Solve a_{n}=4n-2/4+3n

$$a _ { n } = \frac { 4 n - 2 } { 4 + 3 n }$$

Solve for n

$$n=-\frac{2\left(2a_{n}+1\right)}{3a_{n}-4}$$

$$a_{n}\neq \frac{4}{3}$$

Steps for Solving Linear Equation

Variable n cannot be equal to -\frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by 3n+4.

$$a_{n}\left(3n+4\right)=4n-2$$

Use the distributive property to multiply a_{n} by 3n+4.

$$3a_{n}n+4a_{n}=4n-2$$

Subtract 4n from both sides.

$$3a_{n}n+4a_{n}-4n=-2$$

Subtract 4a_{n} from both sides.

$$3a_{n}n-4n=-2-4a_{n}$$

Combine all terms containing n.

$$\left(3a_{n}-4\right)n=-2-4a_{n}$$

The equation is in standard form.

$$\left(3a_{n}-4\right)n=-4a_{n}-2$$

Divide both sides by 3a_{n}-4.

$$\frac{\left(3a_{n}-4\right)n}{3a_{n}-4}=\frac{-4a_{n}-2}{3a_{n}-4}$$

Dividing by 3a_{n}-4 undoes the multiplication by 3a_{n}-4.

$$n=\frac{-4a_{n}-2}{3a_{n}-4}$$

Divide -2-4a_{n} by 3a_{n}-4.

$$n=-\frac{2\left(2a_{n}+1\right)}{3a_{n}-4}$$

Variable n cannot be equal to -\frac{4}{3}.

$$n=-\frac{2\left(2a_{n}+1\right)}{3a_{n}-4}\text{, }n\neq -\frac{4}{3}$$

Solve for a_n

$$a_{n}=-\frac{2\left(1-2n\right)}{3n+4}$$

$$n\neq -\frac{4}{3}$$