Solve x+5/x-3-3/x+4=ax^3+bx+c/(x-3)(x+4)

$$\frac { x + 5 } { x - 3 } - \frac { 3 } { x + 4 } = \frac { a x ^ { 3 } + b x + c } { ( x - 3 ) ( x + 4 ) }$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{-x^{2}+bx-6x+c-29}{x^{3}}\text{, }&x\neq 0\text{ and }x\neq 3\text{ and }x\neq -4\\a\in \mathrm{R}\text{, }&c=29\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x-3\right)\left(x+4\right), the least common multiple of x-3,x+4,\left(x-3\right)\left(x+4\right).

$$\left(x+4\right)\left(x+5\right)-\left(x-3\right)\times 3=ax^{3}+bx+c$$

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

$$x^{2}+9x+20-\left(x-3\right)\times 3=ax^{3}+bx+c$$

Use the distributive property to multiply x-3 by 3.

$$x^{2}+9x+20-\left(3x-9\right)=ax^{3}+bx+c$$

To find the opposite of 3x-9, find the opposite of each term.

$$x^{2}+9x+20-3x+9=ax^{3}+bx+c$$

Combine 9x and -3x to get 6x.

$$x^{2}+6x+20+9=ax^{3}+bx+c$$

Add 20 and 9 to get 29.

$$x^{2}+6x+29=ax^{3}+bx+c$$

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

$$ax^{3}+bx+c=x^{2}+6x+29$$

Subtract bx from both sides.

$$ax^{3}+c=x^{2}+6x+29-bx$$

Subtract c from both sides.

$$ax^{3}=x^{2}+6x+29-bx-c$$

The equation is in standard form.

$$x^{3}a=x^{2}-bx+6x-c+29$$

Divide both sides by x^{3}.

$$\frac{x^{3}a}{x^{3}}=\frac{x^{2}-bx+6x-c+29}{x^{3}}$$

Dividing by x^{3} undoes the multiplication by x^{3}.

$$a=\frac{x^{2}-bx+6x-c+29}{x^{3}}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{29-c+6x+x^{2}-ax^{3}}{x}\text{, }&x\neq 0\text{ and }x\neq 3\text{ and }x\neq -4\\b\in \mathrm{R}\text{, }&c=29\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x-3\right)\left(x+4\right), the least common multiple of x-3,x+4,\left(x-3\right)\left(x+4\right).

$$\left(x+4\right)\left(x+5\right)-\left(x-3\right)\times 3=ax^{3}+bx+c$$

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

$$x^{2}+9x+20-\left(x-3\right)\times 3=ax^{3}+bx+c$$

Use the distributive property to multiply x-3 by 3.

$$x^{2}+9x+20-\left(3x-9\right)=ax^{3}+bx+c$$

To find the opposite of 3x-9, find the opposite of each term.

$$x^{2}+9x+20-3x+9=ax^{3}+bx+c$$

Combine 9x and -3x to get 6x.

$$x^{2}+6x+20+9=ax^{3}+bx+c$$

Add 20 and 9 to get 29.

$$x^{2}+6x+29=ax^{3}+bx+c$$

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

$$ax^{3}+bx+c=x^{2}+6x+29$$

Subtract ax^{3} from both sides.

$$bx+c=x^{2}+6x+29-ax^{3}$$

Subtract c from both sides.

$$bx=x^{2}+6x+29-ax^{3}-c$$

Reorder the terms.

$$bx=-ax^{3}+x^{2}+6x-c+29$$

The equation is in standard form.

$$xb=29-c+6x+x^{2}-ax^{3}$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{29-c+6x+x^{2}-ax^{3}}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{29-c+6x+x^{2}-ax^{3}}{x}$$

Divide -ax^{3}+x^{2}+6x-c+29 by x.

$$b=-ax^{2}+x+\frac{29-c}{x}+6$$