CCSSM Algebra: Equations

CCSSM Algebra: Equations. Reasoning with Equations and Inequalities (A-REI) Eplain each step in solving a simple equation as following from the equalit of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justif a solution method. See page 65 of CCSSM standards. MP5, MP6, MP8. Solve equations and Inequalities in one variable 4. Solve quadratic equations in one variable. a. Use the method of completing the square to transform an quadratic equation in into an equation of the form ( p) = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations b inspection (e.g., for = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. 3 Solving Equations What does 3 5 + 4 = 3 3 mean, where is a real number? That is, what is an equation in a real number? The statement is clearl not true for an real number. For eample, if = then the left side of the equation is 63 54 and the right side of the equation is 3 3. B the cross-multiplication algorithm (CMA), the left side is greater than the right side; the are not equal to each other. 4 3 I will assume that at this point students are familiar with real numbers, but not comple numbers. Therefore, the last sentence of the standard ( Recognize when the quadratic formula gives comple solutions and write them as a±bi for real numbers a and b.) is removed. I will also assume that students have learned about algebraic epressions such as polnomials in. 4 Theorem.: (The version of the CMA that is used here.) Let A, B, C and D be nonzero whole numbers. Then A B > D C AD > BC.

CCSSM ALGEBRA: EQUATIONS 5 We will just sa an equation in, from now on. 6 Theorem.: If A, B and C are real numbers then the following are true.. If A = B then A+C = B+C.. If A = B then AC = BC. The statement is not true for = either. If = then the left side of the equation is 389 54 and the left side is still greater than the right side b CMA. B tring several more numbers ou can see, unless ou are etremel luck, that this statement is false for all numbers that ou tried. So, then what is an equation in a real number? 5 An equation in is an invitation to find a number or numbers that would make the equation a true statement. The net question is, how do we handle this invitation? We know Theorem. from previous grades. 6 I can think of 3 5 3 + 4 as the number A and 3 as the number B. Then in order to use the Theorem., I need to have A = B. Unfortunatel, with all m eperience, I know that A is probabl not equal to B at all. In other words, I cannot use Theorem., for this equation as it stands. We are going to assume that the equation is true for some real number. With that assumption, I can use Theorem.. Do not forget that m goal is to find a number or numbers that would make the equation true. With that in mind, I can add 5 4 to both sides of the equation. (Here, C = 4 5 as in the first part of the Theorem..) Then ( 3 ) ( 5 + + 4 5 4 ) = 3 ( 3 + 5 ) 4 is also true. B using properties of numbers, 3 = 3 ( 3 + 5 ) 4 is also true. B multipling both sides b /(3/), = ( 3 3 + 5 4 3 ) is also true. We have repeatedl used Theorem. based on our assumption that there is a number so that the equation is true. What if,

SOLVING EQUATIONS 3 our assumption is false? Then everthing we have done so far is false. Therefore, We must check and see the number that we have found actuall makes the equation true. The left side of the equation = 3 3 3 + ( 5 4 3 ) + 5 4 = 3 3 Therefore, the number 3 3 the equation true. equation. 3 5 + 4 = 3 3. Therefore, 5 +( 4) 3, when substituted for makes Such a number is called a solution of the 3 5 3 +( 4) 3 is a solution of the equation An equation of the tpe A+B = C+D, where A, B, C, D are fied real numbers and A + C > 0, is called a linear equation in. The equation that we have solved is a linear equation where A = 3, B = 4 5 3, C = 0, and D = 3. Tr to find the solutions of few more linear equations of this tpe. For eample,. Show that 5+3 is a solution of 3 = 5.. Show that 43 7 3.5.3 is a solution of.3+3.5 = 43 7. This discussion leads to the following conjecture. The linear equation in, A+B = D, where A, B, D are fied numbers and A = 0, has a solution D B A. Do not forget to prove this conjecture. This discussion ends with the proof of the following theorem. Theorem.3: Consider a linear equation A + B = C + D, where A, B, C, D are fied numbers and A + C > 0. (i) The equation has a unique solution if A = C and the solution is D B A C. (ii) The equation has no solution if A = C and B = D. (iii) Ever number is a solution if A = C and B = D. Part (i) of Theorem.3 leads to the following corollar.

4 CCSSM ALGEBRA: EQUATIONS Corollar.4: An linear equation A+B = C+D can be written as a+b = 0, if A = C, where a and b are constants and a is non-zero. Therefore, b Theorems and, an linear equation of the form a+ b = 0 has the unique solution b a. For eample, the unique solution of 5 = 0 is 5. Identities An algebraic identit (or more simpl an identit) is a statement that two given epressions are equal for all real numbers, with possible allowance of a small number of eceptions". For eample, the commutative propert of multiplication, ab = ba is an identit for all real numbers a and b. The commutative propert of addition, the associative propert of addition, the associative propert of multiplication, and the distributive propert are identities for all real numbers. The formula k l + m n = kn+ml ln is an identit for all real numbers m, n, k and l provided l = 0 and n = 0. This is an identit which makes allowance for the eceptions of l = 0 and n = 0. Net I will introduce some useful identities. Theorem.: For an two real numbers a and b,. (a+b) = a + ab+b.. (a b) = a ab+b. 3. (a b)(a+b) = a b. You can prove this theorem using properties of real numbers. It is useful to look at these identities backwards. Here is the same theorem written backwards. Theorem.: For an two real numbers a and b,. a + ab+b = (a+b).. a ab+b = (a b). 3. a b = (a b)(a+b).

