Solving Linear Equations - Fractions

Transcription

1 1.4 Solving Linear Equations - Fractions Objective: Solve linear equations with rational coefficients by multiplying by the least common denominator to clear the fractions. Often when solving linear equations we will need to work with an equation with fraction coefficients. We can solve these problems as we have in the past. This is demonstrated in our next example. Example 1. 4 x 7 = Focus on subtraction Add 7 to both sides Notice we will need to get a common denominator to add + 7. Notice we have a ( ) common denominator of. So we build up the denominator, 7 = 1, and we can now add the fractions: 4 x 1 = Same problem, with common denominator Add 1 to both sides 4 x = Reduce to 1 4 x = 1 Focus on multiplication by 4 We can get rid of by dividing both sides by. Dividing by a fraction is the 4 4 same as multiplying by the reciprocal, so we will multiply both sides by 4. ( ) 4 4 x = 1 x = 9 ( ) 4 Multiply by reciprocal Our solution! While this process does help us arrive at the correct solution, the fractions can make the process quite difficult. This is why we have an alternate method for dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid of the fractions for the majority of the problem. We can easily clear the fractions by finding the LCD and multiplying each term by the LCD. This is shown in the next example, the same problem as our first example, but this time we will solve by clearing fractions. 1

2 Example. 4 x 7 = (1) x (1)7 = (1) 4 LCD = 1, multiply each term by 1 Reduce each 1 with denominators ()x ()7 =() 9x 4 = 10 Focus on subtraction by Add 4 to both sides 9x = Focus on multiplication by Divide both sides by 9 x = Our Solution 9 The next example illustrates this as well. Notice the isn t a fraction in the origional equation, but to solve it we put the over 1 to make it a fraction. Example. x = x + 1 () x () = () 1 x + ()1 LCD =, multiply each term by Reduce with each denominator ()x () =()x +(1)1 4x 1 = 9x +1 Notice variable on both sides 4x 4x Subtract 4x from both sides 1 = x +1 Focus on addition of1 1 1 Subtract 1 from both sides 1 =x Focus on multiplication of Divide both sides by 1 = x Our Solution We can use this same process if there are parenthesis in the problem. We will first distribute the coefficient in front of the parenthesis, then clear the fractions. This is seen in the following example. Example 4. ( 9 x ) = Distribute through parenthesis, reducing if possible x + = LCD = 18, multiply each term by 18 9

3 (18) x + (18) = (18) 9 9 ()x +()=(18) 1x +4=4 4 4 Reduce 18 with each denominator Focus on addition of4 Subtract4from both sides 1x = 0 Focus on multiplication by Divide both sides by 1. Reduce on right side. x = 10 Our Solution While the problem can take many different forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term by the LCD and reduce. This will give us a problem with no fractions that is much easier to solve. The following example again illustrates this process. Example. 4 x 1 = 1 ( 4 x + ) 7 4 x 1 = 1 4 x + 7 (4) 4 x (4)1 = (4)1 4 x + (4) (4)7 1 Distribute 1, reduce if possible LCD = 4, multiply each term by 4. Reduce 4 with each denominator (1)x ()1 =(1)1x +(4) ()7 x =x+8 14 Combine like terms 8 14 x =x Notice variable on both sides x x Subtract x from both sides x = Focus on subtraction by + + Addto both sides x= 4 Focus on multiplication by Divide both sides by x= Our Solution World View Note: The Egyptians were among the first to study fractions and linear equations. The most famous mathematical document from Ancient Egypt is the Rhind Papyrus where the unknown variable was called heap Attribution.0 Unported License. (

4 1.4 Practice - Fractions Solve each equation. 1) 1 (1 + p)= 0 ) 0= 4 (x ) ) 4 4 m = ) 7 = ( x) 9) b + 9 = 11 11) (7 n+1)= 1) a ( 8 19 a + 1) = 4 4 1) = ( p ) 17) 1 9 = 4 ( 4 n 4 ) 19) 8 = 4 (r ) 1) 11 + b = (b ) ) ( x )= + x ) n= 7 4 n ) (v + ) = 7 4 v 19 9) x = ( x +1) ) 1 = k + 4) n 8 = 9 1 ) 11 + r = ) 1 9 = 4 ( + n) 10) 7 4 v = 9 8 1) 41 9 = (x + ) 1 x 14) 1 ( 7 10 k +1) k = ) 1 ( x 4 ) 7 x = ) (m + 9 ) 10 = ) 1 1 = 4 x + (x 7 4 ) ) 7 4 n = n+(n + ) 4) r= 7 4 r 4 ( 4 r + 1) ) 7 (a )= a ) 8 1 x = 4 x ( 1 4 x + 1) 0) 1 n + 9 = (4 n + ) Attribution.0 Unported License. ( 4

5 1.4 Answers to Solving with Fractions 1) 4 ) 4 11) 0 1) 4 ) 1 ) ) 1) 4) 9 4 4) 1 ) 19 ) 8 7) 7 9 8) 1 9) 14) 1 1) 4 1) 1 17) 0 18) 19) 1 0) 1 ) 1 ) 1 7) 8) 9) 4 10) 1) 1 0) Attribution.0 Unported License. (