Solve f(x)=x-8/3x-4quada+x=2

$$f ( x ) = \frac { x - 8 } { 3 x - 4 } \quad a + x = 2$$

Solve for a

$$a=-\frac{\left(x-2\right)\left(3x-4\right)}{x-8}$$

$$x\neq 8\text{ and }x\neq \frac{4}{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3x-4.

$$\left(x-8\right)a+\left(3x-4\right)x=2\left(3x-4\right)$$

Use the distributive property to multiply x-8 by a.

$$xa-8a+\left(3x-4\right)x=2\left(3x-4\right)$$

Use the distributive property to multiply 3x-4 by x.

$$xa-8a+3x^{2}-4x=2\left(3x-4\right)$$

Use the distributive property to multiply 2 by 3x-4.

$$xa-8a+3x^{2}-4x=6x-8$$

Subtract 3x^{2} from both sides.

$$xa-8a-4x=6x-8-3x^{2}$$

Add 4x to both sides.

$$xa-8a=6x-8-3x^{2}+4x$$

Combine 6x and 4x to get 10x.

$$xa-8a=10x-8-3x^{2}$$

Combine all terms containing a.

$$\left(x-8\right)a=10x-8-3x^{2}$$

The equation is in standard form.

$$\left(x-8\right)a=-3x^{2}+10x-8$$

Divide both sides by x-8.

$$\frac{\left(x-8\right)a}{x-8}=\frac{\left(2-x\right)\left(3x-4\right)}{x-8}$$

Dividing by x-8 undoes the multiplication by x-8.

$$a=\frac{\left(2-x\right)\left(3x-4\right)}{x-8}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{\sqrt{a^{2}+76a+4}}{6}-\frac{a}{6}+\frac{5}{3}\text{, }&\text{unconditionally}\\x=-\frac{\sqrt{a^{2}+76a+4}}{6}-\frac{a}{6}+\frac{5}{3}\text{, }&a\neq 0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{a^{2}+76a+4}}{6}-\frac{a}{6}+\frac{5}{3}\text{, }&\left(a\neq 0\text{ and }a\geq 12\sqrt{10}-38\right)\text{ or }a\leq -12\sqrt{10}-38\\x=\frac{\sqrt{a^{2}+76a+4}}{6}-\frac{a}{6}+\frac{5}{3}\text{, }&a\geq 12\sqrt{10}-38\text{ or }a\leq -12\sqrt{10}-38\end{matrix}\right.$$