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M098 Carson Elementary and Intermediate Algebra 3e Section 2.3.3 Objectives 1. Use the multiplication principle to clear fractions and decimals from equations. Vocabulary Prior Knowledge Use the addition and multiplication property of equality to solve linear equations. New Concepts 1. Clearing Fractions Notice what happens in this arithmetic problem: 5 8 13 7 7 7 When we multiply each term by the LCD, which is 7, we get an equivalent equation 5 8 13 7 7 7 7 7 7 5 8 13 This is still a true statement but we have eliminated the fractions. This is the process of clearing the fraction. You must multiply EVERY TERM by the LCD, even if the term does not contain a fraction. We can do this only because it is an equation! Example 1: 4 2x 1 1 6 9 2x 1 1 184 18 18 6 9 72 – 3(2x+1) = 2 72 – 6x – 3 = 2 -6x + 69 = 2 -6x = -67 x V Zabrocki 2010 67 6 LCD = 18 Multiply each term by 18 Reduce each fraction - this eliminates the denominators and the problem becomes a regular linear equation Distributive property to remove parentheses. Be careful to distribute the minus sign correctly!! Combine like terms Subtract 69 from both sides. Divide both sides by -6. page 1 M098 Carson Elementary and Intermediate Algebra 3e Section 2.3.3 Example 2: 1 3 2 y 3 y 5 y 5 10 5 1 3 2 10 y 3 10 y 5 10 y 5 10 5 2(y – 3) = 3(y + 5) – 2(2)y 2y – 6 = 3y + 15 – 4y There are 3 terms and the LCD is 10. Multiply each term by 10. Reduce each fraction to eliminate the denominators. Distributive property to remove parentheses. 2y – 6 = -y + 15 Combine like terms. 3y – 6 = 15 Add y to both sides. 3y = 21 Add 6 to both sides. y=7 Divide both sides by 3. 2. Clearing Decimals Decimal numbers are just another way to write a fraction. .3 is 3/10, .05 is 5/100. We can clear decimals from an equation in the same way that we clear fractions – by multiplying each term by the LCD. The nice things about decimals is the LCD is always a power of 10 and to multiply by a power of 10, we just move the decimal point to the right the same number of places as there are zeros in the number. If the LCD is 100, we move the decimal point 2 places to the right. If the LCD is 1000, we move the decimal point 3 places to the right. Example 3: 0.06(25) + 0.27x = 0.3(4 + x) 100(0.06)(25) + 100(0.27)x = 100(0.3)(4 + x) Multiply each term by 100. 6(25) + 27x = 30(4 + x) Move decimal 2 places. 150 + 27x = 120 + 30x Distributive property to remove parentheses. 150 - 3x = 120 Subtract 30x from both sides. -3x = -30 Subtract 150 from both sides. x = 10 V Zabrocki 2010 There are 3 terms and the LCD is 100. Divide both sides by -3. page 2 M098 Carson Elementary and Intermediate Algebra 3e Section 2.3.3 Example 4: 0.5 – (t – 6) = 10.4 – 0.4(1 – t) There are 4 terms and the LCD is 10. Remember that there is a 1 in front of the (t – 6). 10(0.5) – 10(1) (t – 6) = 10(10.4) – 10(0.4)(1 – t) Multiply each term by 10. 5 – 10(t – 6) = 104 – 4(1 – t) Move decimal one place. 5 – 10t + 60 = 104 – 4 + 4t -10t + 65 = 100 + 4t Combine like terms. -14t + 65 = 100 Subtract 4t from both sides. -14t = 35 Subtract 65 from both sides. t t V Zabrocki 2010 Distributive property to remove parentheses. Be careful to distribute the minus sign correctly!! 35 14 5 2.5 2 Divide both sides by -14. Always reduce fractions. The answer can be left in either fraction or decimal form. page 3