Solve a+1/b+frac{1{c}}=15/4

$$a + \frac { 1 } { b + \frac { 1 } { c } } = \frac { 15 } { 4 }$$

Solve for a

$$a=-\frac{-15bc+4c-15}{4\left(bc+1\right)}$$

$$b\neq -\frac{1}{c}\text{ and }c\neq 0$$

Steps for Solving Linear Equation

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

$$a+\frac{1}{\frac{bc}{c}+\frac{1}{c}}=\frac{15}{4}$$

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

$$a+\frac{1}{\frac{bc+1}{c}}=\frac{15}{4}$$

Divide 1 by \frac{bc+1}{c} by multiplying 1 by the reciprocal of \frac{bc+1}{c}.

$$a+\frac{c}{bc+1}=\frac{15}{4}$$

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

$$\frac{a\left(bc+1\right)}{bc+1}+\frac{c}{bc+1}=\frac{15}{4}$$

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

$$\frac{a\left(bc+1\right)+c}{bc+1}=\frac{15}{4}$$

Do the multiplications in a\left(bc+1\right)+c.

$$\frac{abc+a+c}{bc+1}=\frac{15}{4}$$

Multiply both sides of the equation by 4\left(bc+1\right), the least common multiple of bc+1,4.

$$4\left(abc+a+c\right)=15\left(bc+1\right)$$

Use the distributive property to multiply 4 by abc+a+c.

$$4abc+4a+4c=15\left(bc+1\right)$$

Use the distributive property to multiply 15 by bc+1.

$$4abc+4a+4c=15bc+15$$

Subtract 4c from both sides.

$$4abc+4a=15bc+15-4c$$

Combine all terms containing a.

$$\left(4bc+4\right)a=15bc+15-4c$$

The equation is in standard form.

$$\left(4bc+4\right)a=15bc-4c+15$$

Divide both sides by 4bc+4.

$$\frac{\left(4bc+4\right)a}{4bc+4}=\frac{15bc-4c+15}{4bc+4}$$

Dividing by 4bc+4 undoes the multiplication by 4bc+4.

$$a=\frac{15bc-4c+15}{4bc+4}$$

Divide 15bc+15-4c by 4bc+4.

$$a=\frac{15bc-4c+15}{4\left(bc+1\right)}$$

Solve for b

$$b=-\frac{4a+4c-15}{c\left(4a-15\right)}$$

$$a\neq \frac{15}{4}\text{ and }c\neq 0$$