Solve p+1/(q+(1/r))=129/31

$$p + 1 / ( q + ( 1 / r ) ) = 129 / 31$$

Solve for p

$$p=-\frac{-129qr+31r-129}{31\left(qr+1\right)}$$

$$q\neq -\frac{1}{r}\text{ and }r\neq 0$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply q times \frac{r}{r}.

$$p+\frac{1}{\frac{qr}{r}+\frac{1}{r}}=\frac{129}{31}$$

Since \frac{qr}{r} and \frac{1}{r} have the same denominator, add them by adding their numerators.

$$p+\frac{1}{\frac{qr+1}{r}}=\frac{129}{31}$$

Divide 1 by \frac{qr+1}{r} by multiplying 1 by the reciprocal of \frac{qr+1}{r}.

$$p+\frac{r}{qr+1}=\frac{129}{31}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply p times \frac{qr+1}{qr+1}.

$$\frac{p\left(qr+1\right)}{qr+1}+\frac{r}{qr+1}=\frac{129}{31}$$

Since \frac{p\left(qr+1\right)}{qr+1} and \frac{r}{qr+1} have the same denominator, add them by adding their numerators.

$$\frac{p\left(qr+1\right)+r}{qr+1}=\frac{129}{31}$$

Do the multiplications in p\left(qr+1\right)+r.

$$\frac{pqr+p+r}{qr+1}=\frac{129}{31}$$

Multiply both sides of the equation by 31\left(qr+1\right), the least common multiple of qr+1,31.

$$31\left(pqr+p+r\right)=129\left(qr+1\right)$$

Use the distributive property to multiply 31 by pqr+p+r.

$$31pqr+31p+31r=129\left(qr+1\right)$$

Use the distributive property to multiply 129 by qr+1.

$$31pqr+31p+31r=129qr+129$$

Subtract 31r from both sides.

$$31pqr+31p=129qr+129-31r$$

Combine all terms containing p.

$$\left(31qr+31\right)p=129qr+129-31r$$

The equation is in standard form.

$$\left(31qr+31\right)p=129qr-31r+129$$

Divide both sides by 31qr+31.

$$\frac{\left(31qr+31\right)p}{31qr+31}=\frac{129qr-31r+129}{31qr+31}$$

Dividing by 31qr+31 undoes the multiplication by 31qr+31.

$$p=\frac{129qr-31r+129}{31qr+31}$$

Divide 129qr+129-31r by 31qr+31.

$$p=\frac{129qr-31r+129}{31\left(qr+1\right)}$$

Solve for q

$$q=-\frac{31p+31r-129}{r\left(31p-129\right)}$$

$$p\neq \frac{129}{31}\text{ and }r\neq 0$$