Solve {l}{kx^2-7x+12=0}{x=3}{l=k}{text{Solvefor}mtext{where}}{m=l}

$$\left. \begin{array} { l } { k x ^ {2} - 7 x + 12 = 0 }\\ { x = 3 }\\ { l = k }\\ { \text{Solve for } m \text{ where} } \\ { m = l } \end{array} \right.$$

Solve for k, x, l, m

$$m=1$$

Solution Steps

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

$$k\times 3^{2}-7\times 3+12=0$$

Calculate 3 to the power of 2 and get 9.

$$k\times 9-7\times 3+12=0$$

Multiply -7 and 3 to get -21.

$$k\times 9-21+12=0$$

Add -21 and 12 to get -9.

$$k\times 9-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$k\times 9=9$$

Divide both sides by 9.

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

Divide 9 by 9 to get 1.

$$k=1$$

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

$$l=1$$

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

$$m=1$$

The system is now solved.

$$k=1 x=3 l=1 m=1$$