Solve (x+4)(400/x-5)d=(400/x+9)x

$$( x + 4 ) ( \frac { 400 } { x } - 5 ) d = ( \frac { 400 } { x } + 9 ) x$$

Solve for d

$$d=-\frac{x\left(9x+400\right)}{5\left(x-80\right)\left(x+4\right)}$$

$$x\neq -4\text{ and }x\neq 80\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$\left(x+4\right)\left(\frac{400}{x}-5\right)dx=\left(\frac{400}{x}+9\right)xx$$

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

$$\left(x+4\right)\left(\frac{400}{x}-5\right)dx=\left(\frac{400}{x}+9\right)x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{x}{x}.

$$\left(x+4\right)\left(\frac{400}{x}-\frac{5x}{x}\right)dx=\left(\frac{400}{x}+9\right)x^{2}$$

Since \frac{400}{x} and \frac{5x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\left(x+4\right)\times \frac{400-5x}{x}dx=\left(\frac{400}{x}+9\right)x^{2}$$

Express \left(x+4\right)\times \frac{400-5x}{x} as a single fraction.

$$\frac{\left(x+4\right)\left(400-5x\right)}{x}dx=\left(\frac{400}{x}+9\right)x^{2}$$

Express \frac{\left(x+4\right)\left(400-5x\right)}{x}d as a single fraction.

$$\frac{\left(x+4\right)\left(400-5x\right)d}{x}x=\left(\frac{400}{x}+9\right)x^{2}$$

Express \frac{\left(x+4\right)\left(400-5x\right)d}{x}x as a single fraction.

$$\frac{\left(x+4\right)\left(400-5x\right)dx}{x}=\left(\frac{400}{x}+9\right)x^{2}$$

Cancel out x in both numerator and denominator.

$$d\left(x+4\right)\left(-5x+400\right)=\left(\frac{400}{x}+9\right)x^{2}$$

Use the distributive property to multiply d by x+4.

$$\left(dx+4d\right)\left(-5x+400\right)=\left(\frac{400}{x}+9\right)x^{2}$$

Use the distributive property to multiply dx+4d by -5x+400 and combine like terms.

$$-5dx^{2}+380dx+1600d=\left(\frac{400}{x}+9\right)x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{x}{x}.

$$-5dx^{2}+380dx+1600d=\left(\frac{400}{x}+\frac{9x}{x}\right)x^{2}$$

Since \frac{400}{x} and \frac{9x}{x} have the same denominator, add them by adding their numerators.

$$-5dx^{2}+380dx+1600d=\frac{400+9x}{x}x^{2}$$

Express \frac{400+9x}{x}x^{2} as a single fraction.

$$-5dx^{2}+380dx+1600d=\frac{\left(400+9x\right)x^{2}}{x}$$

Cancel out x in both numerator and denominator.

$$-5dx^{2}+380dx+1600d=x\left(9x+400\right)$$

Use the distributive property to multiply x by 9x+400.

$$-5dx^{2}+380dx+1600d=9x^{2}+400x$$

Combine all terms containing d.

$$\left(-5x^{2}+380x+1600\right)d=9x^{2}+400x$$

The equation is in standard form.

$$\left(1600+380x-5x^{2}\right)d=9x^{2}+400x$$

Divide both sides by -5x^{2}+380x+1600.

$$\frac{\left(1600+380x-5x^{2}\right)d}{1600+380x-5x^{2}}=\frac{x\left(9x+400\right)}{1600+380x-5x^{2}}$$

Dividing by -5x^{2}+380x+1600 undoes the multiplication by -5x^{2}+380x+1600.

$$d=\frac{x\left(9x+400\right)}{1600+380x-5x^{2}}$$

Divide x\left(400+9x\right) by -5x^{2}+380x+1600.

$$d=\frac{x\left(9x+400\right)}{-5\left(x-80\right)\left(x+4\right)}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{10\left(-\sqrt{441d^{2}-616d+400}-19d+20\right)}{5d+9}\text{, }&d\neq -\frac{9}{5}\text{ and }d\neq 0\\x=-\frac{10\left(\sqrt{441d^{2}-616d+400}-19d+20\right)}{5d+9}\text{, }&d\neq -\frac{9}{5}\\x=-\frac{720}{271}\text{, }&d=-\frac{9}{5}\end{matrix}\right.$$