Let’s first determine the restricted values of the variable in the equation. Only one rational expression in the equation has a variable in the denominator. Since $$2x+1=0$$ only if $$x=-\dfrac{1}{2}$$, the only restricted value of $$x$$ is $$-\dfrac{1}{2}$$.

We next determine the LCD by finding the LCM of all denominators. Since $$2x+1$$ is a prime polynomial and $$3$$ is a prime number, $$LCM(2x+1, 3)=3\cdot (2x+1)$$. We will next multiply both sides of the equation by the LCD.

[latex]\require{color} \begin{align} \frac{4}{2x+1}{\color{red}{} \cdot 3\cdot (2x+1)} &= \frac{2x}{3}{\color{red}{} \cdot 3\cdot (2x+1)} && {\color{blue}\textsf{multiply by the LCD}}\\[5pt] \frac{4}{{\color{red}\cancel{\color{black}{(2x+1)}}}\cdot 1} \cdot \frac{3\cdot {\color{red}\cancel{\color{black}{(2x+1)}}}}{1} &= \frac{2x}{{\color{red}\cancel{\color{black}{3}}}\cdot 1} \cdot \frac{{\color{red}\cancel{\color{black}{3}}}\cdot (2x+1)}{1} && {\color{blue}\textsf{divide out common factors}}\\[5pt] 12 &= 2x\cdot (2x+1) && {\color{blue}\textsf{simplify}}\\[5pt] \end{align}[/latex]

The resulting equation is quadratic. We solve it by factoring and using the Zero-Product Property.

[latex]\begin{align} {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot 6}}}{\cancel{2}}}} &= {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot x \cdot (2x+1)}}}{\cancel{2}}}} && {\color{blue}\textsf{divide by $2$ and divide out common factors}}\\[5pt] \underset{\large{\color{red}{-6}}}{6} &= 2x^2+x+\underset{\large{\color{red}{-6}}}{0} && {\color{blue}\textsf{distribute $x$ and subtract $6$}}\\[5pt] 0 &= 2x^2+x-6 && {\color{blue}\textsf{simplify}}\\[5pt] 0 &= (2x-3)(x+2) && {\color{blue}\textsf{factor}}\\[5pt] 2x-\underset{\large{\color{red}{+3}}}{3} &= \underset{\large{\color{red}{+3}}}{0}\ \ \textsf{or}\ \ x+\underset{\large{\color{red}{-2}}}{2} = \underset{\large{\color{red}{-2}}}{0} && \color{blue}{\textsf{Zero-Product Property}}\\[5pt] {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot x}}}{\cancel{2}}}} &= {\color{red}{\dfrac{{\color{black}{3}}}{2}}} \ \ \textsf{or}\ \ x = -2 \\[5pt] x &= \dfrac{3}{2} \ \ \textsf{or}\ \ x = -2 \end{align}[/latex]

We must check the obtained solutions against the list of restricted values of the variable!

We determined earlier that the only restricted value of $$x$$ is $$-\dfrac{1}{2}$$. Hence both, $$x = \dfrac{3}{2}$$ and $$x = -2$$ are valid solutions.

It’s always a good idea to verify the solutions.

Substituting $$x={\color{Green}{\dfrac{3}{2}}}$$ in the left hand side (LHS), $$\dfrac{4}{2{\color{Green}{x}}+1}$$, results in

[latex]\begin{align} &\quad\; \frac{4}{\dfrac{{\color{red}\cancel{\color{black}{2}}} \cdot 1}{1}\cdot{\color{Green}{\dfrac{3}{{\color{red}\cancel{\color{Green}{2}}}{\color{black}{} \cdot 1}}}}+1\;}\\[5pt] &= \frac{4}{3+1}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{4}}}\cdot 1}{{\color{red}\cancel{\color{black}{4}}} \cdot 1}\\[5pt] &= 1 \end{align}[/latex]