QUADRATIC EQUATIONS 5 Quadratic Equations Quadratic equations in are equations where the two epressions separated b the equal sign are polnomials; one of them of degree and the other of degree at most in the same number. Some eamples are 4+ = 4, + = 4, = 4, 3+ = 0. Before, we look at finding solutions to quadratic equations, I want to point out that the following theorem is familiar to students from previous grades. Theorem 3.: If A and B are real numbers and AB = 0, then either A = 0 or B = 0. This theorem can be proved b using the trichotom law of real numbers. 7 We will start with simple quadratic equations of the form = a, where a is a real number. 7 The Trichotom Law of Real numbers If A is a real number, then A > 0, or A = 0, or A < 0. For eample, consider the equation = 4. This is an invitation to find the numbers that would make = 4 a true statement. Assume = 4 is true for some number. Then b Theorem., 4 = 0 is also true. Then b identit 3 from Theorem., Then b Theorem 3., ( )(+) = 0 is also true. = 0 is true or + = 0 is true. Now these are two linear equations and their solutions are and. Now we will check both = and = in the equation = 4 to see if our assumption is true. It turns out that both and are solutions of the equation = 4. Now tr to find solutions to several other quadratic equations of this tpe. For eample, consider = 3.

6 CCSSM ALGEBRA: EQUATIONS Assume is a solution of = 3. Then 3 = 0 is true. = ( 3)(+ 3) = 0 is true. = 3 = 0 or + 3 = 0 is true. = = 3 is true or = 3 is true. After checking with the equation, we see that both = 3 and = 3 are solutions of 3 = 0. Tr to find solutions to few more of this tpe of equations before conjecturing and proving the following theorem. Theorem 3.: If a is a fied non-negative real number then the solutions of the equation = a are a and a. Now consider the equation ( 3) = 4. If we look at this equation as a quadratic equation in 3, then b Theorem 3., and are solutions. That is, 3 = and 3 = are solutions of the quadratic equation ( 3) = 4 (in 3). But then 3 = and 3 = are linear equations and the solutions of those two equations are +3 and +3. B checking these numbers with the equation( 3) = 4, we conclude that both +3 and +3 are solutions of the quadratic equation ( 3) = 4 in. Tr to find solutions to several of these equations before proving the following theorem. Theorem 3.3: If a and b are fied real numbers then the solutions of the equation ( + a) = b are a + b and a b. Before we look at the net tpe of quadratic equations, let us look back at the identities and of Theorem... a + ab+b = (a+b) for all real numbers a and b. a ab+b = (a b) for all real numbers a and b. When both a and b are positive numbers, we can provide a geometrical interpretation for identit. Consider a square with sides of length a+b (in some length units). We can partition this square into 4 rectangles; a square with sides a and area a, a square with sides b and area b, and two rectangles with sides a, b and area ab as shown below.

QUADRATIC EQUATIONS 7 b ba b a a ab a b Now consider + h, where both and h are positive numbers. Then geometricall, this epression can be visualized as three rectangles; a square with sides and area, and two rectangles with sides, h and area h put together with no gaps or overlaps as shown below. h h h h As ou can see a piece of the square with sides +h is missing now. We can complete the square if we adjoin a square with sides h and area h, again with no gaps or overlaps as shown below.

8 CCSSM ALGEBRA: EQUATIONS h h h h h Algebraicall, if we add h to + h, then we get a square, namel (+h). This is precisel the identit of Theorem.. That is, + h+h = (+h). Similarl, if we add h to h, once more we get a square, namel ( h). This is precisel the identit of Theorem.. That is, h+h = ( h). As ou ma have noticed, the coefficient of the term in the polnomial + h+h is h. Then h can be identified as the square of the half of the coefficient of ". Therefore, we can alwas complete the square" in an quadratic equation if we add the square of the half of the coefficient of to either + h or h according to identities and of Theorem.. We will use this observation in solving quadratic equations in general. Now Consider the quadratic equation 3 = 0. Assume is a solution of 3 = 0 for some number. Then = 3 is true, b Theorem.. Now to complete the square on the left-hand epression of this equation, we have to add ( ( )) [which is equal to ] to it. However, we want to maintain the truthfulness of the above statement. Therefore, we will use Theorem., and add to both sides of the above true statement. B doing so, we are

QUADRATIC EQUATIONS 9 killing two birds with one stone: we will complete the square and also will maintain the truthfulness of the statement. Therefore, + = 3+ is still true. That is ( ) = 4 is true. B Theorem 3.3, + 4 and 4 are solutions of ( ) = 4. Let us check and see if + 4 and 4 are indeed solutions of 3 = 0. Also, (+ 4) (+ 4) 3 = + 4+4 4 3 = 0 ( 4) ( 4) 3 = 4+4 + 4 3 = 0 Therefore, both + 4 and 4 are solutions of 3 = 0. Find solutions of several more quadratic equations of this tpe. Eample: Solve + 4 7 = 0. Assume is a solution of + 4 7 = 0 for some number. = + 4 = 7 is true. = + 4+ = 7+ is true. = (+) = is true. Therefore, b Theorem 3.3, + and are solutions of (+) =. Since ( + ) + 4( + ) 7 = (4 4 +) 8+4 7 and = 5 8 7 = 0 ( ) + 4( ) 7 = (4+4 +) 8 4 7 = 5 8 7 = 0 both + and are solutions of + 4 7 = 0.

0 CCSSM ALGEBRA: EQUATIONS Eample: Solve + 3+ = 0. Assume is a solution of + 3+ = 0 for some number. = + 3 = is true. = + 3+ ( 3 ) = + ( 3 ) is true. = ( + 3 ) = + 9 4 is true. = ( + 3 ) = 4 is true. Therefore, 3 + and 3 are solutions of ( + 3 ) = 4. Since and ( ) + 3( )+ = 3+ = 0 ( ) + 3( )+ = 4 6+ = 0 both and are solutions of + 3+ = 0. Eample: Solve + 3+3 = 0. Assume is a solution of + 3+3 = 0, for some number. = + 3 = 3 is true. = + 3+ ( 3 ) = 3+ ( 3 ) is true. = ( + 3 ) = 3+ 9 4 is true. = ( + 3 ) = 3 4 is true. This is impossible as ( + ) 3 0. (Wh?) Our assumption that there is a number that would make + 3 + 3 = 0 true leads us to this non-sensical statement. Therefore, our assumption must be invalid. That is, + 3 + 3 = 0 has no solutions. Instead of following our natural instincts of finding a general theorem for solving quadratic equations of the tpe + b+ c = 0, where a and b are real numbers, we will look to find a more general theorem b considering quadratic equations of the form a + b+c = 0, where a, b, c are real numbers and a = 0. Eample : Solve + 3+3 = 0. Assume is a solution of + 3+3 = 0 for some number. = 3 3 = 0 is true, b Theorem..

