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Pre Algebra Section 2.3 Unit 2 - Fractions Section 3 – Adding and Subtracting Adding and Subtracting with common denominators If you have the same denominator then you have fractions that are describing the same size pieces. You can add or subtract the numerators to get your answers. ଷ Adding ଶ To Is as easy as adding 2 +3 ହ The answer is . Example 1) ࢊࢊ ଵ ଵଶ + ଵଶ ଼ = ଵଶ ଶ∙ସ = ଷ∙ସ = ૠ + adding 1+7 you get the new numerator of 8 This answer needs to be reduced The common factor is 4 ଶ ଷ American River College 73 Milano Pre Algebra Section 2.3 Example 2) ࡿ࢛࢈࢚࢘ࢇࢉ࢚ ଵଶ ૠ − ହ − ଵଶ Subtracting the numerators 7-5 = 2 gives the new numerator ଶ = ଵଶ This answer needs to be written in lowest terms. ଵ = Example 3) ࡿ࢛࢈࢚࢘ࢇࢉ࢚ − − This problem can also be seen as ଷ −ସ− ସ = ିଷିଵ = ିସ ଵ − ସ + ቀ− ସቁ ଵ ଷ So the math becomes -3 +(-1) = -4 ସ ସ = −1 Example 4 ) ૡ ଼ ଶ − ૢ − ቀ− ૢቁ − ଽ − ቀ− ଽቁ Recall subtracting a negative numbers becomes adding a positive. ଼ ଶ = −ଽ + ଽ = −ଽ Reduce to lowest terms. ଶ = −ଷ American River College 74 Milano Pre Algebra Section 2.3 Mixed numbers If the problem begins with mixed numbers there are a few ways you can handle it. You can leave them as mixed numbers – add the whole numbers to the other whole numbers and the fraction parts to the other fraction parts. This method might require you adjust your answer if you end up with an incorrect form for a fraction. Or you could change both mixed numbers into improper fractions then add. We will complete the next example both ways Example 5) Using Mixed numbers + Adding the whole number 3+6 = 9 so the new whole number is 9 , adding the numerators 1+2 = 3 and ଷ keeping the common denominator of 5 gives us a new fraction part of . ହ ଵ ଶ 3ହ + 6ହ ଵ ଶ = ሺ3 + 6ሻ + ቀହ + ହቁ ଷ = ሺ9ሻ + ቀହቁ ଷ = 9ହ is the answer as a mixed number. Example 5) Using improper fractions 3 +6 ଵ ଶ = ଵ ଷଶ = ସ଼ + ହ ହ ହ + ହ ହ is the answer as an improper fraction. Notice that since 5 goes into 48 9 times with 3 left over. ସ଼ ହ ଷ = 9 ହ . Unless you are asked to write your answer a specific way either answer is correct. American River College 75 Milano Pre Algebra Section 2.3 Example 6) Using Mixed Numbers ଵ ଶ − + −4 ଷ + 2 ଷ Since the signs are different this is a subtraction problem. We always do the number with the largest absolute value minus the smaller absolute value. So we need ଵ ଶ to do 4 ଷ − 2 ଷ . However the fraction we are taking away from must be larger. So we must borrow from the whole number 4. So ଵ ଵ ଵ ଷ ଵ ସ ସ 4 = 4 + ଷ = 3 + 1 + ଷ = 3 + ଷ + ଷ = 3 + ଷ = 3 ଷ. ଷ ଵ ଶ ସ ଶ Therefore 4 ଷ − 2 ଷ becomes 3 ଷ − 2 ଷ . ଶ ସ ଶ 3ଷ − 2ଷ = 1ଷ ଵ ଶ However the original problem was −4 ଷ + 2 ଷ , so the negative number was larger making the end answer negative. ଶ ଵ ଶ −4 + 2 ଷ = −1 ଷ ଷ Example 6) Using improper