Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 22 By Herbert I. Gross and Richard A. Medeiros next

Problem #1 © 2007 Herbert I. Gross next Answer: d = 7 – 2c next How must d be related to c if when x = c and y = d, it’s true that… 2x + y = 7 4x + 2y = 14

Answer: d = 7 – 2c Solution for #1: If we decide to eliminate y, we can multiply both sides of the top equation by - 2 to obtain the equivalent system… next © 2007 Herbert I. Gross next If we now replace the top equation by the sum of the top and the bottom equation, we obtain the equivalent system… 2x + y = 7 4x + 2y = x - 2y x + y = 7 4x + 2y = = 0

next = 0 4x + 2y = 14 Solution for #1: Since the top equation in the system is a true statement regardless of the values of x and y, the system is equivalent to the single equation… © 2007 Herbert I. Gross We can simplify the equation above by dividing both sides of it by 2 to obtain… 4x + 2y = x + y = 7

Solution for #1: If x = c and y = d satisfies the equation 2x + y = 7, it means that… © 2007 Herbert I. Gross or… 2c + d = 7 d = 7 – 2c next

Notes on #1 What happened in this problem is that the two equations in the system above defined the same constraint. In particular, the bottom equation in the system was obtained by multiplying both sides of the top equation by 2. © 2007 Herbert I. Gross next 2x + y = 7 4x + 2y = 14 In other words, if c + 2d = 7, then it is also true that 2(c + 2d) = 2(7) or 2c + 4d = 14.

next Notes on #1 In terms of graphs, suppose we let L 1 represent the set of points (x,y) for which 2x + y = 7, and we let L 2 represent the set of points (x,y) for which 4x + 2y = 14. Then the point (c,d) is on line L 1 if and only if it is also on line L 2. © 2007 Herbert I. Gross next 2x + y = 7 4x + 2y = 14 So while the two equations in the system are different, they describe the same line.

Problem #2 © 2007 Herbert I. Gross next Answer: ad = bc next No matter how the numbers a, b, c, and d are chosen; x = 0 and y = 0 is a solution of the system of equations… ax + by = 0 cx + dy = 0 How must a, b, c, and d be related in order for the system to have solutions for which neither x nor y is 0? next

Answer: ad = bc Solution for # 2: From a geometric point of view, both equations in the system represent straight lines that pass through the point (0,0). next © 2007 Herbert I. Gross next ax + by = 0 cx + dy = 0 In other words, when we replace x and y by 0, both equations in the system become true statements (namely, 0 = 0).

Solution for # 2: We know that if two lines have different slopes, they will intersect at one and only one point. next © 2007 Herbert I. Gross next ax + by = 0 cx + dy = 0 Hence, in order for the system to have more than this one solution, (ie. x = 0, and y = 0) the two lines have to have the same slope.

Solution for # 2: If we rewrite the two equations in the y = mx + b form we see that… next © 2007 Herbert I. Gross next ax + by = 0 cx + dy = 0 Hence… ax + by = 0 by = - ax y = x + (0) - a b m = - a b cx + dy = 0 Hence… dy = - cx y = x + (0) - c d m = - c d next

Solution for # 2: Comparing the equations ax + by = 0 and cx + dy = 0, we see that for the system to have more than one solution… next © 2007 Herbert I. Gross next If we then multiply both sides of the equation by - 1, we see that… - a b - c d = a b c d = and finally, to eliminate fractions, we multiply both sides of the equation by bd (which often is referred to as cross multiplying) to obtain… ad = bc next

