Solve 1/(x-3)+5c/(x+3)=1

$$\frac { 1 } { ( x - 3 ) } + \frac { 5 c } { ( x + 3 ) } = 1$$

Solve for c (complex solution)

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5\left(x-3\right)}$$

$$x\neq 3\text{ and }x\neq -3$$

Steps for Solving Linear Equation

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

$$x+3+\left(x-3\right)\times 5c=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply x-3 by 5.

$$x+3+\left(5x-15\right)c=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 5x-15 by c.

$$x+3+5xc-15c=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$x+3+5xc-15c=x^{2}-9$$

Subtract x from both sides.

$$3+5xc-15c=x^{2}-9-x$$

Subtract 3 from both sides.

$$5xc-15c=x^{2}-9-x-3$$

Subtract 3 from -9 to get -12.

$$5xc-15c=x^{2}-12-x$$

Combine all terms containing c.

$$\left(5x-15\right)c=x^{2}-12-x$$

The equation is in standard form.

$$\left(5x-15\right)c=x^{2}-x-12$$

Divide both sides by 5x-15.

$$\frac{\left(5x-15\right)c}{5x-15}=\frac{\left(x-4\right)\left(x+3\right)}{5x-15}$$

Dividing by 5x-15 undoes the multiplication by 5x-15.

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5x-15}$$

Divide \left(-4+x\right)\left(3+x\right) by 5x-15.

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5\left(x-3\right)}$$

Solve for c

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5\left(x-3\right)}$$

$$|x|\neq 3$$

Steps for Solving Linear Equation

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

$$x+3+\left(x-3\right)\times 5c=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply x-3 by 5.

$$x+3+\left(5x-15\right)c=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 5x-15 by c.

$$x+3+5xc-15c=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$x+3+5xc-15c=x^{2}-9$$

Subtract x from both sides.

$$3+5xc-15c=x^{2}-9-x$$

Subtract 3 from both sides.

$$5xc-15c=x^{2}-9-x-3$$

Subtract 3 from -9 to get -12.

$$5xc-15c=x^{2}-12-x$$

Combine all terms containing c.

$$\left(5x-15\right)c=x^{2}-12-x$$

The equation is in standard form.

$$\left(5x-15\right)c=x^{2}-x-12$$

Divide both sides by 5x-15.

$$\frac{\left(5x-15\right)c}{5x-15}=\frac{\left(x-4\right)\left(x+3\right)}{5x-15}$$

Dividing by 5x-15 undoes the multiplication by 5x-15.

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5x-15}$$

Divide \left(-4+x\right)\left(3+x\right) by 5x-15.

$$c=\frac{\left(x-4\right)\left(x+3\right)}{5\left(x-3\right)}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{\sqrt{25c^{2}-50c+49}+5c+1}{2}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{25c^{2}-50c+49}+5c+1}{2}\text{, }&c\neq 0\end{matrix}\right.$$