QUADRATIC EQUATIONS At this point ou should realize that this new equation is of the tpe we have been working latel. Therefore, finding the solutions must be routine. 3 3 = 0 is true = 3 = 3 is true. = 3 +( ) 3 4 = 3 + ( ) 3 4 is true. = ( 4) 3 = 3 + 6 9 is true. = ( 4) 3 = 33 3 4 + 6 is true. Therefore, 33 4 and 4 3 33 4 are solutions of ( 4 3 Eercise: Show that 3+ 33 3+3 = 0. 4 and 3 33 ) = 33 6. 4 are solutions of + Do few more eamples of solving quadratic equations of this tpe before proving the following theorems. Theorem 3.4: The solutions of the quadratic equation a + b+ c = 0, where a, b, c are fied real numbers and a = 0 are b+ b 4ac a and b b 4ac a if b 4ac > 0. Theorem 3.5: Consider a quadratic equation a + b+ c = 0, where a, b and c are fied real numbers and a = 0. (i) If b 4ac > 0, then a + b+c = 0 has two solutions. (ii) If b 4ac = 0, then a + b+c = 0 has one solution. (iii) If b 4ac < 0, then a + b+c = 0 has no solutions. Theorem 3.6: Consider a quadratic equation A + B+C = D + E+F, where A, B, C, D, E, F are fied real numbers. 8 (i) If A = D, then this equation can be written as a + b+c = 0, where a and b are fied real numbers and a is non-zero. (ii) If A = D and B = E, then the quadratic equation reduces to a linear equation. (iii) If A = D, B = E and C = F, then the equation has no solution. (iv) If A = D, B = E and C = F, then ever number is a solution of the equation. 8 B the wa, a fied (real) number is also known as a constant.

CCSSM ALGEBRA: EQUATIONS

CCSSM Algebra: Lines. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different was. For eample, compare a distance time graph to a distance time equation to determine which of two moving objects has greater speed. (8 EE 5). Use similar triangles to eplain wh the slope m is the same between an two distinct points on a non vertical line in the coordinate plane; derive the equation = m for a line through the origin and the equation = m+b for a line intercepting the vertical ais at b. (8 EE 6) Proportional Relationship Constant rate Consider the following problem. Eample 4.: Lavona walked from the Grand Central Station on 4nd Street to the Penn Station on 7th Avenue. The total distance is. miles. She finished the walk in 5 minutes. How far did she walk in the first 0 minutes? How could we answer this question? As a matter of fact we cannot answer this question without knowing more information about her walk. Suppose we know the following information about her walk. Lavona walked from Grand Central Station (GCS) along 4nd Street to an ATM machine (in between 5th and 6th avenues) in 8 minutes. She stopped at the ATM machine and spent two minutes there getting some mone. (The distance from GCS to the ATM machine is 0.4 miles.) She reached the corner of 7th Avenue and 4nd Street 3 minutes after she left GCS. At

4 CCSSM ALGEBRA: LINES the corner she met her friend Karega and stopped to chat with him for minutes. (The distance from GCS to 7th Avenue is 0.6 miles.) Then Lavona walked along 7th Avenue until she arrived at the Penn Station. Let us put this information into a table. Time (in Minutes) Distance travelled (in miles) 0 0 8 0.4 0 0.4 3 0.6 5 0.6 5. Table With this information, we can answer the question. She walked eactl 0.4 miles in the first 0 minutes. 9 A definition is a common agreement. Is there anthing ou could have said with the information given in the original wording of the Eample? (That is, without knowing the details of her walk.) We ma be able to salvage some information with the following definition. 9 Definition 4.: Suppose a person walks a distance d (miles) in a given time interval t (minutes). Then the average speed in the given time interval is d t in miles/minutes. So we can calculate Lavona s average speed with the information originall given in eample. Lavona s average speed in the time interval of 5 minutes =. 5 miles/min. Definition 4.: Suppose the average speed of a person is the same constant C for an given time interval. Then we sa that the person is walking at a constant speed C. Now let us reword the Eample 4. as given in Eample 4. below. Eample 4.: Lavona walked from Grand Central Station on 4nd Street to Penn Station on 7th Avenue at a constant speed. The total distance is. miles. She finished the walk in 5 minutes. How far did she walk in the first 0 minutes? Now this is a question we can answer. The following is the detailed answer to the question.

PROPORTIONAL RELATIONSHIP CONSTANT RATE 5 Answer: Lavona s average speed for 5 minutes =. 5 Suppose Lavona walked miles in 0 minutes. Then Lavona s average speed for 0 minutes = 0 Suppose Lavona s constant speed is C miles/minute. Then, b the definition of the constant speed and () above, we get: () (). 5 = C (3) B the definition of constant speed and () above, we get: Now b (3) and (4) we get: 0 = C (4). 5 = 0 B the definition of division for real numbers: 0 = 5. So, Lavona walked miles in an 0 minutes, in particular, 5 in the first 0 minutes. Let us look at this same problem more closel. Eample 4.3: Lavona walked from Grand Central Station on 4nd Street to Penn Station on 7th Avenue at a constant speed. The total distance is. miles. She finished the walk in 5 minutes. How far did she walk in the first minutes? Answer: Lavona s average speed for 5 minutes =. 5 Suppose Lavona walked miles in minutes. Then (5) 0 The last statement is known as a proportional relationship in some circles. Lavona s average speed for 0 minutes = (6) If Lavona s constant speed is C miles/minute, then, b the definition of constant speed and (5) above, we get:. 5 = C (7)

