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Intermediate Algebra – Student Workbook – Second Edition 2013

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Lesson 6a - Exponents and Rational Functions<br />

Mini-Lesson<br />

SOLVE RATIONAL EQUATIONS ALGEBRAICALLY<br />

To solve rational equations algebraically (also called symbolically):<br />

Identify the common denominator for all fractions in the equation.<br />

Clear the fractions by multiplying both sides of the equation by the common<br />

denominator<br />

Take note of the values of x that make the common denominator zero. These x-values<br />

cannot be used as solutions to the equation since we cannot divide by 0.<br />

Solve for x.<br />

Check your work by plugging the value(s) back into the equation or by graphing.<br />

Problem 8 WORKED EXAMPLE <strong>–</strong> SOLVE RATIONAL EQUATIONS<br />

ALGEBRAICALLY<br />

3<br />

Solve 5x<br />

4 algebraically. Round solutions to two decimal places.<br />

x 4<br />

Common denominator for all sides is x-4. Multiply both sides of the equation by (x <strong>–</strong> 4)<br />

and solve for x to get the following:<br />

3<br />

(5x)(<br />

x 4) (4 )( x 4)<br />

x 4<br />

2<br />

3<br />

5x<br />

20x<br />

4( x 4) ( x 4)<br />

x 4<br />

2<br />

5x<br />

20x<br />

4x<br />

16<br />

3<br />

5x<br />

2<br />

20x<br />

4x<br />

13<br />

0<br />

5x<br />

2<br />

5x<br />

2<br />

20x<br />

4x<br />

13<br />

24x<br />

13<br />

0<br />

Notice that we now have a quadratic equation, which can be solved using the methods of last<br />

chapter. Because we are asked to solve our original problem algebraically, let’s continue that<br />

process and not resort to a graphical solution. We will use the Quadratic Formula with a=5, b=-<br />

24, and c=13 to get:<br />

( 24)<br />

<br />

x <br />

( 24)<br />

2(5)<br />

2<br />

4(5)(13) 24 <br />

<br />

576 260<br />

10<br />

24 316<br />

<br />

10<br />

Because we want rounded solutions, I do NOT need to continue reducing my fraction solutions<br />

above but can compute the following directly:<br />

24 316 24 316<br />

x 4.18, x .62<br />

10<br />

10<br />

These solutions match what we found in the graphing example previously.<br />

To check, plug the values back into the original equation (one at a time) or use the graphing<br />

method.<br />

Scottsdale Community College Page 250 <strong>Intermediate</strong> <strong>Algebra</strong>

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