Solve 1/a^2+1/a=b

$$\frac { 1 } { a ^ { 2 } } + \frac { 1 } { a } = b$$

Solve for b

$$b=\frac{a+1}{a^{2}}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by a^{2}, the least common multiple of a^{2},a.

$$1+a=ba^{2}$$

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

$$ba^{2}=1+a$$

The equation is in standard form.

$$a^{2}b=a+1$$

Divide both sides by a^{2}.

$$\frac{a^{2}b}{a^{2}}=\frac{a+1}{a^{2}}$$

Dividing by a^{2} undoes the multiplication by a^{2}.

$$b=\frac{a+1}{a^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{\sqrt{4b+1}-1}{2b}\text{; }a=\frac{\sqrt{4b+1}+1}{2b}\text{, }&b\neq 0\text{ and }b\geq -\frac{1}{4}\\a=-1\text{, }&b=0\end{matrix}\right.$$