Substituting $$x={\color{Green}{\dfrac{3}{2}}}$$ in the right hand side (RHS), $$\dfrac{2{\color{Green}{x}}}{3}$$, results in

[latex]\begin{align} &\quad\; \frac{\;\dfrac{{\color{red}\cancel{\color{black}{2}}} \cdot 1}{1}\cdot {\color{Green}{\dfrac{3}{{\color{red}\cancel{\color{Green}{2}}}{\color{black}{} \cdot 1}}}}\;}{3}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{3}}}\cdot 1}{{\color{red}\cancel{\color{black}{3}}} \cdot 1}\\[5pt] &= 1 \end{align}[/latex]

Since $$1=1$$, LHS $$=$$ RHS. $$\large\checkmark$$

Substituting $$x={\color{Green}{-2}}$$ in the LHS, $$\dfrac{4}{2{\color{Green}{x}}+1}$$, results in

[latex]\begin{align} &\quad\; \frac{4}{2\cdot {\color{Green}{(-2)}}+1}\\[5pt] &= \frac{4}{-3}\\[5pt] &= -\frac{4}{3}\end{align}[/latex]

Substituting $$x={\color{Green}{-2}}$$ in the RHS, $$\dfrac{2{\color{Green}{x}}}{3}$$, results in

[latex]\begin{align} &\quad\; \dfrac{2\cdot{\color{Green}{(-2)}}}{3}\\[5pt] &= \dfrac{-4}{3}\\[5pt] &= -\dfrac{4}{3} \end{align}[/latex]

Since $$-\dfrac{4}{3}=-\dfrac{4}{3}$$, LHS $$=$$ RHS. $$\large\checkmark$$

Let’s start with the restricted values of the variable in the equation. There are two rational expression in the equation with a variable in the denominator. Since $$x-2=0$$ only if $$x=2$$, and $$x+3=0$$ only if $$x=-3$$, the restricted values of $$x$$ are $$-3$$ and $$2$$.

The LCD (the LCM of all denominators) is $$(x-2)(x+3)$$ since both $$x-2$$ and $$x+3$$ are prime polynomials. We’re ready to multiply both sides of the equation by the LCD.

[latex]\begin{align} \left(\frac{x}{x-2}-1\right) {\color{red}{} \cdot (x-2)\cdot (x+3)} &= \frac{1}{x+3}{\color{red}{} \cdot (x-2)\cdot (x+3)}\\[5pt] \frac{x}{{\color{red}\cancel{\color{black}{(x-2)}}} \cdot 1} \cdot \frac{{\color{red}\cancel{\color{black}{(x-2)}}}\cdot (x+3)}{1} - \left(x^2+x-6\right) &= \frac{1}{{\color{red}\cancel{\color{black}{(x+3)}}} \cdot 1} \cdot \frac{(x-2)\cdot {\color{red}\cancel{\color{black}{(x+3)}}}}{1}\\[5pt] x\cdot (x+3) - x^2-x+6 &= x-2\\[5pt] {\color{red}\cancel{\color{black}{x^2}}} + 3x {\color{red}\cancel{\color{black}{- x^2}}} - x + 6 &= x - 2\\[5pt] \underset{\large{\color{red}{-x}}}{2x} + \underset{\large{\color{red}{-6}}}{6} &= \underset{\large{\color{red}{-x}}}{x} - \underset{\large{\color{red}{-6}}}{2}\\[5pt] x &= -8 \end{align}[/latex]

Recall that $$x=-8$$ is so far a proposed solution and must be checked against the list of restricted values of the variable. Since the restricted values of $$x$$ are $$-3$$ and $$2$$, $$x=-8$$ is a valid solution.

We leave it to you to substitute $$x=-8$$ into the original equation to verify that it results in a true statement.

