College Algebra—Winter 2005 Section 1.1 Section 1.1--Linear and ...

College Algebra—Winter 2005 Section 1.1Section 1.1--Linear and Absolute Value EquationsPart 1--Linear EquationsIn algebra we use the term "solve an equation" to refer to the process of finding a valueor values for the variable that make the equation true. For example if 3x − 5 1 thenwhen x 2 the equation is true. 2 is called a solution to the equation.Solving equations uses the concept that as long as we do the same thing to both sides ofthe equation we will not lose any of the solutions. So we can1. Add or subtract the same value to both sides of an equation.2. Multiply or divide both sides of an equation by the same non-zero number.etc.The example we used above, 3x − 5 1 , is an example of a conditional equation. It isonly true on the condition that x 2 . If an equation is never true, for examplex 3 x 5 , we call the equation a contradiction. If the equation is always true, forexample x x 2x , we call the equation an identity.Algorithm for solving linear equations:1. Simplify both sides of the equation by removing parentheses and denominators offractions.2. Move all of the terms containing the variable to the left and the other terms to the rightby adding (or subtracting) the appropriate terms to both sides of the equation. Simplifyboth sides again.3. Multiply (or divide) both sides of the equation by the same non-zero number to leavethe equation in the form variable = number. The number is the solution to the equation.4. Check your solution by substituting it back into the original equation.3Question 9: Solvex 1 2 4 2 3Multiply both sides of the equation by 12 to eliminate the fractions.9x 6 8Subtract 6 from both sides of the equation.9x 2Divide both sides by 9.x 2 9Check the answer:3429 1 2 2 3© W Clarke 1 1/11/2005

College Algebra—Winter 2005 Section 1.1Question 14: Solve 0.04x − 0. 2 0. 07Multiply both sides by 100 to remove the decimal.4x − 20 74x 27x 27 6.754Check: 0.04 6.75 − 0. 2 0.07Part 2--Absolute Value Equations.We work under the realization that |−x| |x| . In other words, any equation involvingaboslute values is in reality two (or more) equations. Therefore expect that you will get atleast two answers for any absolute value equation! I catch a lot of people on the test withthis--Don't be one of them.Question 36: Solve|x − 8| 3̇.The part inside the absolute values can be either 3 or −3 in order for the equation tobe true. So you have to solve two equations:x − 8 3 or x − 8 −3Solve one equation at a time.x − 8 3x 11ORx − 8 −3x 5Check visually.The solution is x 11,5Question 44: Solve |4x − 1| −17An absolute value cannot be equal to a negative number. Therefore there are no solutions.Question 64: Solve |x − 3| −x − 3This is true whenever x − 3 is negative.i.e. x − 3 ≤ 0which is x ≤ 3© W Clarke 2 1/11/2005

College Algebra—Winter 2005 Section 1.1Part 3--Word ProblemsTalking about word problems reminds me of one of my favorite Gary Larson cartoons:So beware about asking me what good word problems are. I might reply, "To prepare youfor the afterlife!"Question 51: Ruben is driving along a highway that passes through Barstow. Hisdistance d in miles from Barstow is given by the equation d |210 − 50t| , where t isthe time, in hours, since the start of his trip and 0 ≤ t ≤ 6 . When will Ruben be exactly60 miles from Barstow?So 60 becomes the distance dThere are two answers.ORThe final solution is|210 − 50t| 60210 − 50t 60−50t −150t 3210 − 50t −60−50t −270t 5.43hours or 5h24m© W Clarke 3 1/11/2005