7.6: Algebraic Analysis on Intersection Points

Learning Objectives

Describe the meaning of solving rational equations
Solve rational equations

The Meaning of Solving Exponential Equations

Intersection points of functions

In chapter 3, we learned that the meaning of solving an equation is to find the intersection point(s) between two functions. The intersection point(s) between the graphs of any two functions $f(x)$ and $g(x)$ can be found algebraically by setting the two functions equal to each other:

$f(x)=g(x)$

When the functions are equal, the value of $x$ is the same for both functions, as is the function value. In other words, $f(x)=g(x)$ means that the two functions have the same input $x$ as well as the same output (i.e $f(x)=g(x)$ ). For example, solving the equation $-\dfrac{1}{x}=\dfrac{2}{x}+3$ is equivalent to determining the intersection point between the two functions $f(x)=-\dfrac{1}{x}$ and $g(x)=\dfrac{2}{x}+3$ . The solution is $x=-1$ (figure 1).

Figure 1. The Intersection of two functions at the point $(-1, 1)$ .

Example 1

Use graphing to solve the rational equation $\dfrac{-1}{x-4}=\dfrac{1}{x^2+2x}$ .

Solution

Let $f(x)=\dfrac{-1}{x-4}$ and $g(x)=\dfrac{1}{x^2+2x}$

Then use Desmos to graph the functions.

Look for the intersection points of the two graphs.

two graphs intersecting

The graphs intersect at $x=-4$ and $x=1$ .

Therefore, the solution of the equation is $x=-4,\,1$ .

Try It 1

Use graphing to solve the rational equation $\dfrac{1}{x^2}=\dfrac{1}{2x+3}$ .

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intersection of two graphs

$x=-1,\,3$

Finding the $x$ -value given a function value

Given the equation $\dfrac{8}{x+1}=2$ , in addition to the interpretation of finding the intersection point between the two functions $f(x)=\dfrac{8}{x+1}$ and $g(x)=2$ , another interpretation of the equation is finding the $x$ value when the function value of $f(x)=\dfrac{8}{x+1}$ is 2 (figure 2).

Figure 2. Find the $x$ coordinate given that the $y$ coordinate is 2.

Graphically, this means finding the $x$ value for a given $y$ -value on a graph. Determine where the $y$ -value meets the graph then move vertically down to the $x$ -axis to determine the corresponding $x$ -value. In figure 2, to solve $\dfrac{8}{x+1}=2$ , we graph the function $f(x)=\dfrac{8}{x+1}$ , look for 2 on the $y$ -axis then determine which $x$ -value has a function value at 2. In this case, $x=3$ .

Example 2

Graphically solve the equation $\dfrac{1}{x+3}=-1$ .

Solution

Let $f(x)=\dfrac{1}{x+3}$ . Use Desmos to graph the function then determine where the $y$ -value equals –1. Then determine the corresponding $x$ -value.

f(x)=1/(x+3)

Answer

$x=-4$

Try It 2

Graphically solve the equation $\dfrac{1}{x-4}=1$ .

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f(x)=1/(x-4)

$x=5$

It’s more likely that the $x$ -value given any function value will not be an integer. This can be difficult to show on a graph. Consequently, it is necessary to use algebraic methods to solve rational equations.

Solving an equation in one variable

Algebraically, $-\dfrac{1}{x}=\dfrac{2}{x}+4$ or $\dfrac{8}{x+1}=2$ is an equation in one variable. When solving an equation in one variable, we find the value of the variable that satisfies the equation (e.g., the $x$ -value). There is no function value to report as the equation is in just one variable.

For example, when solving the equation, $\dfrac{8}{x+1}=2$ , we find the value of $x$ that makes the equation true. The value of $x$ is 3 because $\dfrac{8}{3+1}=\dfrac{8}{4}=2$ . We need algebraic methods to solve equations, including the properties of equality covered in chapter 3.

In summary, when we set functions equal to each other, we are finding the intersection point between two functions (e.g., $f(x)=\dfrac{8}{x+1}$ and $g(x)=2$ ). In this example, the intersection point between the graphs of the two functions $f(x)$ and $g(x)$ is $(3, 2)$ because the $y$ -value is the function value $f(3)=\dfrac{8}{3+1}=2$ and $g(3)=2$ . However, when we are solving an equation algebraically, there is no function in sight so we can just report the value of the variable (e.g., $x=3$ ).