6 CCSSM ALGEBRA: LINES B the definition of constant speed and (6) above, we get: Now b (7) and (8) we get: = C (8). 5 = B the definition of division for real numbers: =. 5. This last equation is an eample of an equation in two variables and. With this equation we can find the distance Lavona walked in an given time. Let us look at another problem. Eample 4.4: Dave lives 5 miles from town A. One da, he was driving at a constant speed of 50 mph from his home awa from (in the opposite direction of) the cit. How far awa is Dave from the town after hours of driving? Answer: Recall the definition of constant speed. That is, if the average speed of a motion over an time interval is the same constant then we sa that the motion is in constant speed. Since Dave is driving at a constant speed, let us assume that this constant speed is m miles/hour. Let the distance between Dave and the town after hours be miles. Then the distance travelled b Dave in hours is 5 miles. Then The average speed of Dave in hours = 5. Since Dave is travelling at a constant speed, the average speed for an time interval is 50 mph. That is, 5 = 50 After using the definition of division for real numbers and collecting like terms, we have: = 50+5

CONSTANT RATE PROBLEMS 7 This is an equation in two variables and. With this equation we can find the distance between Dave and the town for an given time. For eample, how far awa was Dave from the town after one hour? This is an invitation to find when is. Since we know that = 50+5, = 65 when =. Therefore, Dave is 65 miles from town after hour. Eample 4.5: Find how far is Dave after hour, hours, 3 hours, 3.5 hours and 4. hours. Show our answers in a table. Solution: linear equation in (hours) (miles) =50()+5 65 =50()+5 5 3 =50(3)+5 65 3.5 =50(3.5)+5 90 4. =50(4.)+5 0 Table 4. Constant Rate Problems In the last section we learned about constant speed. It turns out constant speed problems are just a special case of a larger variet of problems known as constant rate problems. You have seen some of these problems in grade 7. The constant rate of water pouring out of a faucet into a tub, constant rate of painting a house, constant rate of mowing a lawn, etc. First we will define average rate. Definition 5.: Suppose V gallons of water flow from a faucet in a given time interval t (minutes). Then the average rate of water flow in the given time interval is V t in gallons/minute. Definition 5.: Suppose the average rate of water flow is the same constant C for an given time interval. Then we sa that the water is flowing at a constant rate C.

8 CCSSM ALGEBRA: LINES Definition 5.3: Suppose A square feet of lawn is mowed in a given time interval t (minutes). Then the average rate of lawn mowing in the given time interval is A t in square feet/minute. Definition 5.4: Suppose the average rate of lawn mowing is the same constant C for an given time interval. Then we sa that the lawn is mowed at a constant rate C. Definition 5.5: Suppose A square feet of wall are painted in a given time interval t (minutes). Then the average rate of wallpainting in the given time interval is A t in square feet/minute. Definition 5.6: Suppose the average rate of wall painting is the same constant C for an given time interval. Then we sa that the wall is painted at a constant rate C. As ou can see all these definitions are similar. We will group them all together and call them constant rate problems. A transcription of an of these problems lead to an equation of two variables. Let us look at few of problems. Eample 5. Peter mows a lawn at a constant rate. Suppose he mowed a 35 square foot section of lawn in.5 minutes. What area of lawn does he mow in 0 minutes? in t minutes? Answer: Suppose Peter s constant rate is C square feet per minute. Suppose Peter mows A square feet of lawn in 0 minutes and square feet of lawn in t minutes. Peter s average rate in.5 minutes = 35 square feet/min..5 Peter s average rate in 0 minutes = A 0 square feet/min. Since Peter s average rate in an time interval is the constant C, we have 35.5 = C () and A 0 = C () B () and () we get: 35.5 = A 0. We can find A from the above equation. Finding A is left as an eercise.

CONSTANT RATE PROBLEMS 9 Peter s average rate in t minutes = square feet/min. Therefore, t t = 35.5 B using the definition of division for real numbers we get the following equation of two variables for this problem. ( ) 35 = t..5 Eample 5.: Pumi paints a house at a constant rate. Suppose she paints a 35 square feet section of house in.5 minutes. What area of the house does she paint in 0 minutes? in t minutes? Answer: Suppose Pumi s constant rate is C square feet per minute. Suppose Pumi paints A square feet of the house in 0 minutes and square feet of the house in t minutes. Pumi s average rate in.5 minutes = 35 square feet/min..5 Pumi s average rate in 0 minutes = A square feet/min. 0 Since Pumi s average rate in an time interval is the constant C, we have 35.5 = C () and A 0 = C () B () and () we get: 35.5 = A 0. We can find A from the above equation. Pumi s average rate in t minutes = square feet/min. Therefore, t t = 35.5 B using definition of division for real numbers we get the following equation of two variables for this problem. ( ) 35 = t..5

0 CCSSM ALGEBRA: LINES Eample 5.3: Water flows from a faucet at a constant rate. Suppose the faucet pours 35 gallons of water into a tub in.5 minutes. What is the amount of water pours out of the faucet in 0 minutes? in t minutes? Answer: Suppose constant rate of water flow is C gallons per minute. Suppose V gallons of water flow in 0 minutes and gallons of water flow in t minutes. The average rate of water flow in.5 minutes = 35.5 gallons/min. The average rate of water flow in 0 minutes = A 0 gallons/min. Since average rate of water flow in an time interval is the constant C, we have 35.5 = C () and V 0 = C () B equations () and () we get: 35.5 = V 0. We can find V from the above equation. The average rate of water flow in t minutes = t gallons/min. Therefore, t = 35.5 B using definition of division for real numbers we get the following equation of two variables for this problem. = ( ) 35 t..5 Notice the seemingl identical thought pattern in Eamples 5.-5.3. We will investigate linear equations in two variables net.

