EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)

Note:

• A rational equation is an equation where at least one denominator contains a variable.

• When a denominator contains a variable, there is a restriction on the domain. The variable cannot take on any number that would cause any denominator to be zero.

• The first step is solving a rational equation is to convert the equation to an equation without denominators. This new equation may be equivalent (same solutions as the original equation) or it may not be equivalent (extraneous solutions).

• The next step is to set the equation equal to zero and solve.

• Remember that you are trying to isolate the variable.

• Depending on the problem, there are several methods available to help you solve the problem.

If you would like an in-depth review of fractions, click on Fractions.

Solve for x in the following equation.

Problem 5.5 c:         $$\frac{500}{2x^{2}-9x+1}+1=x-5$$

Answer:     $x\approx 10.23$

Simplify the equation by subtracting 1 from both sides of the equation..

$$\begin{eqnarray*}&& \\ \displaystyle \frac{500}{2x^{2}-9x+1}+1 &=&x-5 \\ && \\... ...displaystyle \frac{500}{2x^{2}-9x+1} &=&x-6 \\ && \\ && \\ && \end{eqnarray*}$$

Multiply both sides of the equation by an expression that represents the lowest common denominator. The expression $\left( 2x^{2}-9x+1\right)$ is the smallest expression because it is the smallest expression that is divisible by all the denominators.

$$\begin{eqnarray*}&& \\ \displaystyle \frac{500}{2x^{2}-9x+1} &=&x-6 \\ && \\ ... ... && \\ && \\ 0 &=&2x^{3}-21x^{2}+55x-506 \\ && \\ && \\ && \end{eqnarray*}$$

This expression is not easily factored. You can solve this equation using iteration or by graphing.

With iteration, choose two real numbers and evaluate f(x)=2x³-21x²+55x-506 at each of the two numbers. You are looking for a sign change. If there is a sign change, then you know that one real solution is located between the two values you choose to test.

$$\begin{eqnarray*}&& \\ f(0) &=&-506 \\ && \\ f\left( 20\right) &=&8194 \\ && \\ && \end{eqnarray*}$$

This indicates that a real solution exists between 0 and 20. Now narrow the range to 10 and 20.

$$\begin{eqnarray*}&& \\ f(10) &=&-56 \\ && \\ f\left( 20\right) &=&8194 \\ && \end{eqnarray*}$$

This indicates that real solution exists between 10 and 20. Narrow the range again. Since -56 is close to 0, choose a number near 10.

$$\begin{eqnarray*}&& \\ f\left( 11\right) &=&220 \\ && \\ f\left( 10\right) &=&-56 \\ && \end{eqnarray*}$$

You know a real solution exists between 10 and 11. Narrow the range again.

$$\begin{eqnarray*}&& \\ f\left( 10.5\right) &=&71.5 \\ && \\ f\left( 10\right) &=&-56 \\ && \end{eqnarray*}$$

You know a real solution exists between 10 and 1.5. Narrow the range again.

$$\begin{eqnarray*}&& \\ f\left( 10.25\right) &=&5.219 \\ && \\ f\left( 10\right) &=&-56 \\ && \end{eqnarray*}$$

You know a real solution exists between 10 and 10.25. Narrow the range again.

$$\begin{eqnarray*}&& \\ f\left( 10.2\right) &=&-7.424 \\ && \\ f\left( 10.25\right) &=&5.219 \\ && \end{eqnarray*}$$

You know a real solution exists between 10.2 and 10.25. You can keep going until you have the accuracy you need in your answer.

You can also check your answer by graphing

$$\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{500}{2x^{2}-9x+1}-x+6 \\ && \end{eqnarray*}$$

$.\quad$(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at one spot, $10.2294572878\approx 10.23$

If you would like to test yourself by working some problems similar to this example, click on problem.

If you would like to go back to the equation table of contents, click on Contents

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