Solve {l}{9+k^2+18k+81=k^2}{l={(-4+1)}+{(k+9)}}{text{Solvefor}mtext{where}}{m=l}

$$\left. \begin{array} { l } { 9 + k ^ {2} + 18 k + 81 = k ^ {2} }\\ { l = {(-4 + 1)} + {(k + 9)} }\\ { \text{Solve for } m \text{ where} } \\ { m = l } \end{array} \right.$$

Solve for k, l, m

$$m=1$$

Solution Steps

Consider the first equation. Add 9 and 81 to get 90.

$$90+k^{2}+18k=k^{2}$$

Subtract k^{2} from both sides.

$$90+k^{2}+18k-k^{2}=0$$

Combine k^{2} and -k^{2} to get 0.

$$90+18k=0$$

Subtract 90 from both sides. Anything subtracted from zero gives its negation.

$$18k=-90$$

Divide both sides by 18.

$$k=\frac{-90}{18}$$

Divide -90 by 18 to get -5.

$$k=-5$$

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

$$l=-4+1-5+9$$

Add -4 and 1 to get -3.

$$l=-3-5+9$$

Subtract 5 from -3 to get -8.

$$l=-8+9$$

Add -8 and 9 to get 1.

$$l=1$$

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

$$m=1$$

The system is now solved.

$$k=-5 l=1 m=1$$