LINEAR EQUATIONS IN TWO VARIABLES Linear Equations in two variables An equation of the form a+b = c is called a linear equation in two variables, where a, b and c are constants and at least one of a and b is not zero. Once again keep in mind that and are numbers. Eample 6.: 50 + = 5 is a linear equation in and. As ou can easil see, an pair of numbers and does not make 50+ = 5 a true statement. For eample, if = 5 and = then 5 50() = 5. A solution of this equation is an ordered pair of numbers (, ) so that and make 50+ = 5 a true statement. (, 65) is a solution of 50 + = 5 since 50()+65 = 5. Similarl, (, 5), (3, 65), (3.5, 80) and (4., 0) are all solutions of 50+ = 5. We can obtain man more solutions as follows. Fied the first number. Sa = 5. Then the resulting equation 50(5)+ = 5 is a linear equation in. Since we know how to solve linear equations in one variable, b solving it, we get = 65 as the solution of the equation 50(5) = 5. Therefore, (5, 65) is a solution of 50+ = 5. We can find solutions to 50+ = 5 b fiing the second number instead of the first one. For eample, let = 75. Then the resulting equation 50+ 75 = 5 is a linear equation in and the solution is = 5 6. Therefore, ( 6 5, 75) is a solution of 50+ = 5. Eample 6.: Find five solutions of the linear equation in two variables + 3 5 =. Answer: Let us pick few values for and make a table.

CCSSM ALGEBRA: LINES linear equation in ()+ 3 5 = 5 ()+ 3 5 = 5 3 3 (3)+ 3 5 = 5 6 4 (4)+ 3 5 = 0 5 (5)+ 3 5 = 5 6 Table 6. Therefore, (, 5 ), (, 3 5), (3, 6 5), (4, 0), and (5, 5 6 ) are solutions of + 5 3 =. Eample 6.3: Plot the five solutions of the linear equation in two variables + 3 5 = that ou have found in eample. on a coordinate plane. Solution: We can visualize these solutions using a coordinate plane. Let A be the point with coordinates(, 5 ), B be the point with coordinates (, 5 3 ), C be the point with coordinates (3, 6 5), D be the point with coordinates (4, 0), and E be the point with coordinates (5, 5 6 ), Then the five solutions we obtained for the equation + 3 5 =, that is, the five ordered pairs (, 5), (, 5 3 ), (3, 6 5), (4, 0) and (5, 5 6 ), can be visualized as points on the coordinate plane. 3 A B C D 0 3 4 5 E Figure 6.

GRAPH OF A LINEAR EQUATION IN TWO VARIABLES 3 Graph of a Linear Equation in Two Variables We have seen in eample 6.3 that the solutions of a linear equation in two variables can be plotted on a coordinate plane as points. The collection of all points (, ) in the -plane so that each (, ) is a solution of a+ b = c is called the graph of a+ b = c. Obviousl, we cannot plot all solutions of a linear equation of two variables on a coordinate plane. That means we cannot draw the graph of a linear equation. What can we do? We can plot a few points of the graph of an equation of two variables and then make predictions of what the graph should look like. With that in mind let us plot some points of a graph of a linear equation. Eample 7.: Find a few points on the graph of 3 = and plot them on a coordinate plane. Answer: Let =. Then () 3 =. Solve this equation for. Then = 0. Therefore, (, 0) is a point on the graph of () 3 =. We will continue this process for few more values of. When =, () 3 = and = 3 and (, 3 ) is a point on the graph of () 3 =. Let us make a table to gather these points on the graph. 0 3 3 4 3 4 5 8 3 Now let us plot these points of the graph of 3 = on a coordinate sstem.

4 CCSSM ALGEBRA: LINES 3 0 3 4 5 Figure 7. Can ou predict the shape of the graph of this linear equation based on the meager information we have so far? It looks like the points lie on a line, but, for all we know, the graph of the equation can be some curve passing through the five points we have, such as the following curve. 3 0 3 4 5 Figure 7. With the hope that more points on the graph would shed more light let us choose few more points on the graph. The following are few additional points on the graph.

GRAPH OF A LINEAR EQUATION: HORIZONTAL AND VERTICAL LINES5 3 3 5 7 9 We will add the plots of these additional points to the same coordinate sstem. 4 3 7 3 3 0 3 4 5 Figure 7.3 Can ou make a prediction about the graph of 3 =? Graph of a Linear Equation: Horizontal and Vertical Lines It looks like the graph of a linear equation of two variables (at least the ones that we looked at so far) are lines. Is this prediction actuall true? Let us investigate to see if we can prove this claim. That is, we want to prove, not just predict, that the graph of a linear equation in two variables is a line. Let us start our investigation with two special cases of linear equations of two variables.

6 CCSSM ALGEBRA: LINES Consider a linear equation of two variables a+b = c, where a, b and c are constants and both a and b cannot be 0. Case : a = and b = 0. In this case, the linear equation becomes +0 = c. What can we sa about the graph of + 0 = c? Eample 8.: Let us pick a particular c, Consider the linear equation +0 = 5. What is the collection of all solutions of +0 = 5? We are looking for ordered pairs (, ) so that + 0 = 5 is true. Can be an number other than 5 in a solution (, ) of + 0 = 5? For eample, can(4.5, ) be a solution of + 0 = 5? Let us check. The left side of the equation = (4.5)+0 = 4.5. The right side of the equation = 5. Since, 4.5 = 5, (4.5, ) cannot be a solution of +0 = 5, for an number. In general, (, ) cannot be a solution of +0 = 5, if = 5, for an number. But if = 5 then (5, ) is a solution of +0 = 5, because 5+0 = 5, for an number. Therefore, the solutions of +0 = 5 are eactl all pairs of the form (5, ), where is an number. The points with coordinates (5, ) alwas lie on the vertical line passing through the point (5, 0). Since can be an number, the collection of all points (5, ) is the complete vertical line passing through (5, 0).

