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Beginning and Intermediate Algebra - Wallace Math Courses ...

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missing on each fraction.<br />

Example 351.<br />

� �<br />

2<br />

2<br />

6<br />

4(2a+1)<br />

6 3a<br />

+<br />

8a +4 8<br />

Factor denominators to find LCD<br />

4(2a+1) 8 LCD is 8(2a+1), build up each fraction<br />

+ 3a<br />

8<br />

� �<br />

2a +1<br />

2a +1<br />

12<br />

8(2a +1) + 6a2 + 3a<br />

8(2a +1)<br />

6a 2 +3a + 12<br />

8(2a+1)<br />

Multiply first fraction by 2, second by 2a+1<br />

Add numerators<br />

Our Solution<br />

With subtraction remember to add the opposite.<br />

Example 352.<br />

x +1<br />

x − 4 −<br />

x + 1<br />

x2 − 7x + 12<br />

Add the opposite (distribute negative)<br />

x +1 −x − 1<br />

+<br />

x − 4 x2 − 7x + 12<br />

Factor denominators to find LCD<br />

x − 4 (x − 4)(x − 3) LCD is (x − 4)(x − 3), build up each fraction<br />

� �<br />

x − 3 x +1 −x − 1<br />

+<br />

x − 3 x − 4 x2 − 7x + 12<br />

x2 − 2x − 3<br />

(x − 3)(x − 4) +<br />

−x − 1<br />

(x − 3)(x − 4)<br />

x 2 − 3x − 4<br />

(x − 3)(x − 4)<br />

(x − 4)(x +1)<br />

(x − 3)(x − 4)<br />

x + 1<br />

x − 3<br />

Only first fraction needs to be multiplied by x − 3<br />

Add numerators, combine like terms<br />

Factor numerator<br />

Divide out x − 4 factor<br />

Our Solution<br />

260

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