Solve {l}{b+c=110}{b=a+10}{c=b+3}

$$\left. \begin{array} { l } { b + c = 110 } \\ { b = a + 10 } \\ { c = b + 3 } \end{array} \right.$$

Solve for b, c, a

$$b = \frac{107}{2} = 53\frac{1}{2} = 53.5$$

$$c = \frac{113}{2} = 56\frac{1}{2} = 56.5$$

$$a = \frac{87}{2} = 43\frac{1}{2} = 43.5$$

Short Steps Using Substitution

Reorder the equations.

$$b=a+10 b+c=110 c=b+3$$

Substitute a+10 for b in the second and third equation.

$$a+10+c=110 c=a+10+3$$

Solve these equations for c and a respectively.

$$c=-a+100 a=-13+c$$

Substitute -a+100 for c in the equation a=-13+c.

$$a=-13-a+100$$

Solve a=-13-a+100 for a.

$$a=\frac{87}{2}$$

Substitute \frac{87}{2} for a in the equation c=-a+100.

$$c=-\frac{87}{2}+100$$

Calculate c from c=-\frac{87}{2}+100.

$$c=\frac{113}{2}$$

Substitute \frac{87}{2} for a in the equation b=a+10.

$$b=\frac{87}{2}+10$$

Calculate b from b=\frac{87}{2}+10.

$$b=\frac{107}{2}$$

The system is now solved.

$$b=\frac{107}{2} c=\frac{113}{2} a=\frac{87}{2}$$