Solve 1+ay/1-ay=c/d

$$\frac { 1 + a y } { 1 - a y } = \frac { c } { d }$$

Solve for c

$$c=-\frac{d\left(ay+1\right)}{ay-1}$$

$$\left(y=0\text{ or }a\neq \frac{1}{y}\right)\text{ and }d\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by d\left(ay-1\right), the least common multiple of 1-ay,d.

$$-d\left(1+ay\right)=\left(ay-1\right)c$$

Use the distributive property to multiply -d by 1+ay.

$$-d-day=\left(ay-1\right)c$$

Use the distributive property to multiply ay-1 by c.

$$-d-day=ayc-c$$

Swap sides so that all variable terms are on the left hand side.

$$ayc-c=-d-day$$

Combine all terms containing c.

$$\left(ay-1\right)c=-d-day$$

The equation is in standard form.

$$\left(ay-1\right)c=-ady-d$$

Divide both sides by -1+ay.

$$\frac{\left(ay-1\right)c}{ay-1}=-\frac{d\left(ay+1\right)}{ay-1}$$

Dividing by -1+ay undoes the multiplication by -1+ay.

$$c=-\frac{d\left(ay+1\right)}{ay-1}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{d-c}{y\left(c+d\right)}\text{, }&d\neq 0\text{ and }y\neq 0\text{ and }\left(c=d\text{ or }|c|\neq |d|\right)\\a\in \mathrm{R}\text{, }&d=c\text{ and }y=0\text{ and }c\neq 0\end{matrix}\right.$$