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Section 7.5 Equations that Include Algebraic Fractions

Subsection 1. Solve equations graphically

If we can't solve an equation algebraically, we may be able use a graph to find at least an approximation for the solution.

Subsubsection Examples

Example 7.52.

Use a graph to solve the equation \(~2x^3+9x^2-8x+36=0\)

Solution

We graph the equation \(~y=2x^3+9x^2-8x-36~\) and look for the points where \(~y=0~\) (the \(x\)-intercepts).

cubic

From the graph, we estimate the solutions at \(~x=-4.5,~x=-2,\) and \(~x=2\text{.}\) By substituting each of these values into the original equation, you can verify that they are indeed solutions.

Example 7.53.

Use a graph to solve the equation \(~x^2+2x+3 = 15-2x\)

Solution

We graph the equations \(~y_1=x^2+2x+3~\) and \(~y_2=15-2x~\) and look for points on the two graphs where the coordinates are equal (intersection points).

parabola and line

From the graph, we see that the points with \(~x=-6~\) and \(~x=2~\) have the same \(y\)-coordinate on both graphs. In other words, \(~y_1=y_2~\) when \(~x=-6~\) or \(~x=2~\text{,}\) so \(~x=-6~\) and \(~x=2~\) are the solutions.

Subsubsection Exercises

Use a graph to solve the equation \(~2x^3+7x^2-7x-12=0\)

Answer
\(x = -4,~ -1,~ \dfrac{3}{2}\)

Use a graph to solve the equation \(~\dfrac{24}{x+4}=11+2x-x^2\)

Answer
\(x=-1,~4\)

Subsection 2. Solve proportions

Cross-multiplying is a short-cut method for clearing the fractions from a proportion. Remember that it works only on proportions, not on other types of equations or operations on fractions!

Property of Proportions.

We can clear the fractions from a proportion by cross-multiplying.

\begin{equation*} \text{If}~~~~\dfrac{a}{b} = \dfrac{c}{d},~~~~\text{then}~~~~ad=bc. \end{equation*}

Subsubsection Examples

Example 7.56.

Solve \(~\dfrac{2.4}{1.5} = \dfrac{8.4}{x}\)

Solution

We apply the property of proportions and cross-multiply to get

\begin{align*} 2.4x \amp = 1.5(8.4) \amp \amp \blert{\text{Divide both sides by 2.4.}}\\ x \amp = \dfrac{1.5(8.4)}{2.4} \amp \amp \blert{\text{Simplify the right side.}}\\ x \amp = 5.25 \end{align*}
Example 7.57.

If 3 pounds of coffee beans makes 225 cups, how many pounds of coffee beans will you need to make 3000 cups of coffee?

Solution

We assume that the number of cups is proportional to the amount of coffee beans. That is, the ratio of cups to coffee beans is constant. So

\begin{align*} \dfrac{225~ \text{cups}}{3~ \text{pounds}} \amp = \dfrac{3000~ \text{cups}}{x~ \text{pounds}}\\ \dfrac{225}{3} \amp = \dfrac{3000}{x} \end{align*}

Cross-multiplying, we find

\begin{align*} 225x \amp = 3(3000)\\ x \amp = \dfrac{3(3000)}{225}= 40 \end{align*}

You will need 40 pounds of coffee beans.

Subsubsection Exercises

Solve \(~\dfrac{182}{65} = \dfrac{21}{w}\)

Answer
\(w=7.5\)

A cinnamon bread recipe calls for \(1 \frac{1}{4}\)tablespoons of cinnamon and 5 cups of flour. Write and solve a proportion to discover how much cinnamon would be needed with 8 cups of flour.

Answer

\(\dfrac{1.25}{5} = \dfrac{x}{8};~~2~\) tablespoons

Subsection 3. Solve quadratic equations

Once we have cleared the fractions from an equation, we may have a quadratic equation to solve. We can choose the easiest method to solve: factoring, extracting roots, or the quadratic formula.

Subsubsection Example

Example 7.60.

Solve each quadratic equation by the easiest method.

  1. \(\displaystyle 2x^2-2x=3\)
  2. \(\displaystyle (2x-1)^2=3\)
  3. \(\displaystyle 2x^2-x=3\)
Solution

  1. Because \(~2x^2-2x-3~\) does not factor, we use the quadratic formula.

    \begin{equation*} x = \dfrac{2 \pm \sqrt{(-2)^2-4(2)(-3)}}{2(2)} = \dfrac{2\pm \sqrt{28}}{4} = \dfrac{1 \pm \sqrt{7}}{2} \end{equation*}

  2. We use extraction of roots.

    \begin{align*} 2x-1 \amp = \pm \sqrt{3}\\ x \amp = \dfrac{1 \pm \sqrt{3}}{2} \end{align*}

  3. We write the equation in standard form and factor the left side.

    \begin{align*} 2x^2-x-3 \amp = 0\\ (2x-3)(x+1) \amp = 0\\ 2x-3=0~~~~x+1 \amp = 0\\ x = \dfrac{3}{2}~~~~x \amp = -1 \end{align*}

Subsubsection Exercises

Solve each equation by the easiest method.

  1. \(\displaystyle 3x^2+10x=8\)
  2. \(\displaystyle x^2+6x+9=8\)
  3. \(\displaystyle 81x^2-18x+1=0\)
  4. \(\displaystyle 9x^2+18x=27\)
Answer
  1. \(\displaystyle x=-4,~\dfrac{2}{3}\)
  2. \(\displaystyle x=-2\pm2\sqrt{2}\)
  3. \(\displaystyle x=\dfrac{1}{9},~\dfrac{1}{9}\)
  4. \(\displaystyle x=-3,~1\)