Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-2 Copyright © 2005 Pearson Education, Inc. Definitions Algebra: a generalized form of arithmetic. Variables: used to represent numbers Algebraic expression: a collection of variables, numbers, parentheses, and operation symbols.  Examples:

Slide 6-3 Copyright © 2005 Pearson Education, Inc. Order of Operations 1. First, perform all operations within parentheses or other grouping symbols (according to the following order). 2.Next, perform all exponential operations (that is, raising to powers or finding roots). 3.Next, perform all multiplication and divisions from left to right. 4.Finally, perform all additions and subtractions from left to right.

Slide 6-4 Copyright © 2005 Pearson Education, Inc. Example: Evaluating an Expression Evaluate the expression x 2 + 4x + 5 for x = 3. Solution: x 2 + 4x + 5 = (3) + 5 = = 26

Slide 6-5 Copyright © 2005 Pearson Education, Inc. Example: Substituting for Two Variables Evaluate when x = 3 and y = 4. Solution:

Copyright © 2005 Pearson Education, Inc. Linear Equations in One Variable

Slide 6-7 Copyright © 2005 Pearson Education, Inc. Definitions Like terms are terms that have the same variables with the same exponents on the variables. Unlike terms have different variables or different exponents on the variables.

Slide 6-8 Copyright © 2005 Pearson Education, Inc. Properties of the Real Numbers Equality Principles Associative property of multiplication (ab)c = a(bc) Associative property of addition (a + b) + c = a + (b + c) Commutative property of multiplication ab = ba Commutative property of addition a + b = b + a Distributive propertya(b + c) = ab + ac

Slide 6-9 Copyright © 2005 Pearson Education, Inc. Example: Combine Like Terms 8x + 4x = (8 + 4)x = 12x 5y  6y = (5  6)y =  y x + 15  5x + 9 = (1  5)x + (15+9) =  4x x y  4 + 7x = (3 + 7)x + 6y + (2  4) = 10x + 6y  2

Slide 6-10 Copyright © 2005 Pearson Education, Inc. Solving Equations Addition Property of Equality If a = b, then a + c = b + c for all real numbers a, b, and c. Find the solution to the equation x  9 = 24. x  = x = 33 Check: x  9 =  9 = 24 ? 24 = 24 true

Slide 6-11 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Subtraction Property of Equality If a = b, then a  c = b  c for all real numbers a, b, and c. Find the solution to the equation x + 12 = 31. x + 12  12 = 31  12 x = 19 Check: x + 12 = = 31 ? 31 = 31 true

Slide 6-12 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Multiplication Property of Equality If a = b, then a c = b c for all real numbers a, b, and c, where c  0. Find the solution to the equation

Slide 6-13 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Division Property of Equality If a = b, then for all real numbers a, b, and c, c  0. Find the solution to the equation 4x = 48.

Slide 6-14 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations If the equation contains fractions, multiply both sides of the equation by the lowest common denominator (or least common multiple). This step will eliminate all fractions from the equation. Use the distributive property to remove parentheses when necessary. Combine like terms on the same side of the equal sign when possible.

Slide 6-15 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations continued Use the addition or subtraction property to collect all terms with a variable on one side of the equal sign and all constants on the other side of the equal sign. It may be necessary to use the addition or subtraction property more than once. This process will eventually result in an equation of the form ax = b, where a and b are real numbers.

Slide 6-16 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations continued Solve for the variable using the division or multiplication property. This will result in an answer in the form x = c, where c is a real number.

Slide 6-17 Copyright © 2005 Pearson Education, Inc. Example: Solving Equations Solve 3x  4 = 17.

Slide 6-18 Copyright © 2005 Pearson Education, Inc. Solve 21 = 6 + 3(x + 2)

Slide 6-19 Copyright © 2005 Pearson Education, Inc. Solve 8x + 3 = 6x + 21

Slide 6-20 Copyright © 2005 Pearson Education, Inc. Solve 6(x  2) + 2x + 3 = 4(2x  3) + 2 False, the equation has no solution. The equation is inconsistent.

Slide 6-21 Copyright © 2005 Pearson Education, Inc. Solve 4(x + 1)  6(x + 2) =  2(x + 4) True, 0 = 0 the solution is all real numbers.

Slide 6-22 Copyright © 2005 Pearson Education, Inc. Proportions A proportion is a statement of equality between two ratios. Cross Multiplication  If then ad = bc, b  0, d  0.

Slide 6-23 Copyright © 2005 Pearson Education, Inc. To Solve Application Problems Using Proportions Represent the unknown quantity by a variable. Set up the proportion by listing the given ratio on the left-hand side of the equal sign and the unknown and other given quantity on the right-hand side of the equal sign. When setting up the right-hand side of the proportion, the same respective quantities should occupy the same respective positions on the left and right.

Slide 6-24 Copyright © 2005 Pearson Education, Inc. To Solve Application Problems Using Proportions continued For example, an acceptable proportion might be Once the proportion is properly written, drop the units and use cross multiplication to solve the equation. Answer the question or questions asked.

Slide 6-25 Copyright © 2005 Pearson Education, Inc. Example A 50 pound bag of fertilizer will cover an area of 15,000 ft 2. How many pounds are needed to cover an area of 226,000 ft 2 ? 754 pounds of fertilizer would be needed.

Slide 6-26 Copyright © 2005 Pearson Education, Inc. Translating Words to Expressions 3x  9 The sum of three times a number decreased by 9. 2x + 8Eight more than twice a number 2x2xTwice a number x  5 Five less than a number x + 10Ten more than a number Mathematical ExpressionPhrase

Slide 6-27 Copyright © 2005 Pearson Education, Inc. To Solve a Word Problem Read the problem carefully at least twice to be sure that you understand it. If possible, draw a sketch to help visualize the problem. Determine which quantity you are being asked to find. Choose a letter to represent this unknown quantity. Write down exactly what this letter represents. Write the word problem as an equation. Solve the equation for the unknown quantity. Answer the question or questions asked. Check the solution.

Slide 6-28 Copyright © 2005 Pearson Education, Inc. Example The bill (parts and labor) for the repairs of a car was $ The cost of the parts was $339. The cost of the labor was $45 per hour. How many hours were billed? Let h = the number of hours billed Cost of parts + labor = total amount h =

Slide 6-29 Copyright © 2005 Pearson Education, Inc. Example continued The car was worked on for 3.5 hours.

Slide 6-30 Copyright © 2005 Pearson Education, Inc. Example Sandra Cone wants to fence in a rectangular region in her backyard for her lambs. She only has 184 feet of fencing to use for the perimeter of the region. What should the dimensions of the region be if she wants the length to be 8 feet greater than the width?

Slide 6-31 Copyright © 2005 Pearson Education, Inc. continued, 184 feet of fencing, length 8 feet longer than width Let x = width of region Let x + 8 = length P = 2l + 2w x + 8 x The width of the region is 42 feet and the length is 50 feet.