Let’s start with the restricted values of the variable in the equation. There are three rational expression in the equation with a variable in the denominator. Since $$x-1=0$$ only if $$x=1$$, and $$x+1=0$$ only if $$x=-1$$, $$x=-1$$ and $$x=1$$ are restricted. Since $$x^2-1 = (x+1)(x-1)$$, $$x^2-1 = 0$$ only if $$x=-1$$ or $$x=1$$ by the Zero-Product Property. Hence all restricted values of $$x$$ are $$-1$$ and $$1$$.

Since $$x^2-1 = (x+1)(x-1)$$, The LCD is $$(x-1)(x+1)$$. We’re ready to multiply both sides of the equation by the LCD.

[latex]\begin{align} \left(\frac{x}{x-1}-\frac{x+3}{x^2-1}\right){\color{red}{} \cdot (x-1)\cdot (x+1)} &= \frac{1}{x+1}{\color{red}{} \cdot (x-1)\cdot (x+1)}\\[5pt] \frac{x}{{\color{red}\cancel{\color{black}{(x-1)}}} \cdot 1}\cdot \frac{{\color{red}\cancel{\color{black}{(x-1)}}}\cdot (x+1)}{1} - \frac{x+3}{{\color{red}\cancel{\color{black}{(x-1)\cdot (x+1)}}}\cdot 1}\cdot \frac{{\color{red}\cancel{\color{black}{(x-1)\cdot (x+1)}}}\cdot 1}{1} &= \frac{1}{{\color{red}\cancel{\color{black}{(x+1)}}}} \cdot \frac{(x-1)\cdot {\color{red}\cancel{\color{black}{(x+1)}}}}{1}\\[5pt] x\cdot (x+1)-(x+3) &= x-1\\[5pt] x^2 + x - \underset{\large{\color{red}{-x}}}{x} -\underset{\large{\color{red}{+1}}}{3} &= \underset{\large{\color{red}{-x}}}{x}-\underset{\large{\color{red}{+1}}}{1}\\[5pt] x^2-x-2 &= 0\\[5pt] (x-2)(x+1) &= 0\\[5pt] x-\underset{\large{\color{red}{+2}}}{2}=\underset{\large{\color{red}{+2}}}{0} \ \ \textsf{or}\ \ x+\underset{\large{\color{red}{-1}}}{1} &= \underset{\large{\color{red}{-1}}}{0}\\[5pt] x = 2 \ \ \textsf{or}\ \ x &= -1 \end{align}[/latex]

Of course, $$x=2$$ and $$x=-1$$ are so far proposed solutions and each must be checked against the list of restricted values of the variable. Since the restricted values of $$x$$ are $$-1$$ and $$1$$, $$x=2$$ is a valid solution while $$x=-1$$ is not. In particular, $$x=-1$$ is an extraneous solution.

We leave it to you to substitute $$x=2$$ into the original equation to verify that it results in a true statement. Also note what happens when you substitute the extraneous solution.

Notice that once we substitute the rational expressions defining $$f(x)$$ and $$g(x)$$ into the equation $$f(x)=g(x)$$, we obtain the rational equation

[latex]\dfrac{16}{x+4} = \dfrac{x^2}{x+4}[/latex]

For the restricted values of $$x$$, since $$x+4=0$$ only if $$x=-4$$, the only restricted values of $$x$$ is $$-4$$.

Although it’s tempting to cross multiply, since the denominators of the rational expressions on both sides are the same, it will be more convenient to multiply both sides by that common denominator and simplify.

