Sect Least Common Denominator
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1 4 Sect.3 - Least Common Denominator Concept #1 Writing Equivalent Rational Expressions Two fractions are equivalent if they are equal. In other words, they are equivalent if they both reduce to the same fraction. For instance, the fractions 30 1 and 0 3 are equivalent because they both reduce to 10. We can also build fractions so that they have a given denominator by multiplying the top and bottom by the same non-zero number. If we want to write 10 so that it has a denominator of 3, we would multiply top and bottom by since goes into 3 five times: (1) We can do the same thing with rational expressions. Write the following as an equivalent rational expression with the indicated denominator: Ex. 1 Ex. Ex. 3 3 xy x yz Take the new denominator divide by the old denominator: x yz xy xz. So, we need to multiply top and bottom by xz: 3 xy 3 xy (1) 3 xy xz xz 1xz x yz x 3y 1y 4 Since 1y 4 3y y 3, we need to multiply top and bottom by y 3 : x 3y x 3y (1) x 3y y3 y 3 w 10w 31w+1 49xy 3 1y 4 Since 10w 31w + 1 (w )(w 3) and
2 (w )(w 3) (w ) (w 3), then we need to multiply top and bottom by (w 3): w (w 3) (w 3) (w 3) (w )(w 3) 10w 6 (w )(w 3) Notice that we left the denominator in factored form. In general, we usually leave the denominator of a rational expression in factored form. Ex x 16 81x x 3 19x 6 9x 16 (3x 4)(3x + 4) 81x x 3 19x 6 (81x x 3 ) + ( 19x 6) x 3 (3x + 4) 64(3x + 4) (3x + 4)(x 3 64) (3x + 4)(3x 4)(9x + 1x + 16) Since (3x + 4)(3x 4)(9x + 1x + 16) (3x 4)(3x + 4) (9x + 1x + 16), then we need to multiply top and bottom by (9x + 1x + 16): 8 9x 16 8 (3x 4)(3x+4) (9x +1x+16) (9x +1x+16) x +96x+18 (3x 4)(3x+4)(9x +1x+16) 8(9x +1x+16) (3x 4)(3x+4)(9x +1x+16) Concept # Least Common Denominator When adding two fractions with different denominators, we had to first find the least common denominator. To find the LCD of two fractions, we first find the prime factorization of each denominator. Next, we write down the product of each factor that appears in the prime factorizations. Finally, we choose the highest power of the factor in the prime factorizations. The resulting product is our L.C.D. For instance, if our two denominators are 18 and 60, the prime factorization of each is: The factors, 3, and appear in the prime factorizations so we write their product: 3
3 But the highest power of the factors of is and the highest power of the factors of 3 is, so we write: 3 The principle is the same for finding the L.C.D. of rational expressions: To Find the L.C.D. of Two or More Rational Expressions: 1. Factor the denominator of each rational expression.. Write down as a product each factor that appears in at least one of the factored denominators. 3. Raise each factor in step # to the highest power of that factor that appears in the factored denominators. Find the L.C.D. of the following rational expression: Ex. Ex. 6 1, 1 a 3 b 8ab c We do not need to factor the denominators since they each have one term. The least common multiple of and 8 is 8. For the variable part, we write the product of each factor that appears: 8abc The highest power of a is 3, of b is, and c is 1. Thus, L.C.D. is 8a 3 b c x, x 4x 4 18x +1x 6 4x 4 4(x 1) 4(x 1)(x + 1) 18x + 1x 6 6(3x + x 1) 6(3x 1)(x + 1) The least common multiple of 4 and 6 is 1. For the variable part, we write the product of each factor that appears: 1(x 1)(x + 1)(3x 1) The highest power of each factor is one. Thus, L.C.D. 1(x 1)(x + 1)(3x 1) 6
4 Ex. Ex. 8, 1a 3 +48a +48a 4a +0a+4 1a a + 48a 1a(a + 4a + 4) 1a(a + ) 4a + 0a + 4 4(a + a + 6) 4(a + 3)(a + ) The least common multiple of 4 and 1 is 1. For the variable part, we write the product of each factor that appears: 1a(a + )(a + 3) The highest power of each factor is one except of the factor of (a + ); its highest power is Thus, L.C.D. 1a(a + ) (a + 3) y, 41y 1000y y 3 300y 90y y 3 + (10y) 3 + (3) 3 (10y + 3)(100y 30y + 9) 1000y 3 300y 90y + (1000y 3 300y ) + ( 90y + ) 100y (10y 3) 9(10y 3) (10y 3)(100y 9) (10y 3)(10y 3)(10y + 3) (10y 3) (10y + 3) We write the product of each factor that appears: (10y + 3)(10y 3)(100y 30y + 9) The highest power of each factor is one except of the factor of (10y 3); its highest power is Thus, L.C.D. (10y + 3)(10y 3) (100y 30y + 9) Concept #3 Writing Rational Expressions with the L.C.D. Now, let s put concepts #1 and # together. Find the L.C.D. of each pair of rational expressions and then write each one as an equivalent rational expression with the L.C.D. for the denominator: Ex. 9, 11 14a b 1ab 3 The L.C.D. of 14a b, 11 1ab 3 is 4a b 3.
