18

3c

MODULE 3. FRACTIONS

Solving Equations

While it is possible to solve these equations as is, usually it is preferable to clear all of the fractions prior to solving.

Clearing Fractions from the Equation Clearing Fractions from the Equation. To clear all fractions from an equation, multiply both sides of the equation by the least common denominator of the fractions that appear in the equation. Let’s put this idea to work. You Try It! EXAMPLE 1. Solve the following equation for x: x−

5 1 = . 6 3

Solution. First clear all fractions from the equation. To achieve this, multiply both sides of the equation by the least common denominator of all the fractions appearing in the equation. 1 5 = 6 ! " 3! " 5 1 6 x− =6 6 3 ! " ! " 5 1 6x − 6 =6 6 3 x−

6x − 5 = 2

Original equation. Multiply both sides by 6. Distribute the 6. On each side, multiply first. ! " ! " 5 1 6 = 5 and 6 = 2. 6 3

Note that the equation is now entirely clear of fractions, making it a much simpler equation to solve. 6x − 5 + 5 = 2 + 5 6x = 7 7 6x = 6 6 7 x= 6

Add 5 to both sides. Simplify both sides. Divide both sides by 6. Simplify.

19

3C. SOLVING EQUATIONS

Checking the Solution. Substitute 7/6 for x in the original equation and simplify. 5 6 7 5 − 6 6 2 6 1 3 x−

1 3 1 = 3 1 = 3 1 = 3 =

Original equation. Substitute 7/6 for x. Subtract. Reduce.

Because the last statement is true, we conclude that 7/6 is a solution of the equation x − 5/6 = 1/3.

Answer: 13/15 !

Let’s solve another one. You Try It! EXAMPLE 2. Solve the following equation for x. 8 5 − x= 9 18 As before, solve this equation by first clearing all fractions. Solution. Multiply both sides of the equation by the least common denominator. 8 5 − x= 9 18 ! " ! " 8 5 18 − x = 18 9 18 −16x = 5

Original equation. Multiply both sides by 18. On each side, cancel and multiply. " ! " ! 5 8 = −16 and 18 = 5. 18 − 9 18

Note that the equation is now entirely free of fractions. Continuing, −16x 5 = −16 −16 5 x=− 16

Divide both sides by −16. Simplify.

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MODULE 3. FRACTIONS

Checking the Solution. Substitute −5/16 for x in the original equation and simplify. 8 − x= 9" ! 8 5 − − = 9 16 40 = 144 5 = 18 Answer: 6/5

5 18 5 18 5 18 5 18

Original equation. Substitute -5/16 for x. Multiply numerators; multiply denominators. Reduce both sides to lowest terms.

Because this last statement is true, we conclude that −5/16 is a solution of the equation (−8/9)x = 5/18. !

A couple of key points to keep in mind are: • Both sides of the equation need to be multiplied by the LCD. • We must use the distributive property when ever addition or subtraction show up on in an equation.

Let’s work another example. You Try It! Solve for s: 3 2s s 1 − = − 2 5 3 5

EXAMPLE 3. Solve for x:

x 1 2 3x − = − . 3 4 2 8

Solution. Multiply both sides of the equation by the least common denominator. 2 3x x 1 − = − 4" 2 ! 8 " ! 3 x 1 2 3x = 24 − − 24 3 4 2 8 ! " ! " ! " #x$ 2 3x 1 24 − 24 = 24 − 24 3 4 2 8 16 − 18x = 12x − 3

Original equation. Multiply both sides by 24. On both sides, distribute 24. ! " ! " 3x 2 = 16, 24 = 18x. Left: 24 3 4 ! " #x$ 1 Right: 24 = 3. = 12x, 24 2 8

21

3C. SOLVING EQUATIONS

Note that the equation is now entirely free of fractions. We need to isolate the terms containing x on one side of the equation. 16 − 18x − 12x = 12x − 3 − 12x 16 − 30x = −3

Subtract 12x from both sides. Left: −18x − 12x = −30x. Right: 12x − 12x = 0. Subtract 16 from both sides.

16 − 30x − 16 = −3 − 16 −30x = −19

Left: 16 − 16 = 0. Right: −3 − 16 = −19.

−19 −30x = −30 −30 19 x= 30

Divide both sides by −30. Simplify both sides.

Readers are encouraged to check this solution in the original equation.

Answer: 51/22 !

Applications We close out this review with an application problem. You Try It! EXAMPLE 4. In the third quarter of a basketball game, announcers informed the crowd that attendance for the game was 12,250. If this is two-thirds of the capacity, find the full seating capacity for the basketball arena. Solution. We follow the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let F represent the full seating capacity. Note: It is much better to use a variable that “sounds like” the quantity that it represents. In this case, letting F represent the full seating capacity is much more descriptive than using x to represent the full seating capacity. 2. Set up an Equation. Two-thirds of the full seating capacity is 12,250.

Two-thirds

of

Full Seating Capacity

is

12,250

2 3

·

F

=

12,250

Hence, the equation is

2 F = 12250. 3

Attendance for the Celtics game was 9,510. If this is 3/4 of capacity, what is the capacity of the Celtics’ arena?

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MODULE 3. FRACTIONS 3. Solve the Equation. Multiply both sides by 3 to clear fractions, then solve. 2 F = 12250 ! 3" 2 F = 3(12250) 3 3 2F = 36750 36750 2F = 2 2 F = 18375

Original equation. Multiply both sides by 3. Simplify both sides. Divide both sides by 2. Simplify both sides.

4. Answer the Question. The full seating capacity is 18,375. 5. Look Back. The words of the problem state that 2/3 of the seating capacity is 12,250. Let’s take two-thirds of our answer and see what we get. 2 2 18375 · 18375 = · 3 3 1 2 3 · 6125 = · 3 1 2 !3 · 6125 = · 1 3! = 12250 Answer: 12,680

This is the correct attendance, so our solution is correct. !