Solve d/5k/5-frac{b{r}k}=x/k

$$\frac { \frac { d } { 5 } k } { 5 - \frac { b } { r } k } = \frac { x } { k }$$

Solve for d

$$d=-\frac{5x\left(bk-5r\right)}{rk^{2}}$$

$$k\neq 0\text{ and }r\neq 0\text{ and }\left(b=0\text{ or }r\neq \frac{bk}{5}\right)\text{ and }\left(b=0\text{ or }k\neq \frac{5r}{b}\right)$$

Steps for Solving Linear Equation

Multiply both sides of the equation by k.

$$k\left(5r-bk\right)^{-1}r\times \frac{d}{5}k=x$$

Multiply k and k to get k^{2}.

$$k^{2}\left(5r-bk\right)^{-1}r\times \frac{d}{5}=x$$

Express k^{2}\times \frac{d}{5} as a single fraction.

$$\frac{k^{2}d}{5}\left(5r-bk\right)^{-1}r=x$$

Express \frac{k^{2}d}{5}\left(5r-bk\right)^{-1} as a single fraction.

$$\frac{k^{2}d\left(5r-bk\right)^{-1}}{5}r=x$$

Express \frac{k^{2}d\left(5r-bk\right)^{-1}}{5}r as a single fraction.

$$\frac{k^{2}d\left(5r-bk\right)^{-1}r}{5}=x$$

Reorder the terms.

$$\frac{\frac{1}{-bk+5r}drk^{2}}{5}=x$$

Express \frac{1}{-bk+5r}d as a single fraction.

$$\frac{\frac{d}{-bk+5r}rk^{2}}{5}=x$$

Express \frac{d}{-bk+5r}r as a single fraction.

$$\frac{\frac{dr}{-bk+5r}k^{2}}{5}=x$$

Express \frac{dr}{-bk+5r}k^{2} as a single fraction.

$$\frac{\frac{drk^{2}}{-bk+5r}}{5}=x$$

Reorder the terms.

$$\frac{drk^{2}}{5\left(-bk+5r\right)}=x$$

Multiply both sides of the equation by 5\left(5r-bk\right).

$$drk^{2}=x\times 5\left(5r-bk\right)$$

Use the distributive property to multiply x\times 5 by 5r-bk.

$$drk^{2}=25rx-5xbk$$

The equation is in standard form.

$$rk^{2}d=25rx-5bkx$$

Divide both sides by rk^{2}.

$$\frac{rk^{2}d}{rk^{2}}=\frac{25rx-5bkx}{rk^{2}}$$

Dividing by rk^{2} undoes the multiplication by rk^{2}.

$$d=\frac{25rx-5bkx}{rk^{2}}$$

Divide 25xr-5xbk by rk^{2}.

$$d=\frac{5x\left(5r-bk\right)}{rk^{2}}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{r\left(25x-dk^{2}\right)}{5kx}\text{, }&\left(d=\frac{25x}{k^{2}}\text{ or }d\neq 0\right)\text{ and }k\neq 0\text{ and }r\neq 0\text{ and }x\neq 0\\b\neq \frac{5r}{k}\text{, }&r\neq 0\text{ and }x=0\text{ and }d=0\text{ and }k\neq 0\end{matrix}\right.$$