Solve {l}{a+b=8}{a+c=13}{c-a=6}

$$\left. \begin{array} { l } { a + b = 8 } \\ { a + c = 13 } \\ { c - a = 6 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$b = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$c = \frac{19}{2} = 9\frac{1}{2} = 9.5$$

Short Steps Using Substitution

Solve a+b=8 for a.

$$a=-b+8$$

Substitute -b+8 for a in the second and third equation.

$$-b+8+c=13 c-\left(-b+8\right)=6$$

Solve these equations for b and c respectively.

$$b=-5+c c=-b+14$$

Substitute -5+c for b in the equation c=-b+14.

$$c=-\left(-5+c\right)+14$$

Solve c=-\left(-5+c\right)+14 for c.

$$c=\frac{19}{2}$$

Substitute \frac{19}{2} for c in the equation b=-5+c.

$$b=-5+\frac{19}{2}$$

Calculate b from b=-5+\frac{19}{2}.

$$b=\frac{9}{2}$$

Substitute \frac{9}{2} for b in the equation a=-b+8.

$$a=-\frac{9}{2}+8$$

Calculate a from a=-\frac{9}{2}+8.

$$a=\frac{7}{2}$$

The system is now solved.

$$a=\frac{7}{2} b=\frac{9}{2} c=\frac{19}{2}$$