Solve {l}{1/2b=1/4s+1/4d}{b-5=s+d-35}{d+3=s+b+7}

$$\left\{ \begin{array} { l } { \frac { 1 } { 2 } b = \frac { 1 } { 4 } s + \frac { 1 } { 4 } d } \\ { b - 5 = s + d - 35 } \\ { d + 3 = s + b + 7 } \end{array} \right.$$

Solve for b, s, d

$$s=13$$

$$b=30$$

$$d=47$$

Short Steps Using Substitution

Reorder the equations.

$$b-5=s+d-35 \frac{1}{2}b=\frac{1}{4}s+\frac{1}{4}d d+3=s+b+7$$

Solve b-5=s+d-35 for b.

$$b=-30+s+d$$

Substitute -30+s+d for b in the second and third equation.

$$\frac{1}{2}\left(-30+s+d\right)=\frac{1}{4}s+\frac{1}{4}d d+3=s-30+s+d+7$$

Solve these equations for d and s respectively.

$$d=60-s s=13$$

Substitute 13 for s in the equation d=60-s.

$$d=60-13$$

Calculate d from d=60-13.

$$d=47$$

Substitute 47 for d and 13 for s in the equation b=-30+s+d.

$$b=-30+13+47$$

Calculate b from b=-30+13+47.

$$b=30$$

The system is now solved.

$$b=30 s=13 d=47$$