1$$

Multiply 100 and -\frac{3}{100} to get -3.

$$l=-3+1$$

Add -3 and 1 to get -2.

$$l=-2$$

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

$$m=-2$$

The system is now solved.

$$k=-\frac{3}{100} l=-2 m=-2$$