Solving Rational Equations

We have already learned to add/subtract and multiply/divide rational expressions. We will use these skills to clear fractions (or rational expressions) in a rational equation by multiplying both sides by the lowest common denominator. After clearing the rational expressions, we will no longer have denominators to deal with and it will be easier for us to solve the resulting equation for the variable.

Example 3

Find the intersection point between the two functions $f(x)=\dfrac{1}{x+3}$ and $g(x)=\dfrac{1}{3x-5}$ .

Solution

To find the intersection point of $f(x)$ and $g(x)$ , we set the functions equal to one another, $f(x)=g(x)$ , and then solve for $x$ . The value of $x$ will be the $x$ -coordinate of the intersection point. We will then substitute the variable $x$ into either $f(x)$ or $g(x)$ to determine the $y$ coordinate of the intersection point.

$\begin{aligned}f(x)&=g(x)\\\\\dfrac{1}{x+3}&=\dfrac{1}{3x-5}\end{aligned}$

Before we solve the equation for $x$ , we need to discuss the restrictions on the variable. Restrictions on the domain of a rational function occur when the denominator equals zero: In $f(x)$ , $x+3=0$ solves to $x=-3$ . In $g(x)$ , $3x-5=0$ solves to $x=\dfrac{5}{3}$ . Therefore, $-3, \dfrac{5}{3}$ are restricted values of $x$ .

Now, we will start to solve the equation by clearing out the fractions. The LCM of $(x+3)$ and $(3x-5)$ is $(x+3)(3x-5)$ , so we will multiply both sides of the equation by $(x+3)(3x-5)$ :

$\begin{aligned}\dfrac{1}{x+3}&=\dfrac{1}{3x-5}\\\\ \color{blue}{(x+3)(3x-5)}\times\dfrac{1}{(x+3)}&=\dfrac{1}{3x-5}\times\color{blue}{(x+3)(3x-5)}&&\text{Multiply buy the LCM on both sides}\\\\\cancel{(x+3)}(3x-5)\times\dfrac{1}{\cancel{x+3}}&=\dfrac{1}{\cancel{3x-5}}\times(x+3)\cancel{(3x-5)}&&\text{Cancel common factors}\\\\3x-5&=x+3&&\text{This is a linear equation}\\\\3x-5\color{blue}{-x}&=x+3\color{blue}{-x}&&\text{Subtract }x\text{ from both sides}\\\\2x-5&=3\\\\2x-5\color{blue}{+5}&=3\color{blue}{+5}&&\text{Add }5\text{ to both sides}\\\\2x&=8\\\\\dfrac{2x}{\color{blue}{2}}&=\dfrac{2}{\color{blue}{2}}&&\text{Divide both sides by }2\\\\x&=4\end{aligned}$

$x=4$ is not a restricted value, so we have our $x$ -coordinate of the intersection point.

Now, that we know the value for $x$ at the intersection point, we evaluate either $f(4)$ or $g(4)$ to find the $y$ -coordinate.

$\begin{aligned}f(4)&=\dfrac{1}{4+3}\\\\&=\dfrac{1}{7}\end{aligned}$

Therefore, the intersection point between the two functions $f(x)=\dfrac{1}{x+3}$ and $g(x)=\dfrac{1}{3x-5}$ is $\left(4, \dfrac{1}{7}\right)$ .

Try It 3

Find the intersection point between the two functions $f(x)=\dfrac{2}{x-4}$ and $g(x)=\dfrac{1}{2x-6}$ .

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$\left(\dfrac{8}{3},\,-\dfrac{3}{2}\right)$

Example 4

Solve the rational equation $2+\dfrac{4}{y-2}=\dfrac{8}{y^2-2y}$ .

Solution

Before we solve the equation for $y$ , let’s determine the restrictions on the variable by setting each denominator to zero: $y-2=0$ solves to $y=2$ , and $y^2-2y=0$ factors to $y(y-2)=0$ then solves to $y=0,\,2$ . Therefore, $y\neq0, 2$ .

Now, we will start to solve the equation by clearing the fractions by multiplying by the LCM $y(y-2)$ .

