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

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

Solve for D, E, F

$$D=-1$$

$$E=-3$$

$$F=0$$

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 equation 1+4-D+2E+F=0.

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

Solve the second equation for D and the third equation for F.

$$D=-1-\frac{1}{5}F F=-1-D$$

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

$$F=-1-\left(-1-\frac{1}{5}F\right)$$

Solve F=-1-\left(-1-\frac{1}{5}F\right) for F.

$$F=0$$

Substitute 0 for F in the equation D=-1-\frac{1}{5}F.

$$D=-1-\frac{1}{5}\times 0$$

Calculate D from D=-1-\frac{1}{5}\times 0.

$$D=-1$$

Substitute -1 for D and 0 for F in the equation E=-5-2D-F.

$$E=-5-2\left(-1\right)-0$$

Calculate E from E=-5-2\left(-1\right)-0.

$$E=-3$$

The system is now solved.

$$D=-1 E=-3 F=0$$