Equations with Fractions, Translating & Word Problems Formulas Chapter 6 Sections & 6.8

Copyright © Cengage Learning. All rights reserved. Equations with Fractions 6.4

To Solve an Equation with Fractions 1.Find the least common denominator (LCD) of all the fractional terms on both sides of the equation. 2. Multiply all terms of the equation by the LCD. (If this step has been done correctly, no fractions should now appear in the resulting equation.) 3. Solve the resulting equation from Step 2 using the methods of basic rules. Equations with Fractions

Solve: The LCD of 4 and 20 is 20; therefore, multiply both sides of the equation by 20. » Note: When you multiply you can use 20/1 to help you see the canceling. Example 1

15x = 45 x = 3 Example 1 Divide both sides by 15. cont’d

When the variable appears in the denominator of a fraction in an equation, multiply both members by the LCD which will have the variable in it. Be careful that the replacement for the variable does not make the denominator zero. Equations with Fractions

Solve: Example 6 Multiply both sides by the LCD, x. Divide both sides by 2.

Check: Thus, the root is Example 6 cont’d True

Copyright © Cengage Learning. All rights reserved. Translating Words into Algebraic Symbols 6.5

The ability to translate English words into algebra is very important for solving “applied” problems. To help you, we provide the following table of common English words for the common mathematical symbols: Translating Words into Algebraic Symbols

Translate into algebra: One number is four times another, and their sum is twenty. Let x = first number 4x = four times the number x + 4x = their sum Sentence in algebra: x + 4x = 20 Example 1

Example 2 George is 5 times as old as his son. In 15 years, he will be only twice as old as his son. How old will his son be in 15 years? Let x = the son 5x = George In 15 years: x + 15 = the son 5x + 15 = George

Example 2 Continued Once all variables are defined set up an equation. “In 15 years, he will be only twice as old as his son” (he=George) (In 15 years means use the second set of variables) 5x +15 (George in 15 years) = 2( x + 15) (twice as old as the son)

Example 2 Continued Now Solve 5x +15 = 2( x + 15) 5x + 15 = 2x + 30 (distribute the 2!) -2x 3x +15 = x = 15 Divide both sides by 3 So x = 5

Copyright © Cengage Learning. All rights reserved. Applications Involving Equations 6.6

An applied problem can often be expressed mathematically as a simple equation. The problem can then be solved by solving the equation. To solve such an application problem, we suggest the following steps. Solving Application Problems Step 1: Read the problem carefully at least twice. Step 2: If possible, draw a diagram. This will often help you to visualize the mathematical relationship needed to write the equation. Applications Involving Equations

Step 3: Choose a letter to represent the unknown quantity in the problem, and write what it represents. Step 4: Write an equation that expresses the information given in the problem and that involves the unknown. Step 5: Solve the equation from Step 4. Step 6: Check your solution both in the equation from Step 4 and in the original problem itself. Applications Involving Equations

You need to tile the floor of a rectangular room with a wooden outer border of 6 in. The floor of the room is 10 ft by 8 ft 2 in. How many rows of 4-in.-by-4-in. tiles are needed to fit across the length of the room? The sketch shown in Figure 6.1 is helpful in solving the problem. Example 1 Figure 6.1

Let x = the number of tiles across the length of the room 4x = the number of inches in x tiles The total length of the rectangular room is then 4x = 120 4x + 12 = 120 4x = 108 x = 27 So there are 27 rows of tiles. Example 1 cont’d 10 ft = 120 in. Subtract 12 from both sides. Divide both sides by 4.

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