Solve {l}{4+16-2D+4E+F=0}{1+9-D+3E+F=0}{4+36+2D+6E+F=0}

$$\left. \begin{array} { l } { 4 + 16 - 2 D + 4 E + F = 0 } \\ { 1 + 9 - D + 3 E + F = 0 } \\ { 4 + 36 + 2 D + 6 E + F = 0 } \end{array} \right.$$

Solve for D, E, F

$$D=0$$

$$E=-10$$

$$F=20$$

Short Steps Using Substitution

Solve 4+16-2D+4E+F=0 for F.

$$F=-20+2D-4E$$

Substitute -20+2D-4E for F in the second and third equation.

$$1+9-D+3E-20+2D-4E=0 4+36+2D+6E-20+2D-4E=0$$

Solve these equations for E and D respectively.

$$E=-10+D D=-5-\frac{1}{2}E$$

Substitute -10+D for E in the equation D=-5-\frac{1}{2}E.

$$D=-5-\frac{1}{2}\left(-10+D\right)$$

Solve D=-5-\frac{1}{2}\left(-10+D\right) for D.

$$D=0$$

Substitute 0 for D in the equation E=-10+D.

$$E=-10+0$$

Calculate E from E=-10+0.

$$E=-10$$

Substitute -10 for E and 0 for D in the equation F=-20+2D-4E.

$$F=-20+2\times 0-4\left(-10\right)$$

Calculate F from F=-20+2\times 0-4\left(-10\right).

$$F=20$$

The system is now solved.

$$D=0 E=-10 F=20$$