Solve 1/+1=ax/(2x+1)^2=

$$\frac { 1 } { + 1 } = \frac { a x } { ( 2 x + 1 ) ^ { 2 } } =$$

Solve for a

$$a=\frac{\left(2x+1\right)^{2}}{x}$$

$$x\neq 0\text{ and }x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(2x+1\right)^{2}.

$$\left(2x+1\right)^{2}=ax$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=ax$$

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

$$ax=4x^{2}+4x+1$$

The equation is in standard form.

$$xa=4x^{2}+4x+1$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{\left(2x+1\right)^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{\left(2x+1\right)^{2}}{x}$$

Solve for x

$$x=\frac{-\sqrt{a\left(a-8\right)}+a-4}{8}$$

$$x=\frac{\sqrt{a\left(a-8\right)}+a-4}{8}\text{, }a\geq 8\text{ or }a<0$$