Solve {l}{a+d=1}{F+2d=1}{3a+d+3F=2}

$$\left\{ \begin{array} { l } { a + d = 1 } \\ { F + 2 d = 1 } \\ { 3 a + d + 3 F = 2 } \end{array} \right.$$

Solve for a, d, F

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

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

$$F=0$$

Short Steps Using Substitution

Solve a+d=1 for a.

$$a=-d+1$$

Substitute -d+1 for a in the equation 3a+d+3F=2.

$$3\left(-d+1\right)+d+3F=2$$

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

$$d=-\frac{1}{2}F+\frac{1}{2} F=\frac{2}{3}d-\frac{1}{3}$$

Substitute -\frac{1}{2}F+\frac{1}{2} for d in the equation F=\frac{2}{3}d-\frac{1}{3}.

$$F=\frac{2}{3}\left(-\frac{1}{2}F+\frac{1}{2}\right)-\frac{1}{3}$$

Solve F=\frac{2}{3}\left(-\frac{1}{2}F+\frac{1}{2}\right)-\frac{1}{3} for F.

$$F=0$$

Substitute 0 for F in the equation d=-\frac{1}{2}F+\frac{1}{2}.

$$d=-\frac{1}{2}\times 0+\frac{1}{2}$$

Calculate d from d=-\frac{1}{2}\times 0+\frac{1}{2}.

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

Substitute \frac{1}{2} for d in the equation a=-d+1.

$$a=-\frac{1}{2}+1$$

Calculate a from a=-\frac{1}{2}+1.

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

The system is now solved.

$$a=\frac{1}{2} d=\frac{1}{2} F=0$$