Solve 1/43x(x+1)=mx^2

$$\frac { 1 } { 4 } 3 x ( x + 1 ) = m x ^ { 2 }$$

Solve for m

$$\left\{\begin{matrix}m=\frac{3}{4}+\frac{3}{4x}\text{, }&x\neq 0\\m\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply \frac{1}{4} and 3 to get \frac{3}{4}.

$$\frac{3}{4}x\left(x+1\right)=mx^{2}$$

Use the distributive property to multiply \frac{3}{4}x by x+1.

$$\frac{3}{4}x^{2}+\frac{3}{4}x=mx^{2}$$

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

$$mx^{2}=\frac{3}{4}x^{2}+\frac{3}{4}x$$

The equation is in standard form.

$$x^{2}m=\frac{3x^{2}+3x}{4}$$

Divide both sides by x^{2}.

$$\frac{x^{2}m}{x^{2}}=\frac{3x\left(x+1\right)}{4x^{2}}$$

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

$$m=\frac{3x\left(x+1\right)}{4x^{2}}$$

Divide \frac{3x\left(1+x\right)}{4} by x^{2}.

$$m=\frac{3}{4}+\frac{3}{4x}$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=-\frac{3}{3-4m}\text{, }&m\neq \frac{3}{4}\end{matrix}\right.$$