Algebra 1 Section 2.8

Possible Color Choices for this slide:

Equations with Fractions To eliminate fractions from an algebraic equation, you can multiply both sides of the equation by the least common multiple of the denominators.

Solving Equations Containing Fractions Find the LCD. Eliminate fractions by multiplying both sides by the LCD. Solve the resulting equation. Check the solution.

( ) Example 1 Solve + x – = 3. x 4 1 3 12 + 12(x) – 12 = 12(3) x 4 1 3 ( ) 12 + 12(x) – 12 = 12(3) x 4 1 3 The LCD is 12. 3x + 12x – 4 = 36

( ) ( ) Example 2 Solve – = . x + 2 6 x – 1 3 1 9 18 – 18 = 18 x + 2 6 ( ) 18 – 18 = 18 ( ) x + 2 6 x – 1 3 1 9 The LCD is 18. 3(x + 2) – 6(x – 1) = 2 3x + 6 – 6x + 6 = 2

Example 3 The width of the field is one- third of the length. Let x = the length of the field Let ⅓x = the width of the field

Example 3 The perimeter of the field is found by adding all the lengths of its sides. P = l + l + w + w 1 3 x + x + x + x = 280

( ) Example 3 1 3 x + x + x + x = 280 3x + 3x + 3 x + 3 x = 3(280) 1 3 ( ) 3x + 3x + 3 x + 3 x = 3(280) 1 3 The LCD is 3. 3x + 3x + x + x = 840 8x = 840

Example 3 8x = 840 x = 105 1 x = 35 3 The length is 105 ft. The width is 35 ft.

Solving Equations Containing Decimals In these equations, you can multiply both sides of the equation by a power of ten.

Coins Coin Value Number Value Pennies Nickels Dimes Quarters $0.01 $0.05 $0.10 $0.25 Number p n d q Value 0.01p 0.05n 0.10d 0.25q

Example 5 q 0.25q 17 – q 0.10(17 – q) 0.25q + 0.10(17 – q) = 2.45 Number of coins Value of coins Coin Quarters q 0.25q Dimes 17 – q 0.10(17 – q) 0.25q + 0.10(17 – q) = 2.45

Solving Coin Problems 1. Read. Find the main unknown and assign a variable to it; then express any other unknowns in terms of it.

Solving Coin Problems 2. Plan and organize by making a table indicating the number and the total value of each type of coin.

Solving Coin Problems 3. Solve an equation obtained by using the information in the value column and any other information from the problem.

Solving Coin Problems 4. Check that you have answered all the questions and that your answers are reasonable.

Example 6 n + 4 0.25(n + 4) 2n 0.10(2n) n 0.05n 3n + 5 0.01(3n + 5) Number of coins Value of coins Coin Quarters n + 4 0.25(n + 4) Dimes 2n 0.10(2n) Nickels n 0.05n Pennies 3n + 5 0.01(3n + 5)

Example 6 n = 40 0.25(n + 4) + 0.10(2n) + 0.05n + 0.01(3n + 5) = 22.25 44 quarters 40 nickels 80 dimes 125 pennies

Homework: pp. 87-89