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Example

Example. Add. Simplify the result, if possible. a) b) Solution a) b). Combining like terms. Combining like terms in the numerator. Factoring. The parentheses are needed to make sure that we practice safe math. Example. Subtract and, if possible, simplify: a) b) Solution a).

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Example

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  1. Example Add. Simplify the result, if possible. a) b) Solution a) b) Combining like terms Combining like terms in the numerator Factoring

  2. The parentheses are needed to make sure that we practice safe math. Example Subtract and, if possible, simplify: a) b) Solution a) Removing the parentheses and changing the signs (using the distributive law) Combining like terms

  3. Example continued b) Removing the parentheses (using the distributive law) Factoring, in hopes of simplifying Removing the clever form of 1

  4. 16a is a factor of the LCM 24b is a factor of the LCM Example For each pair of polynomials, find the least common multiple. a) 16a and 24b b) 24x4y4 and 6x6y2 c) x2 4 and x2  2x  8 Solution a) 16a = 2  2  2  2  a 24b = 2  2  2  3  b The LCM =2  2  2  2  a 3  b The LCM is 24  3  a  b, or 48ab

  5. x2 4 is a factor of the LCM x2 2x  8 is a factor of the LCM Example continued • b) 24x4y4 = 2  2  2  3  x  x  x  x  y  y  y  y • 6x6y2 = 2  3  x  x  x  x  x  x  y  y • LCM = 2  2  2  3  x  x  x  x  y  y  y  y x  x • Note that we used the highest power of each factor. The LCM is 24x6y4 • c) x2 4 = (x  2)(x + 2) • x2  2x  8 = (x + 2)(x  4) • LCM = (x  2)(x + 2)(x  4)

  6. Example For each group of polynomials, find the least common multiple. a) 15x, 30y, 25xyz b) x2 + 3, x + 2, 7 Solution a) 15x = 3  5  x 30y = 2  3  5  y 25xyz = 5  5  x  y  z LCM = 2  3  5  5  x  y  z The LCM is 2  3  52  x  y  z or 150xyz b) Since x2 + 3, x + 2, and 7 are not factorable, the LCM is their product: 7(x2 + 3)(x + 2).

  7. Example Add: Solution 1. First, we find the LCD: 9 = 3  3 12 = 2  2  3 2. Multiply each expression by the appropriate number to get the LCD. LCD = 2  2  3  3 = 36 =

  8. Example Add: Solution First, we find the LCD: a2  4 = (a  2)(a + 2) a2  2a = a(a  2) We multiply by a form of 1 to get the LCD in each expression: LCD = a(a 2)(a + 2). 3a2 + 2a + 4 will not factor so we are done.

  9. Example Subtract: When subtracting a numerator with more than one term, parentheses are important, practice safe math. Solution First, we find the LCD. It is just the product of the denominators: LCD = (x + 4)(x + 6). We multiply by a form of 1 to get the LCD in each expression. Then we subtract and try to simplify. Multiplying out numerators

  10. Example Add: Solution Adding numerators

  11. Continued

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