University:
Purdue University
Course:
MA 15300 | College Algebra
Academic year:

2024
Views:

158

Pages:

3
Author:

Delphinixxc3

Question 1 Solve Equations with Fraction or Decimal Coefficients. In the following exercises, solve each equation by clearing the fractions. 1 3 𝑥 + 1 5 𝑥 = 8 3 1 x+ 5 1 x=8 Answer 1 Consider the equation: 1 3 𝑥 + 1 5 𝑥 = 8 3 1 x+ 5 1 x=8 In order to find the solution, follow the steps below: First, find the least common denominator of all the fractions. Then, multiply both sides of the equation by the least common denominator. Then, use the distributive property, which states that for any real numbers 𝑎 a, 𝑏 b, and 𝑐 c, 𝑎 ( 𝑏 + 𝑐 ) = 𝑎 𝑏 + 𝑎 𝑐 . a(b+c)=ab+ac. Then simplify. The least common denominator is 15. So, 15 ( 1 3 𝑥 + 1 5 𝑥 ) = 15 ( 8 ) 15( 3 1 x+ 5 1 x)=15(8) 15 ( 1 3 ) 𝑥 + 15 ( 1 5 ) 𝑥 = 15 ( 8 ) 15( 3 1 )x+15( 5 1 )x=15(8) 5 𝑥 + 3 𝑥 = 120 5x+3x=120 8 𝑥 = 120 8x=120 Solve further to get, 𝑥 = 120 8 = 15 x= 8 120 =15 Hence, 𝑥 = 15 x=15 is the solution of 1 3 𝑥 + 1 5 𝑥 = 8 3 1 x+ 5 1 x=8. Question 2 Solve the following equation by clearing the fractions: 1 3 𝑥 + 2 5 = 1 − 2 5 𝑥 3 1 x+ 5 2 =1− 5 2 x Answer 1 The equation 1 3 𝑥 + 2 5 = 1 − 2 5 𝑥 3 1 x+ 5 2 =1− 5 2 x can be solved as follows. The least common multiple of 3, 5, and 5 is 15. So, multiply each term by 15: 15 × 1 3 𝑥 + 15 × 2 5 = 15 × 1 − 15 × 2 5 𝑥 15× 3 1 x+15× 5 2 =15×1−15× 5 2 x Cancel the common factors: 5 𝑥 + 6 = 15 − 6 𝑥 5x+6=15−6x Solve the equation: 5 𝑥 + 6 𝑥 = 15 − 6 5x+6x=15−6 11 𝑥 = 9 11x=9 𝑥 = 9 11 x= 11 9 Question 3 Find the least common denominator for the fractions 9 28 28 9 and 21 32 32 21 . Answer 1 Calculation: We will factor each of the denominators 28 and 32 into primes. Prime factors of these numbers are shown below: 28 = 2 × 2 × 7 28=2×2×7 32 = 2 × 2 × 2 × 2 × 2 32=2×2×2×2×2 Now, we will take common numbers along with the remaining numbers. The common number is 2, which occurs twice. Therefore, LCD = 2 × 2 × 2 × 2 × 7 = 224 LCD=2×2×2×2×7=224 Question 4 In the following exercises, find the least common denominator (LCD) for each set of fractions. 9 16 7 12 16 9 12 7 Answer 1 Calculation: We will factor each of the denominators 16 and 12 into primes. Prime factors of these numbers are shown below: 16 = 2 × 2 × 2 × 2 16=2×2×2×2 12 = 2 × 2 × 3 12=2×2×3 Now, we will take common numbers along with the remaining numbers. The common number is 2 twice. Therefore, LCD = 2 × 2 × 2 × 3 = 48 LCD=2×2×2×3=48 Question 5 Find three fractions equivalent to 4 5 5 4 . Answer 1 Concept used: Two fractions 𝑎 𝑏 b a and 𝑥 𝑦 y x are equivalent if and only if they have a common multiple in numerator and denominator, and can be reduced to the same fraction by removing the common multiple. Calculation: Multiply the given fraction 4 5 5 4 by 2, 3, and 4 in numerator and denominator. Multiplying by 2: 4 5 × 2 2 = 8 10 5 4 × 2 2 = 10 8 Multiplying by 3: 4 5 × 3 3 = 12 15 5 4 × 3 3 = 15 12 Multiplying by 4: 4 5 × 4 4 = 16 20 5 4 × 4 4 = 20 16