Chapters 4/5 Class Notes Intermediate Algebra, MAT1033C Palm Beach State College
Class Notes 4.1 Professor Burkett 4.1 Systems of Linear Equations in Two Variables A system of equations is a set of two or more equations with a same set of unknowns. When solving a system of equations, we try to find values for each of the unknowns (an ordered pair, [x, y]) that will satisfy every equation in the system. Example 1: Is the given ordered pair ( 4, 2) a solution of the given system? 2x + y = 6 x + 3y = 2 So all we do is plug in the given point (-4, 2) into each equation and see if we get a true statement. 2( 4) + 2 = 6? 8 + 2 = 6? 6 = 6 True. ( 4) + 3(2) = 2? 4 + 6 = 2? 2 = 2 True. Therefore, the point (-4, 2) is a solution to the given system because when we plug it in to both equations we get a true statement for each equation. Now, if you re not given an ordered pair to plug in and test, you ll be asked to find the ordered pair; this is called solving the system. There are three ways to solve a system of equations, and each way will give you the same answer. The three ways are graphing, solving by substitution, and solving by elimination. Example 2: Graphing Solve the system by graphing. y = 3x + 2 y = 2x 3 The first step is to graph both of these lines. The point where they intersect is your solution. 1
Class Notes 4.1 Professor Burkett Sometimes you ll be given equations that end up being parallel lines; in this case, there is no solution to the system. Other times you will end up plotting the same line, one on top of the other; in this case, there are infinite solutions. The next way to solve is by substitution. 2
Class Notes 4.1 Professor Burkett Example 3: Solve by substitution. When solving by substitution, we have to ask ourselves two questions. First, is one of the equations already solved for x or y, or if not, is one of the equations easily solvable for x or y? Let s say we re given these two equations and asked to solve by substitution: y = 5x 1 and 2y = 3x + 12 I see that the first equation is already solved for y. So I take that first equation and plug it into the second equation everywhere I see a y: Now that I ve solved for x, I take that value and plug it back into my first original equation everywhere I see an x, so that I can solve for y. The final way is by elimination: 3
Class Notes 4.1 Professor Burkett When we solve by elimination, the first thing you want to do is decide whether you want to eliminate your x or your y from the given equations. To do this, write the equations stacked on top of each other to make it easier to see which is more easily eliminated. Example 4: Solve by elimination. 3x + 5y = 16 and 2x + 3y = 9 Now that we have solved for y, we take that value and plug it back in to one of the original equations to solve for x. 2x + 3( 5) = 9 2x 15 = 9 2x = 6 x = 3 Therefore, the solution to this system, which we solved by elimination, is (3, 5). 4
Class Notes 4.3 Professor Burkett 4.3 Applications of Systems of Linear Equations In this section, all we re doing is taking word problems and using the information given in the word problems to come up with equations and then solving, just like we did in 4.1. Steps to solving a word problem: 1) READ the problem, several times if necessary. What information is given? What is to be found? This is often stared in the last sentence. 2) ASSIGN VARIABLES to represent the unknown values. Use a sketch, diagram, or table to help you, as needed. 3) WRITE A SYSTEM OF EQUATIONS using the variable expressions. 4) SOLVE the system of equations by either graphing, substitution, or elimination. 5) STATE THE ANSWER to the problem. Label it appropriately. Does it seem reasonable? 6) CHECK the answer in the words of the original problem. Example 1: A rectangular parking lot has a length that is 10 ft. more than twice its width. The perimeter of the parking lot is 620 ft. What are the dimensions of the lots? Step 1) READ the problem again. Step 2) ASSIGN VARIABLES. It will be helpful if we draw a diagram: 2W+10 (length) W (width) W 2W+10 We re told the length of the parking lot is 10 feet more than twice its width. In the original word problem, they tell us the perimeter (the sum of all sides) is 620 feet. Therefore, we use the perimeter formula (2L+2W=P) to write our equation: 2(2W + 10) + 2(W) = 620 Twice the length plus twice the width gives us the perimeter. 4W + 20 + 2W = 620 Simplify and combine like terms. 6W + 20 = 620 Solve for W to find the distance of the width. 6W = 600 Subtract 20 from both sides. W = 100 Divide by 6 on both sides. The width of the parking lot is 100 feet. 1
