Solve {l}{14+3b+8b=250}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { 14 + 3 b + 8 b = 250 }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for b, c, d

$$d = \frac{236}{11} = 21\frac{5}{11} \approx 21.454545455$$

Solution Steps

Consider the first equation. Combine 3b and 8b to get 11b.

$$14+11b=250$$

Subtract 14 from both sides.

$$11b=250-14$$

Subtract 14 from 250 to get 236.

$$11b=236$$

Divide both sides by 11.

$$b=\frac{236}{11}$$

Consider the second equation. Insert the known values of variables into the equation.

$$c=\frac{236}{11}$$

Consider the third equation. Insert the known values of variables into the equation.

$$d=\frac{236}{11}$$

The system is now solved.

$$b=\frac{236}{11} c=\frac{236}{11} d=\frac{236}{11}$$