Solve {l}{4+1+2D+E+F=0}{4+9-2D-3E+F=0}{1+4-D+2E+F=0}

$$\left. \begin{array} { l } { 4 + 1 + 2 D + E + F = 0 } \\ { 4 + 9 - 2 D - 3 E + F = 0 } \\ { 1 + 4 - D + 2 E + F = 0 } \end{array} \right.$$

Solve for D, E, F

$$D=\frac{1}{2}=0.5$$

$$E = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$F = -\frac{15}{2} = -7\frac{1}{2} = -7.5$$

Short Steps Using Substitution

Solve 4+1+2D+E+F=0 for E.

$$E=-5-2D-F$$

Substitute -5-2D-F for E in the second and third equation.

$$4+9-2D-3\left(-5-2D-F\right)+F=0 1+4-D+2\left(-5-2D-F\right)+F=0$$

Solve these equations for D and F respectively.

$$D=-7-F F=-5-5D$$

Substitute -7-F for D in the equation F=-5-5D.

$$F=-5-5\left(-7-F\right)$$

Solve F=-5-5\left(-7-F\right) for F.

$$F=-\frac{15}{2}$$

Substitute -\frac{15}{2} for F in the equation D=-7-F.

$$D=-7-\left(-\frac{15}{2}\right)$$

Calculate D from D=-7-\left(-\frac{15}{2}\right).

$$D=\frac{1}{2}$$

Substitute \frac{1}{2} for D and -\frac{15}{2} for F in the equation E=-5-2D-F.

$$E=-5-2\times \frac{1}{2}-\left(-\frac{15}{2}\right)$$

Calculate E from E=-5-2\times \frac{1}{2}-\left(-\frac{15}{2}\right).

$$E=\frac{3}{2}$$

The system is now solved.

$$D=\frac{1}{2} E=\frac{3}{2} F=-\frac{15}{2}$$