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Beginning and Intermediate Algebra - Wallace Math Courses ...

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1.4<br />

Solving Linear Equations - Fractions<br />

Objective: Solve linear equations with rational coefficients by multiplying<br />

by the least common denominator to clear the fractions.<br />

Often when solving linear equations we will need to work with an equation with<br />

fraction coefficients. We can solve these problems as we have in the past. This is<br />

demonstrated in our next example.<br />

Example 69.<br />

3 7 5<br />

x − =<br />

4 2 6<br />

+ 7<br />

2<br />

+ 7<br />

2<br />

Focus on subtraction<br />

7<br />

Add to both sides<br />

2<br />

Notice we will need to get a common denominator to add 5 7<br />

+<br />

6 2<br />

common denominator of 6. So we build up the denominator, 7<br />

�<br />

3<br />

2 3<br />

can now add the fractions:<br />

3 21 5<br />

x − =<br />

4 6 6<br />

+ 21<br />

6<br />

+ 21<br />

6<br />

3 26<br />

x =<br />

4 6<br />

3 13<br />

x =<br />

4 3<br />

We can get rid of 3<br />

4<br />

Same problem, with common denominator6<br />

Add 21<br />

6<br />

Reduce 26<br />

6<br />

to both sides<br />

to 13<br />

3<br />

Focus on multiplication by 3<br />

4<br />

. Notice we have a<br />

�<br />

= 21<br />

, <strong>and</strong> we<br />

6<br />

3<br />

by dividing both sides by . Dividing by a fraction is the<br />

4<br />

same as multiplying by the reciprocal, so we will multiply both sides by 4<br />

3 .<br />

� 4<br />

3<br />

� 3<br />

4<br />

� �<br />

13 4<br />

x =<br />

3 3<br />

x = 52<br />

9<br />

Multiply by reciprocal<br />

Our solution!<br />

While this process does help us arrive at the correct solution, the fractions can<br />

make the process quite difficult. This is why we have an alternate method for<br />

dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid<br />

of the fractions for the majority of the problem. We can easily clear the fractions<br />

43

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