Solve {l}{1/25+1/5+1/33026=1/R}{a=R}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { \frac{1}{25} + \frac{1}{5} + \frac{1}{33026} = \frac{1}{R} }\\ { a = R }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for R, a, b

$$b = \frac{825650}{198181} = 4\frac{32926}{198181} \approx 4.166141053$$

Solution Steps

Consider the first equation. Variable R cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 825650R, the least common multiple of 25,5,33026,R.

$$825650R\times \frac{1}{25}+825650R\times \frac{1}{5}+825650R\times \frac{1}{33026}=825650$$

Multiply 825650 and \frac{1}{25} to get 33026.

$$33026R+825650R\times \frac{1}{5}+825650R\times \frac{1}{33026}=825650$$

Multiply 825650 and \frac{1}{5} to get 165130.

$$33026R+165130R+825650R\times \frac{1}{33026}=825650$$

Combine 33026R and 165130R to get 198156R.

$$198156R+825650R\times \frac{1}{33026}=825650$$

Multiply 825650 and \frac{1}{33026} to get 25.

$$198156R+25R=825650$$

Combine 198156R and 25R to get 198181R.

$$198181R=825650$$

Divide both sides by 198181.

$$R=\frac{825650}{198181}$$

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

$$a=\frac{825650}{198181}$$

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

$$b=\frac{825650}{198181}$$

The system is now solved.

$$R=\frac{825650}{198181} a=\frac{825650}{198181} b=\frac{825650}{198181}$$