GRAPH OF A LINEAR EQUATION: HORIZONTAL AND VERTICAL LINES7 5 (5, 5) 4 0 (5, 0 ) 3 0 3 4 5 6 (5, ) 3 Figure 8.: The vertical line passing through (5, 0) The above argument does not depend on the particular value of c. If we replace, 5 with 3 then ou will come to the conclusion that the collection of all solutions of + 0 = 3 is the vertical line passing through the point (3, 0). Notice that +0 = c is equivalent to = c. Therefore, we come to the following general conclusion. (In other words we have proved the following theorem.) Theorem 8.: The graph of = c is the vertical line passing through (c, 0), where c is a constant. Now let us look at vertical lines. We claim that an vertical line has to pass through the -ais at some point (c, 0) for some constant c. The following is the reason (proof) for the above statement. If a vertical line is not passing through the -ais, then b the definition of parallel lines, this vertical line is parallel to the -ais. But, being vertical, it is parallel to the -ais as well. Then this implies that -ais is parallel to the -ais and we all know that that is not true. Therefore, a vertical line has to pass

8 CCSSM ALGEBRA: LINES through the -ais. Consider a vertical line passing through the point (c, 0), for some constant c. We claim that this is the onl vertical line passing through (c, 0). The following is the reason (proof) for the above claim. The vertical line passing through (c, 0) is parallel to the -ais. There is eactl one line passing through (c, 0) and parrallel to -ais b the Parallel Postulate. B putting the two claims together, we have: There is eactl one vertical line passing through (c, 0). But we have shown that this vertical line is the graph of the linear equation = c. With this, we have a second conclusion. Theorem 8.: Ever vertical line is the graph of = c, where (c, 0) is the point of intersection of the line and the -ais. Case : a = 0 and b =. In this case, the linear equation becomes 0 + = c or = c. What can ou sa about the graph of = c? Theorem 8.3: The graph of = c is the horizontal line passing through the point (0, c), where c is a constant. Theorem 8.4: Ever horizontal line is the graph of = c, where (c, 0) is the point of intersection of the line with the -ais. The proofs of above theorems are left as eercises. Case 3: a = 0 and b = 0. In this case, the linear equation becomes a + 0 = c and after solving for, = a c. Since a c is a constant, we come to the same conclusion as in case. Case 4: a = 0 and b = 0. In this case, the linear equation becomes 0 +b = c and after solving for, = b c. Since b c is a constant, we come to the same conclusion as in case.

THE SLOPE OF A NON-VERTICAL LINE 9 The Slope of a Non-Vertical Line In the last lesson, we have seen that: The graph of the linear equation = c is the vertical line passing through the point (c, 0) and ever vertical line is the graph of the equation = c, where (c, 0) is the point of intersection of the line and the -ais. We have also seen that for a special non-vertical line, namel an horizontal line, the conclusion is the "same". That is, The graph of the linear equation = c is the horizontal line passing through the point (0, c), and ever horizontal line is the graph of the equation = c, where (0, c) is the point of intersection of the line and the -ais. We would like to prove a similar statement for other non-vertical lines based on our eperiments, if possible. However, this task requires that we develop some tools first. The following is a list of tools that we will develop to complete the task.. Define a number for each non-vertical line that can be used as a measure of "steepness" or "slant" of a non-vertical line. Once defined, this number will be called the slope of the line.. Show that an two points on a non-vertical line can be used to find the slope of that line. 3. Show that the line joining two points on the graph of a linear equation of the form = m+k has the slope m. 4. Show that there is onl one line passing through a given point with a given slope. In this section and net few sections we will develop the said tools. With this in mind, let us look closel at non-vertical lines. Having ecluded horizontal lines, there are two other tpes of non-vertical lines. The following figures show the two tpes of lines in a coordinate sstem. (The coordinate sstem is not necessar; it is included onl for clarit.) The two lines are namedl andl. For lack of better terms we will call the line l a left-to-right inclining line and the line l a left-to-right declining line.

30 CCSSM ALGEBRA: LINES l l Figure 9. We would like to associate a number with an non-vertical line which indicates the amount of its "slant" or "steepness" of a line. Our definition of such a number should assign 0 to a horizontal line as it has no slant or steepness. Our definition of such a number should also indicate that a vertical line has the ultimate steepness" even though we are not defining this number for vertical lines. We can formall achieve this b doing the following. Let O be the origin of the coordinate sstem. Consider the translation T of the plane along the vector OQ. Then the translated -ais is l. That is, T(-ais) = l. As a result, l is a number line with 0 at Q, positive numbers above Q and negative numbers below Q. Suppose a non-vertical line l is given in the coordinate plane. Let P be an point on l. Draw a line l parallel to the -ais through P. Locate the point Q one unit to the right of P on l. Draw a line l parallel to the -ais through Q. We will make l a number line with 0 at Q, positive numbers above Q and negative numbers below Q.

THE SLOPE OF A NON-VERTICAL LINE 3 l l m R P Q l Figure 9. The line l, being non-vertical, will intersect l at some point R on l. Let m be the number at R on l. We will call this number m the slope of l, and this is the number we want to associate with an non-vertical line. In the above figure l is a "left-to-right inclining" line and b definition, the slope of such a line is a positive number. Eample 9.: Consider the following left-to-right inclining line. After going through the steps of the definition of the slope, we can sa that the slope of this line is. 3 R P Q 0 Figure 9.3

3 CCSSM ALGEBRA: LINES Now consider a left-to-right declining" line l. (See the given figure below.) B the definition of the slope of a line, the slope of such a line is negative as seen in the following figure. l P Q l Figure 9.4 m R l Eample 9.: Consider the following left-to-right inclining line. After going through the steps of the definition of the slope, we can sa that the slope of this line is 3. P Q 0 3 R Figure 9.5 If l is a horizontal line, then l is same as l in the definition. Therefore, b definition, the slope of a horizontal line is 0. (See the following figure.)

THE SLOPE OF A NON-VERTICAL LINE 33 l P m Q = R l = l Figure 9.6 Even though our definition of the slope is for non-vertical lines onl, if we choose l to be a vertical line and follow the definition of the slope anwa, then l and l are parallel line and therefore, do not intersect. That is, there is no number on l that we can assign as a slope of l. Therefore, we sa that the slope of a vertical line does not eist. (See the figure below.) l l P Q l Figure 9.7 How does the slope relate to the steepness" of a line? Consider the two intersecting left-to-right inclining lines L, L in the following figure. and let P be the point of intersection. We will use P and the definition of slope to get the slopes m and m of the two lines respectivel. B construction, the line L is steeper" than L and it turns out that m > m.