fractions ଵ ଶ − + −4 ଷ + 2 ଷ =− ଵଷ ଷ ହ Write both fractions as improper fractions ଼ +ଷ Recall -13 +8 = -5 ଶ = − ଷ or −1 ଷ Adding and subtracting with uncommon denominators In order to add or subtract fractions they must have the same denominator. So if they do not we must create the common denominator. American River College 76 Milano Pre Algebra Section 2.3 Example 7) +ૡ ଵ ହ The LCD or Lowest Common Denominator is the Least Common Multiple between 6 and 8. The smallest number we can turn them both into is 24. In other words we need to add a factor of 4 to the 6, and a factor of 3 to the 8. +଼ ଵ∙ସ ହ∙ଷ = ∙ସ + ଼∙ଷ ଵହ ସ = ଶସ + ଶସ ଵଽ = ଶସ Example 8) − − ଵଶ ହ =− =− ૠ − ૡ − ଵ଼ ହ∙ଷ ଵଶ∙ଷ ଵହ ଷ ∙ଶ − ଵ଼∙ଶ ଵସ − ଷ Sometimes it helps to look at the prime factors when looking for the LCD 12 = 2 ∙ 2 ∙ 3 18 = 2 ∙ 3 ∙ 3 The LCD needs to contain every factor above two 2’s and two 3’s. ଶଽ = − ଷ Example 9) ହ ଵ − + To change the denominator we multiply both the numerator and denominator by the factors that were missing to obtain the LCD. − ଶଵ + ହ ଵ =− ଷ∙ + ଶ∙ଷ ହ∙ଶ ଵ∙ = − ଷ∙∙ଶ + ଶ∙ଷ∙ ଵ = − ସଶ + ସଶ ଷ = − ସଶ Always reduce if you can. ଵ = − ଵସ American River College 77 Milano Pre Algebra Section 2.3 Example 10) − ସ 6 − 3 ଵଵ ଵଵ We need to borrow 1 from the 6 to get a fraction to take away from. ସ = 5 ଵଵ − 3 ଵଵ ଵଵ ଵଵ 1 = ଵଵ So 6 = 5 ଵଵ. = 2 ଵଵ Example 11) ࢞ ଷ ௫ − ଵ −ଶ ଵ∙௫ ଷ∙ଶ = ௫∙ଶ − ଶ∙௫ = = ଶ௫ ௫ − ଶ௫ The LCD has to have an x because the first fraction has a denominator of x. It also must have a 2 since the second fraction has a denominator of 2. The LCD is 2x. We multiply in the factor that is missing. ି௫ ଶ௫ American River College 78 Milano Pre Algebra Section 2.3 Perimeter The perimeter of a shape can be found by adding together all the sides . The perimeter is the distance around a shape. Example 12) Find the Perimeter of the triangle below ଷ ଶ Perimeter = add up all 3 sides = 1 + 2 + ଷ ସ ଷ ଶ ଶ 1ସ + ଵ + ଷ ଶ ଶ =ସ+ଵ+ଷ = ∙ଷ ସ∙ଷ ଶଵ ଶ∙ଵଶ ଶ∙ସ + ଵ∙ଵଶ + ଷ∙ସ ଶସ ଼ = ଵଶ + ଵଶ + ଵଶ ହଷ = ଵଶ The Perimeter is ହଷ ଵଶ ହ ݅݊ or 4 ଵଶ ݅݊. American River College 79 Milano Pre Algebra Section 2.3 This Page was intentionally left Blank. American River College 80 Milano Pre Algebra Section 2.3 Exercise 2.3 NAME:___________________________________ Add or subtract 4. ଷ ଵ 1. ଷ ସ ଷ −ହ ହ ଵ −ସ 5. ଵ ଵ ଵ 16. ଷ ଵ −ସ ଷ − ቀ− ቁ ସ ହ ଽ ଵ ଵ +ଽ 6. ଵ ଵ American River College ଶ 17. − ହ + ହ − ቀ− ଵଶቁ ସ ଷ ଵ 14. − ହ − ଷ ଵ ଵ ଵଶ ଶ 9. 3 ହ − 1 ହ 11. 2 ସ − 7 ସ ସ ଶ 3. − ଽ − ଽ 8. 2 ସ − 1 ସ 10. −2 ହ − 3 ହ ଷ ଶ ଷ 7. 3 ଷ + 2 ଷ 13. ଶ 2. − + ଵ ଵଶ 81 ଵ ହ 12. −3 − ቀ−4 ቁ 15. 18. ଷ − ቀ− ହቁ ଶ − ସ ଷ ଵ ଵ Milano Pre Algebra Section 2.3 ଶ ଶ 19. 5 + 3 ହ 20. 3 − 5 ଷ ଷ ଵ ଵ 22. −5 ଶ + 2 ହ 23. 2 ହ − 6 ଶ ଶ 21. 6 ହ + 2 ଷ ଶ ହ 24. −2 − 4 Find the perimeter of the following shapes 5݉ 25. ଵ ଵ 2ସ݉ 2 ݉ ସ 5݉ ଵ 5 ସ ݅݊ 26. ଶ 3݅݊ 2 ଷ ݅݊ 2݅݊ American River College 82 Milano
