Solve {l}{1+9+D+3E+F=0}{16+4+4D+2E+F=0}{1+49+D-7E+F=0}

$$\left\{ \begin{array} { l } { 1 + 9 + D + 3 E + F = 0 } \\ { 16 + 4 + 4 D + 2 E + F = 0 } \\ { 1 + 49 + D - 7 E + F = 0 } \end{array} \right.$$

Solve for D, E, F

$$D=-2$$

$$E=4$$

$$F=-20$$

Short Steps Using Substitution

Solve 1+9+D+3E+F=0 for D.

$$D=-10-3E-F$$

Substitute -10-3E-F for D in the second and third equation.

$$16+4+4\left(-10-3E-F\right)+2E+F=0 1+49-10-3E-F-7E+F=0$$

Solve these equations for F and E respectively.

$$F=-\frac{20}{3}-\frac{10}{3}E E=4$$

Substitute 4 for E in the equation F=-\frac{20}{3}-\frac{10}{3}E.

$$F=-\frac{20}{3}-\frac{10}{3}\times 4$$

Calculate F from F=-\frac{20}{3}-\frac{10}{3}\times 4.

$$F=-20$$

Substitute -20 for F and 4 for E in the equation D=-10-3E-F.

$$D=-10-3\times 4-\left(-20\right)$$

Calculate D from D=-10-3\times 4-\left(-20\right).

$$D=-2$$

The system is now solved.

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