34 CCSSM ALGEBRA: LINES l L m L P m Q l Figure 9.8 Now consider the two intersecting left-to-right declining lines L 3, L 4 in the following figure. and let P be the point of intersection. Again, we will use P and the definition of slope to get the slopes m 3 and m 4 of the lines, respectivel. B construction, the line L 3 is steeper" than L 4 and it turns out that m 3 > m 4. l P Q l m 4 m 3 L 4 Figure 9.9 L 3 As a result of these observations, we can conclude the following. Given two lines with slopes m and n, if m > n then the line with slope m is steeper than the the line with slope n.

THE COMPUTATION OF THE SLOPE OF A NON-VERTICAL LINE 35 The Computation of the Slope of a Non-Vertical Line Now we will make a bold claim. If the coordinates of an two distinct points P = (p, p ) and R = (r, r ) on a nonvertical line are given, then we can compute the slope of that line. As a matter of fact, we claim that the slope of the line is p r. p r Notice that (p r ) = (r p ) and (p r ) = (r p ). Then p r = (r q ) p r (r p ) = (r p ) (r p ). Therefore, the order of using P and R to calculate the slope would not matter, if the claim is correct. Let us handle the eas case first. Suppose the given non-vertical line is horizontal. P R Then p = r and Figure 0. p r p r = 0 p r. Since P and R are distinct points, p = r. Therefore, p r p r = 0.

36 CCSSM ALGEBRA: LINES Now consider a left-to-right inclining line. Suppose the slope of the line is m. Let us call this given line L for convenience. Let us assume that P is to the left of R and use P to get the slope of the line b construction according to the definition of the slope. Let Q be the point one unit to the right of P on the horizontal line l through P and let R be the point of intersection of L and the vertical line through Q. Let Q be the point of intersection of l and the vertical line through R. (See the following figure.) R L R P Q Q l Figure 0. Now consider the triangles PQ R and PQR. The angles Q and Q are right angles b construction. The angle P is common to both triangles. Therefore, b the AA criterion, two triangles are similar. Let the length of the segment PQ be PQ, the length of the segment RQ be RQ, the length of the segment PQ be PQ and the length of the segment R Q be R Q. Then, since PQ R PQR, R Q RQ = PQ PQ. () B the definition of slope, R Q = m, PQ =. Since the coordinates of P and Q are (p, p ) and (r, r ), RQ = r p

THE COMPUTATION OF THE SLOPE OF A NON-VERTICAL LINE 37 and PQ = r p. Therefore, b (): m r p = r p. B the definition of the division for real numbers: ( ) m = (r r p ). p or, equivalentl, m = r p r p. Finall, consider a left-to-right declining line. Once again suppose the slope of the line is m and the point P is to the left of R on the line. The following construction follows the same sequence of thoughts. P Q Q R R L Figure 0.3 Now consider the triangles PQ R and PQR again. The angles Q and Q are right angles b construction. The angle P is common to both triangles. Therefore, b the AA criterion, the triangles are similar and R Q RQ = PQ PQ. () B definition of slope, R Q = m, PQ =. Since the coordinates of P and Q are (p, p ) and (r, r ), RQ = p r

38 CCSSM ALGEBRA: LINES and PQ = r p. Then b (): Then or, equivalentl, m p r = r p. m (r p ) = r p m r p = r p. B the definition of division for real numbers: ( ) m = (r r p ). p or m = r p r p. Therefore, we have proved the following theorem. Theorem 0.: Given an two distinct points P = (p, p ) and R = (r, r ) on a non-vertical line L, the slope of L is r p r p. The Line Joining Two Distinct Points of the Graph of = m+k has Slope m Recall our goal. We want to prove that the graph of a linear equation of two variables a + b = c, where a, b, and c are constants and a = 0 and b = 0 is a non-vertical line. We can write a+b = c in an equivalent form as = a b + c b. The following is the reason for this. Assume (, ) is a solution of a+b = c. That is, we assume that a+b = c is true. Then a+ b = c is true = a+b a = c a is true. = b = a+c is true. = = a b + c is true, since b = 0. b

THE LINE JOINING TWO DISTINCT POINTS OF THE GRAPH OF Y = MX+K HAS SLOPE M39 Now assume that (, ) is a solution of = b a + b c. That is, we assume that = b a + b c is true. Then = a b + c b is true = b = a+c is true, since b = 0 = b+a = a+ c+a is true. = a+b = c is true. Therefore, an solution of a + b = c is a solution of = b a + b c and an solution of = b a + b c is a solution of a+b = c. Let m = b a and k = b c. m = 0 since a = 0. Then the above linear equation becomes: We can restate our goal as = m+k, m = 0. We want to prove that the graph of a linear equation of two variables = m+k, where m = 0 and k are constants, is a non-vertical line. In the last two sections we took one step towards this goal b defining the slope of a non-vertical line and learning how to calculate the slope given an two points on the line. Now we will take another step towards our goal b showing the following. Theorem.: The line joining an two distinct points of a graph of = m+k, m = 0 has the slope m. Let us look at an eample first. Eample.: Show that the slope of a line joining an two distinct points of the graph of = +3 has the slope. Let P = (p, p ) and Q = (q, q ) be an two distinct points of the graph of = +3. Since P is on the graph of = +3, p = p + 3. Similarl, since Q is on the graph of = + 3, q = q + 3. B the Theorem 0., the slope of the line passing through P

40 CCSSM ALGEBRA: LINES and Q is: p q = (p + 3) (q + 3) p q p q = p + 3 q 3 p q = p q p q = (p q ) p q =, b the cancellation law for real numbers, since p = q. Therefore, the slope of the line passing through P and Q is. Now we will handle the general proof of the Theorem.. Remember that we do not know the "shape" of the graph of = m + k at this point. We guess that it is a line but we have no proof. (Getting the proof is our goal.) Therefore, let us not assume the graph of = m+k is a line until we have a proof. In the following figure we will draw the graph of = m+k as "some" curve; not a line. Let P and Q be two points on the graph of = m + k so that P = (p, p ) and Q = (q, q ). Q l P graph of = m+k Figure. P is a point on the graph of = m + k. Therefore, p = mp + k and the coordinates of P are (p, mp + k). The point Q is also on the graph of = m + k. Therefore, the coordinates of Q are (q, mq + k). Let l be the line passing through P and Q. Since P and Q are