$\begin{aligned}2+\dfrac{4}{y-2}&=\dfrac{8}{y^2-2y}\\\\2+\dfrac{4}{y-2}&=\dfrac{8}{y(y-2)}&&\text{Factor denominators}\\\\ \color{blue}{y(y-2)} \times \left(2+\dfrac{4}{y-2}\right)&=\dfrac{8}{y(y-2)} \times\color{blue}{y(y-2)}&&\text{Multiply both sides by the LCM}\\\\2\cdot\color{blue}{y(y-2)}+\dfrac{4}{(y-2)}\cdot\color{blue}{y(y-2)}&=\dfrac{8}{y(y-2)}\cdot\color{blue}{y(y-2)}&&\text{Distribute on the left side}\\\\2y(y-2) + y\cancel{(y-2)}\cdot\dfrac{4}{\cancel{(y-2)}}&=\dfrac{8}{\cancel{y}\cancel{(y-2)}} \cdot \cancel{y}\cancel{(y-2)}&&\text{Cancel common factors}\\\\2y^2-4y+4y&=8&&\text{Simplify. We have a quadratic equation.}\\\\2y^2-8&=0&&\text{Subtract 8 from both sides}\\\\2\left(y^2-4\right)&=0&&\text{Factor}\\\\2(y-2)(y+2)&=0&&\text{Factor: difference of two squares}\\\\y-2=0\text{ or }y+2&=0&&\text{Zero product property}\\\\y&=\pm 2&&\text{Solve}\end{aligned}$

Since $2$ is a restricted value, the solution is $y=-2$ .

Try It 4

Solve the rational equation $\dfrac{4}{x-4}-\dfrac{3}{x-3}=1$ . Remember to check the restrictions on the variable.

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$x=2,\,6$

Example 5

Solve the rational equation $\dfrac{7}{y^2+y-12}-\dfrac{4y}{y^2+7y+12}=\dfrac{6}{y^2-9}$ .

Solution

Let’s start by determining the restrictions on the variable. $y^2+y-12=0$ factors to $(y+4)(y-3)=0$ . So, $y=-4, 3$ . Next, $y^2+7y+12=0$ factors to $(y+4)(y+3)=0$ . So, $y=-4, -3$ . Finally, $y^2-9=0$ factors to $(y-3)(y+3)=0$ . So, $y=3, -3$ . Therefore, $y\neq\pm3,\,-4$ .

Additionally, the LCM is $(y+4)(y-3)(y+3)$ .

Now, we will start to solve the equation by multiplying both sides by the LCM to clear the fractions.

$\begin{aligned}\dfrac{7}{y^2+y-12}-\dfrac{4y}{y^2+7y+12}&=\dfrac{6}{y^2-9}\\\\\dfrac{7}{(y+4)(y-3)}-\dfrac{4y}{(y+4)(y+3)}&=\dfrac{6}{(y+3)(y-3)}&&\text{Factor}\\\\\color{blue}{(y+4)(y+3)(y-3)\cdot}\left(\dfrac{7}{(y+4)(y-3)}-\dfrac{4y}{(y+4)(y+3)}\right)&=\dfrac{6}{(y+3)(y-3)}\color{blue}{\cdot(y+4)(y+3)(y-3)}&&\text{Multiply} \\&&&\text{by LCM}\\\\\color{blue}{(y+4)(y+3)(y-3)}\cdot\dfrac{7}{(y+4)(y-3)}&-\color{blue}{(y+4)(y+3)(y-3)}\cdot\dfrac{4y}{(y+4)(y+3)}\\\\&=\dfrac{6}{(y+3)(y-3)}\cdot(y+4)(y+3)(y-3)&&\text{Distribute}\\\\\cancel{(y+4)}(y+3)\cancel{(y-3)}\cdot\dfrac{7}{\cancel{(y+4)}\cancel{(y-3)}}&-\cancel{(y+4)}\cancel{(y+3)}(y-3)\cdot\dfrac{4y}{\cancel{(y+4)}\cancel{(y+3)}}\\\\&=\dfrac{6}{\cancel{(y+3)}\cancel{(y-3)}}\cdot(y+4)\cancel{(y+3)}\cancel{(y-3)}&&\text{Cancel}\\\\7(y+3)-4y(y-3)&=6(y+4)&&\text{Simplify}\\\\7y+21-4y^2+12y&=6y+24&&\text{Distribute}\\\\4y^2-13y+3&=0&&\text{Simplify}\\\\(4y-1)(y-3)&=0&&\text{Factor}\\\\4y-1=0\text{ or }y-3&=0&&\text{Zero product}\\&&&\text {property}\\\\y&=\dfrac{1}{4},\,3\end{aligned}$

However, $y=3$ is a restricted value since $y\neq\pm3,\,-4$ .

Consequently, the only solution is $y=\dfrac{1}{4}$ .

Try It 5

Solve the rational equation $\dfrac{x+11}{x^2-5x+4}=\dfrac{5}{x-4}-\dfrac{3}{x-1}$ .

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Chapter 7: Rational Functions