[latex]\begin{align} \frac{16}{{\color{red}\cancel{\color{black}{(x+4)}}}\cdot 1}{\color{red}{} \cdot \dfrac{\cancel{(x+4)}\cdot 1}{1}} &= \frac{x^2}{{\color{red}\cancel{\color{black}{(x+4)}}}\cdot 1}{\color{red}{} \cdot \dfrac{\cancel{(x+4)}\cdot 1}{1}}\\[5pt] \underset{\large{\color{red}{-16}}}{16} &= x^2+\underset{\large{\color{red}{-16}}}{0}\\[5pt] 0 &= x^2-16\\[5pt] 0 &= (x+4)(x-4)\\[5pt] x+\underset{\large{\color{red}{-4}}}{4} &= \underset{\large{\color{red}{-4}}}{0}\ \ \textsf{or}\ \ x-\underset{\large{\color{red}{+4}}}{4} = \underset{\large{\color{red}{+4}}}{0}\\[5pt] x &= -4 \ \ \textsf{or}\ \ x = 4 \end{align}[/latex]

Since $$-4$$ is a restricted value of $$x$$, $$x = 4$$ is the only valid solution and the only value of $$x$$ for which $$f(x)=g(x)$$.

The restricted values of $$x$$ are $$-3$$ and $$2$$.

The LCD of the LHS is $$x-2$$, so

[latex]\begin{align} \frac{x}{x-2}-1&=\frac{x}{x-2}-\frac{1{\color{red}{} \cdot (x-2)}}{1{\color{red}{} \cdot (x-2)}}\\[5pt] &= \frac{x-(x-2)}{x-2}\\[5pt] &= \frac{x-x+2}{x-2}\\[5pt] &= \frac{2}{x-2} \end{align}[/latex]

Thus the original equation is so far equivalent to

[latex]\dfrac{2}{x-2} = \dfrac{1}{x+3}[/latex]

The LCD of both sides is $$(x-2)(x+3)$$. Now, [latex]\dfrac{2{\color{red}{} \cdot (x+3)}}{(x-2){\color{red}{} \cdot (x+3)}} = \dfrac{2x+6}{(x-2)(x+3)}[/latex], and [latex]\dfrac{1{\color{red}{} \cdot (x-2)}}{(x+3){\color{red}{} \cdot (x-2)}} = \dfrac{x-2}{(x-2)(x+3)}[/latex], so the original equation is equivalent to

[latex]\dfrac{2x+6}{(x-2)(x+3)} = \dfrac{x-2}{(x-2)(x+3)}[/latex]

We can now equate the numerators to obtain

[latex]2x+6 = x-2[/latex]

We solve accordingly to obtain $$x=-8$$. Since it’s not a restricted value, it’s a valid solution of the original equation.

Refer to the Example further up the page how to identify the restricted values of $$x$$: $$-1$$ and $$1$$.

To isolate $$0$$ on the RHS, we start by subtracting $$\dfrac{1}{x+1}$$ from both sides of the equation and proceed to rewrite the LHS as a simplified single rational expression.

[latex]\begin{align} \frac{x}{x-1}-\frac{x+3}{x^2-1}{\color{red} \: - \: \frac{1}{x+1}} &= \frac{1}{x+1}{\color{red} \: - \: \frac{1}{x+1}}\\[5pt] \frac{x{\color{red} \: \cdot \: (x+1)}}{(x-1){\color{red} \: \cdot \: (x+1)}}-\frac{x+3}{(x-1)\cdot(x+1)}-\frac{1{\color{red} \: \cdot \: (x-1)}}{(x+1){\color{red} \: \cdot \: (x-1)}} &= 0\\[5pt] \frac{x^2+x-(x+3)-(x-1)}{(x-1)\cdot(x+1)}&=0\\[5pt] \frac{x^2+x-x-3-x+1}{(x-1)\cdot(x+1)}&=0\\[5pt] \frac{x^2-x-2}{(x-1)\cdot(x+1)}&=0\\[5pt] \frac{(x-2)\cdot{\color{red}\cancel{\color{black}{(x+1)}}}}{(x-1)\cdot{\color{red}\cancel{\color{black}{(x+1)}}}}&=0\\[5pt] \frac{x-2}{x-1}&=0\\[5pt] \end{align}[/latex]

We now solve $$x-2=0$$ resulting in $$x=2$$. It’s not a restricted value of $$x$$, so its a valid (only) solution of the original equation.