THERE IS ONLY ONE LINE PASSING THROUGH A GIVEN POINT WITH A GIVEN SLOPE4 two distinct points on l, we can use the coordinates of P and Q to find the slope of l. (Theorem 0..) slope of l = (mp +k) (mq + k) p q = mp +k mq k p q = mp mq p q = m(p q ) p q = m. b using cancellation law (for real numbers), since p q = 0. There is Onl One Line Passing Through a Given Point with a Given Slope Let us recall our goal. We want to prove that the graph of a linear equation of two variables = m+k, where m = 0 and k are constants, is a non-vertical line. In the last section, we developed the following tool. Given an two points P, Q on the graph of a linear equation = m+k, where m and k are constants, the line through P and Q has the slope m. In this section we will develop the last tool that we need to achieve our goal. We want to show that there is onl one line passing through a given point with a given slope. Theorem.: If two straight lines have the same slope and pass through the same point, then the are the same line. Proof: Let l and l be two lines with the same slope m passing through a point P. Since m is a number, either m > 0, m = 0 or m < 0. We will prove the theorem for m > 0 and leave the the other two cases as eercises. Since m > 0, both l and l are left-to-right inclining lines.

4 CCSSM ALGEBRA: LINES Our task is to show that l and l are the same line. As usual we will not assume the theorem is true until proven. Therefore, we will treat l and l as two distinct lines as given below. Let P be the point common to both lines. Pick the point Q one unit from P on the line passing through P and parallel to the -ais. Draw the line L passing through Q and parallel to the -ais. Then (b the definition of the slope of a line) L is a number line with 0 at Q. Let R be the point of intersection of L and l and let R be the point of intersection of L and l. Both R and R lie on the positive side of the number line L. R R l l P Q Figure. L The slope of l is m. Therefore, the number at R on the number line L is m. The slope of l is also m. That is, the number at R on the number line is m as well. But we know that if two points on a number line have the same number then those two points are the same. Therefore, R = R. This implies that both l and l pass through the same two distinct points P and R. But there is eactl one line passing through two given distinct points. Therefore, l and l are the same line. Eercises.:. Prove the Theorem. for horizontal lines.. Prove the Theorem. for left-to-right declining lines. 3. In the proof of Theorem. we used the following claim.

GRAPH OF A LINEAR EQUATION IN TWO VARIABLES IS A LINE 43 R and R lie on the positive side of the number line L. How do we know this? Eplain. Graph of a Linear Equation in Two Variables is a Line Recall our goal. We want to prove that the graph of a linear equation of two variables = m+k, where m = 0 and k are constants, is a non-vertical line. Now we are read to prove this claim. We will prove the claim in the following novel wa. Let P, Q be two distinct points on the graph of = m+k. Let l be the line passing through P and Q. We will show that () an point on the graph of = m+k is a point on l and () an point on l is a point on the graph of = m+k. Wh do we need to show both () and ()? Let the graph of = m + k be denoted b G. If G is the collection of red segments as given in the following figure, then () is true but () is not true. That is wh we need to show both () and (). 3 Q G l P 3 3 4 5 6 7 8 0 Figure 3. Therefore, if we can prove both () and () then the graph of = m+k is the line l and we are done.

44 CCSSM ALGEBRA: LINES Let us look at an eample first to see how to prove these two claims before proving them in general. Eample 3.: Show that the graph of = + 3 is a nonvertical line. Solution: The points P(0, 3) and Q(, 5) are on the graph of = +3. (Verif) Consider the line l passing through P and Q. The slope of l is, b Theorem 7.. Therefore, l is a non-vertical left-toright inclining line. We claim that the graph of = + 3 is l. We will prove this claim b using the following strateg. () Show that an point on the graph of = +3 is on l. () Show that an point onlis on the graph of = + 3. If both () and () are true then the graph of = +3 is l, as all points on the graph of = + 3 is on l and all points of l are on the graph of = + 3. We will prove () first. Let R be an arbitrar point on the graph of = + 3. We want to show that R is on l. If R is either P or Q then there is nothing to prove. (We alread know that P and Q are on l.) Let l be the line passing through P and R. Then the slope of l is, b Theorem 6.. Then l and l are the same line b Theorem 8. as both l and l pass through P and have the same slope. Therefore, R is on l. We are done with the proof of (). The time we spent developing necessar tools is well worth it! Now we will prove (). Let S be an arbitrar point on l. If S is either P or Q then there is nothing to prove.(we know that the points P and Q are on the graph of = +3.) Let the coordinates of S be ( 0, 0 ). Let us use the coordinates of P and S to calculate the slope of l. slope of l = 0 3 0 0. That is = 0 3 0.

GRAPH OF A LINEAR EQUATION IN TWO VARIABLES IS A LINE 45 B the definition of division for real numbers, we get 0 = 0 3. B adding 3 to both sides of the above equation, we get 0 = 0 + 3. Therefore, the point R( 0, 0 ) is on the graph of = +3. We are done with the proof of (). In general we have the following theorem. Theorem 3.: The graph of a linear equation = m + k is a non-vertical line with slope m and passing through (0, k), where m = 0 and k are constants. The proof of the Theorem 3. is similar to the proof of Eample 3.. Proof of Theorem 3.: Let P, Q be two distinct points on the graph of = m+k. Let l be the line passing through P and Q. As mentioned earlier, we will show that () an point on the graph of = m+k is on l. () an point on l is on the graph of = m+k. Let us prove (). Let R be an point on the graph of = m+k. We want to show that R is on l. If R is either P or Q then there is nothing to prove as both P and Q are on l. Therefore, assume that R is a point distinct from both P and Q. The current situation in our thinking process is given in the following figure. Let us remind ourselves that we still do not know that the graph of the equation = m+k is a line. (That is what we are tring to prove.) Therefore, we will draw some curve" to represent the graph of = m+k.