Solve {l}{0=a-d+c}{c=-5/2}{0=25a+5d+c}

$$\left. \begin{array} { l } { 0 = a - d + c } \\ { c = - \frac { 5 } { 2 } } \\ { 0 = 25 a + 5 d + c } \end{array} \right.$$

Solve for a, d, c

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

$$d=-2$$

$$c = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Short Steps Using Substitution

Reorder the equations.

$$c=-\frac{5}{2} 0=a-d+c 0=25a+5d+c$$

Substitute -\frac{5}{2} for c in the second and third equation.

$$0=a-d-\frac{5}{2} 0=25a+5d-\frac{5}{2}$$

Solve these equations for d and a respectively.

$$d=a-\frac{5}{2} a=\frac{1}{10}-\frac{1}{5}d$$

Substitute a-\frac{5}{2} for d in the equation a=\frac{1}{10}-\frac{1}{5}d.

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

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

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

Substitute \frac{1}{2} for a in the equation d=a-\frac{5}{2}.

$$d=\frac{1}{2}-\frac{5}{2}$$

Calculate d from d=\frac{1}{2}-\frac{5}{2}.

$$d=-2$$

The system is now solved.

$$a=\frac{1}{2} d=-2 c=-\frac{5}{2}$$