Class Notes 4.3 Professor Burkett Now you can take that value and plug it into the expression we wrote that describes our length to find out what the distance of the length is. So if the length equals 2W + 10 and W=100: 2(100) + 10 = Length 200 + 10 = Length 210 = Length Therefore, the dimensions of this parking lot are 100 ft. x 210 ft. Example 2: Board Problems A beam of wood is 9 feet long. It is divided into two sections, where the second piece is twice as long as the first piece. How long is each section of wood? These, and similar word problems, are called board problems. To solve, we use a helpful diagram: x +/- what the problem tells you x x 2x Total And now that we ve filled out our diagram, we have our equation to solve: 2x+x=9 9 3x=9 X=3 Therefore, the lengths of the two piece are 3ft. and 6ft, for a total of 9ft. The next set of word problems we re going to solve are mixture problems. For mixture problems, we use the below diagram to help us set up our equations: Always X here Always Y here Total of what is being made, produced, combined Price/percent/weight for x Price/percent/weight for y Price/percent/weight of newly created mixture Multiply across to get this Multiply across to get this Multiply across to get this 2
Class Notes 4.3 Professor Burkett Example 3: A grocer has some $4 per lb. coffee and some $8 per lb. coffee, which he will mix to make 50 lbs. of $5.60 per lb. coffee. How many pounds of each should be used? X 4 Y 8 50 5.60 Once you have the first two columns filled in, multiply across: X 4 4x Y 8 8y 50 5.60 280 Now that you ve multiplied across, you have your two equations to form your system. X 4 4x Y 8 8y 50 5.60 280 x + y = 50 4x + 8y = 280 Now use either elimination or substitution to solve. See 4.1. Example 4: Ticket/Interest Problems Rachel received a lottery winning of $60,000. She invested the winnings into two separate savings accounts, one that earned 10% interest and the other that earned 11% interest. At the end of the year, Rachel earned $6,100 in interest. How much was invested into each account? X.1.1x Y.11.11y 60,000 6100 For interest/ticket problems, you do not use the bottom middle box. Also, when filling in the boxes with interest percentages, we always rewrite the percent in decimal form (so 35% would be written.35 and 16% would be written as.16). Again, once the boxes are filled in, you have your two equations of the system and you can solve as we did using elimination or substitution. x + y = 60,000. 1x +.11y = 6,100 3
Class Notes 4.3 Professor Burkett Example 4: D R = D R Distance over rate equals distance over rate. A train travels 600 miles in the same amount of time that a truck travels 520 miles. Find the rate of each vehicle if the train s average rate is 8 mph faster than the truck s. 600 x + 8 = 520 x Once your proportion is set up, cross multiply. 520(x + 8) = 600(x) 520x+= 600x 4160 = 80x x = 52 There, the truck is traveling at 52 mph, whereas the train is traveling 8 mph faster, or 60 mph. 4
Class Notes 5.1 Professor Burkett 5.1 Integer Exponents Examples: 29 0 = 1 ( 29) 0 = 1 ( 29) 0 = 1 ( 15p 5 ) 0 = 1 1
Class Notes 5.1 Professor Burkett 29 0 = 1 ( 3 4 ) 3 = ( 4 3 )3 = 43 3 3 = 64 27 5 3 5 7 = 5 10 y 8 y 12 = 1 y 4 s 4 s8 = s 8 s 4 = s8 4 1 = s4 (5t) 3 = 5 3 t 3 = 1 5 3 t 3 = 1 125t 3 2
Class Notes 5.2 Professor Burkett Adding polynomials: 5.2 Adding and Subtracting Polynomials Subtracting polynomials: 1
Class Notes 5.4 Professor Burkett 5.4 Multiplying Polynomials You ve been multiplying polynomials for a long time, you just didn t know it! Remember FOILing? That s multiplying polynomials! Example 1: x 2 + 3x + 2 Example 2: 1
Class Notes 5.4 Professor Burkett Example 3: Example 4: 2
Class Notes 5.5 Professor Burkett Example 1: 5.5 Dividing Polynomials Rewrite the denominator under each numerator, then move from left to right dividing each term. Example 2: Example 3: 1
Class Notes 5.5 Professor Burkett Example 4: Example 5: 2
Class Notes 5.5 Professor Burkett Example 6: 4 3x 1 3