REMINDER: When you go to the open lab next door in 203, please make sure you sign in on the log sheet and enter your instructor’s name and your section number. We need to collect this information to document lab usage and ensure future funding for tutors.Emergency Log InPowerPoint PresentationSlide 4Any questions on the Section 1.4 homework that was due today?Slide 6Section 2.1Slide 8Example:A linear equation in one variable can be written in the form ax + b = c, a  0.Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Reminder:Slide 25REMINDER:When you go to the open lab next door in 203, please make sure you sign in on the log sheet and enter your instructor’s name and your section number. We need to collect this information to document lab usage and ensure future funding for tutors.Removable stickers with tutoring schedulesare still available – If you haven’t gotten one yet, get one from the TA now and stick it on your laptop.Emergency Log InThe next few slides has the information for an emergency log in site to access homework, practice quizzes, etc. Since the Course Compass features are being converted to Pearson MyLab and Mastering, a new emergency log in site may be available later in the semester. For now we will use the following.Any questions on the Section 1.4 homework that was due today?Now pleaseCLOSE YOUR LAPTOPSand turn off and put away your cell phones.Sample Problems Page Link(Dr. Bruce Johnston)Section 2.1Linear Equations in One VariableSection 2.1An algebraic equation is a statement that two expressions have equal value.Solving algebraic equations involves finding values for a variable that make the equation true.Equivalent equations are equations with the same solutions.Example:x – 1 = 5 and 3x + 1 = 19 are equivalent equationsbecause plugging in x = 6 makes both of them trueA linear equation in one variable can be written in the form ax + b = c, a  0.Examples:3x + 2 = 8-5x + 7 = -14/298y + π = 7112z = 100Solving linear equations in one variable1) Multiply to clear fractions (if there are any).2) Simplify each side of equation by distributing where necessary and then combining like terms.3) Get all variable terms on one side and number terms on the other side of equation (addition property of equality).4) Get variable alone (multiplication property of equality).5) ALWAYS check solution by substituting into original problem.z = -16 (Simplify both sides)NOW CHECK: (plug z = -16 into the ORIGINAL equation) 8 + z = -8 8 + (-16) = -8  Example 1: Solve 8 + z = -8Sample problems link8 + -8 + z = -8 + -8 (Add –8 to each side)Solve 3a + 2a = 7 + 4a Example 2:5a = 7 + 4a (Simplify the left side)a = 7 (Simplify both sides) NOW CHECK!5a +(-4a) = 7+4a+(-4a) (Add –4a to each side)3p + -2p – 11 = 2p + -2p – 18 (Add –2p to both sides)3p – 11 = 2p – 18 (Simplify both sides)p = -7 (Simplify both sides)p – 11 = -18 (Simplify both sides)p – 11 + 11 = -18 + 11 (Add 11 to both sides)NOW CHECK! (Plug in -7 everywhere there’s a p in the original equation, and see if both sides simplify to the same number.) 4p – 11 – p = 2 + 2p – 205(3 + z) – (8z + 9) = -4z15 + 5z – 8z – 9 = -4z (Use distributive property)6 – 3z = -4z (Simplify left side)6 + z = 0 (Simplify both sides)z = -6 (Simplify both sides) NOW CHECK!6 – 3z + 4z = -4z + 4z (Add 4z to both sides)6 + -6 + z = 0 + -6 (Add –6 to both sides)-y = 8 y = -8 (simplify both sides) NOW CHECK!(-1)(-y) = 8(-1) (multiply both sides by –1)9571x795717x(multiply both sides by 7)(simplify both sides) NOW CHECK!935xSample problem from today’s homework:Make sure you check your answer! This one can be checked fairly quickly using a calculator.Make sure you put parentheses around expressions in numerators before you divide by the denominator because calculators follow the order of operations exactly!469Answer:Sample problem from today’s homework:5x – 5 = 2(x + 1) + 3x – 75x – 5 = 2x + 2 + 3x – 7 (use distributive property)5x – 5 = 5x – 5 (simplify the right side)Both sides of the equation are identical. Since this equation will be true for every x that is substituted into the equation, the solution is “all real numbers.” This equation is an example of an identity.Note that if you continued to solve this equation by subtracting 5x from both sides and adding 5 to both sides you would come up with 0 = 0. Whenever you get this result, the answer is “all real numbers”, which is NOT the same thing as “x = 0”.3x – 7 = 3(x + 1)3x – 7 = 3x + 3 (use distributive property)3x + (-3x) – 7 = 3x + (-3x) + 3 (add –3x to both sides)-7 = 3 (simplify both sides)Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.” This equation is an example of a contradiction.Sample problem from today’s homework:Answer: RReminder:This homework assignment on section 2.1 is due at the start of next class period.You may now OPEN your LAPTOPSand log in to