5 Since 4a b 3 14a b 3b and 4a b 3 1ab 3 a, multiply the top and bottom of the first fraction by 3b and the top and bottom of the second by a: 3b 14a b 3b 11 a 1ab 3 a 1b 4a b 3 a 4a b 3 So, our fractions are 1b and a. 4a b 3 4a b 3 Ex. 10, 64x x 64x 3 1 (4x) 3 (1) 3 (4x 1)(16x + 4x + 1) 1 16x 16x + 1 1(16x 1) 1(4x 1)(4x + 1) The question becomes what to do with the negative one? But, the following are equivalent: 1 16x 1(4x 1)(4x+1) (4x 1)(4x+1) So, if we get a 1 in the denominator as part of the product, we can get rid of it by changing the sign of the numerator: So, our denominators are (4x 1)(16x + 4x + 1) & (4x 1)(4x + 1). We write the product of each factor that appears: (4x 1)(4x + 1)(16x + 4x + 1) The highest power of each factor is one. Thus, L.C.D. (4x 1)(4x + 1)(16x + 4x + 1) Since (4x 1)(4x + 1)(16x + 4x + 1) (4x 1)(16x + 4y + 1) (4x + 1) and (4x 1)(4x + 1)(16x + 4x + 1) (4x 1)(4x + 1) (16x + 4x + 1), multiply the top and bottom of the first fraction by (4x + 1) and the top and bottom of the second by (16x + 4x + 1): (4x+1) 64x 3 1 (4x 1)(16x +4x+1) (4x 1)(16x +4x+1) (4x+1) (4x+1) 8x+ (4x 1)(4x+1)(16x +4x+1) (4x 1)(4x+1)(16x +4x+1) 1 16x (4x 1)(4x+1) (4x 1)(4x+1) (16x +4x+ 1) (16x +4x+ 1) (16x +4x+ 1) (4x 1)(4x+1)(16x +4x+ 1) 80x 0x (4x 1)(4x+1)(16x +4x+ 1) 8
6 9 So, our fractions are 8x+ (4x 1)(4x+1)(16x +4x+1) & 80x 0x. (4x 1)(4x+1)(16x +4x+ 1) x Ex. 11, 6x +11x 3 8x +10x 63 6x + 11x 3 (3x )(x + ) 8x + 10x 63 (4x 9)(x + ) We write the product of each factor that appears: (3x )(x + )(4x 9) The highest power of each factor is one. Thus, L.C.D. (3x )(x + )(4x 9) Since (3x )(x + )(4x 9) (3x )(x + ) (4x 9) and Ex. 1 (3x )(x + )(4x 9) (4x 9)(x + ) (3x ), multiply the top and bottom of the first fraction by (4x 9) and the top and bottom of the second by (3x ): x 6x +11x 3 x (3x )(x+) x(4x 9) (3x )(x+)(4x 9) x (3x )(x+) (4x 9) (4x 9) 8x 18x (3x )(x+)(4x 9) 8x +10x 63 (4x 9)(x+) (4x 9)(x+) (3x ) (3x ) (3x ) (3x )(x+)(4x 9) So, our fractions are Since 9 x, 11 x 9 1x (3x )(x+)(4x 9) 8x 18x (3x )(x+)(4x 9) and 9 x x+9 (x 9) L.C.D. Our fractions are x 9 x 9 and 11 x 9. 1x (3x )(x+)(4x 9)., we do not need to find the
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