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Section 1.4 Solving with 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 1.4.1.

34x72=56 Focus on subtraction +72+72 Add 72 to both sides 

Notice we will need to get a common denominator to add 56+72 . Notice we have a common denominator of 6. So we build up the denominator, 72(33)=216, and we can now add the fractions:

34x216=56 Same problem, with common denominator 6+216+216 Add 216to both sides 34x=266 Reduce 266 to 13334x=133 Focus on multiplication by 34

We can get rid of 34 by dividing both sides by 34. Dividing by a fraction is the same as multiplying by the reciprocal, so we will multiply both sides by43.

(43)34x=133(43) Multiply by reciprocal x=529 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.

Example 1.4.2.

34x72=56LCD=12, multiply each term by 12(12)34x(12)72=(12)56Reduce each 12 with denominators (3)3x(6)7=(2) Multiply out each term 9x42=10 Focus on subtraction by 42+42+42 Add 42 to both sides 9x=52 Focus on multiplication by 99x9=529 Divide both sides by 9x=529 Our Solution 

The next example illustrates this as well. Notice the 2 fraction in the original equation, but to solve it we put the 2 over 1 to make it a fraction.

Example 1.4.3.

23x2=32x+16 LCD = 6, multiply each term by 6(6)23x(6)21=(6)32x+(6)16 Reduce 6 with each denominator (2)2x(6)2=(3)3x+(1)1 Multiply out each term 4x12=9x+1 Notice variable on both sides 4x4x Subtract 4x from both sides 12=5x+1 Focus on addition of 1 11 Subtract 1 from both sides 13=5x Focus on multiplication of 5135=5x5 Divide both sides by 5135=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 1.4.4.

32(59x+427)=3Distribute 32 through parenthesis, reducing if possible 56x+29=3 LCD = 18, multiply each term by 18(18)56x+(18)29=(18)39 Reduce 18 with each denominator (3)5x+(2)2=(18) Multiply out each term 15x+4=54 Focus on addition of 444 Subtract 4 from both sides 15x=50 Focus on multiplication by 1515x15=5015 Divide both sides by 15x=103Our 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 1.4.5.

34x12=13(34x+6)72 Distribute 13 , reduce if possible 34x12=14x+272 LCD =4, multiply each term by 4(4)34x(4)12=(4)14x+(4)21(4)72 Reduce 4 with each denominator (1)3x(2)1=(1)1x+(4)2(2)7 Multiply out each term 3x2=x+814 Combine like terms 8143x2=x6Notice variable on both sides xx Subtract x from both sides 2x2=6 Focus on subtraction by 2+2+2 Add 2 to both sides 2x2=42 Divide both sides by 2x=2Our 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".

Exercises Exercises - Fractions

Exercise Group.

Solve each equation.

3.

0=54(x65)

4.

32n83=2912

5.

3454m=11324

6.

114+34r=16332

7.

63572=52(114+x)

8.

169=43(53+n)

10.

3274v=98

11.

32(73n+1)=32

12.

419=52(x+23)13x

13.

a54(83a+1)=194

14.

13(74k+1)103k=138

15.

556=52(32p53)

16.

12(23x34)72x=8324

17.

169=43(43n43)

18.

23(m+94)103=5318

19.

58=54(r32)

20.

112=43x+53(x74)

21.

113+32b=52(b53)

22.

7643n=32n+2(n+32)

23.

(52x32)=32+x

24.

14916113r=74r54(43r+1)

25.

4516+32n=74n1916

26.

72(53a+13)=114a+258

27.

32(v+32)=74v196

28.

8312x=43x23(134x+1)

29.

479+32x=53(52x+1)

30.

13n+296=2(43n+23)