Notes on #2 The fact that ad = bc means that any solution of one equation is also a solution of the other equation. More specifically, suppose we have found values of x and y for which it is true that… © 2007 Herbert I. Gross next If we then multiply both sides of this equation by d, we obtain the equation… ax + by = 0 adx + bdy = 0

next Notes on #2 And since ad = bc, in the equation adx + bdy = 0, we may replace ad by bc to obtain the equivalent equation… © 2007 Herbert I. Gross next Finally, since b is a common factor of bcx and bdy we may rewrite the equation bcx + bdy = 0 in the form… bcx + bdy = 0 b(cx + dy) = 0

next Notes on #2 The only way that the product of two numbers can be 0 is if at least one of the two numbers is zero. Therefore, if we assume that b ≠ 0, it means that cx + dy = 0 1. © 2007 Herbert I. Gross next Thus, we have just shown that whenever ad = bc, if it’s true that ax + by = 0, then it is also true that cx + dy = 0. b(cx + dy) = 0 1 If b = 0, the equation ax + by = 0 becomes ax = 0; and since we are assuming that x ≠ 0, it means that a = 0. However, if both a and b are zero, the equation ax + by = 0 “vanishes” in the sense that it becomes the true statement 0 = 0; in which case we now have only the single equation cx + dy = 0. next

Notes on #2 This problem is a generalization of Problem #1. Namely, recall that in Problem #1 the system of equations was… © 2007 Herbert I. Gross next …in which case a = 2, b = 1, c = 4, and d = 2; and therefore ad = bc = 4. 2x + y = 7 4x + 2y = 14

next Notes on #2 If ad = bc, then ad – bc = 0. Recall that in our lesson we called ad – bc the determinant of the system of equations… © 2007 Herbert I. Gross next ax + by = e cx + dy = f At that time we showed that if the determinant of a system was not equal to 0 there was only one set of values for x and y that satisfied the given system.

next Notes on #2 So in terms of the determinant of the system © 2007 Herbert I. Gross next (which we usually write as ) ax + by = 0 cx + dy = 0 a b c d we showed that the system had more than 1 solution if and only if… a b c d = 0 next

Notes on #2 If there is a problem with our solution, it is only that we used a geometric proof. The trouble with this is that if we had been asked the same question about a linear system in which there were more than 3 equations, there would not be a geometric version of the proof. © 2007 Herbert I. Gross

next Notes on #2 On the other hand, the algebraic solution whereby we replace one system by a simpler equivalent system of equations will always work. © 2007 Herbert I. Gross next …one way to solve this system is by eliminating x. ax + by = 0 cx + dy = 0 So in the case of…

next Notes on #2 With this in mind we multiply both sides of the top equation by d © 2007 Herbert I. Gross next If we now subtract the bottom equation from the top equation we obtain the equation… ax + by = 0 cx + dy = 0 and the bottom equation by b to obtain the equivalent system… adx + bdy = 0 bcx + bdy = 0 adx – bcx = 0 next

Notes on #2 may in turn be written as… © 2007 Herbert I. Gross next From the above equation, we see that unless ad – bc = 0, we can divide both sides of the equation by ad – bc and conclude that… adx – bcx = 0 (ad – bc)x = 0 x = 0 ad – bc = 0

next Notes on #2 However, if ad – bc = 0 the equation (ad – bc)x = 0 becomes 0x = 0; and that means x can be any number. © 2007 Herbert I. Gross next No matter what the constants a, b, c, and d are; x = 0 and y = 0 will always be a solution of… ax + by = 0 cx + dy = 0 For this reason we refer to the solution x = 0 and y = 0 as being the trivial solution of the system. next

Notes on #2 In summary, if ad – bc ≠ 0, then the system… © 2007 Herbert I. Gross next has only the trivial solution (that is, x = 0 and y = 0). ax + by = 0 cx + dy = 0

next Notes on #2 However if ad – bc = 0 we can pick either x or y at random and then solve uniquely for the other variable. © 2007 Herbert I. Gross next For example, the determinant of the system is 0. 2x + y = 0 4x + 2y = 0 So suppose that we want a solution for which x = 3. We simply replace x by 3 in either the top or bottom equation and then solve for y.

next Notes on #2 For example, we may replace x by 3 in the top equation and obtain 2(3)+ y = 0; from which we conclude that y = - 6. © 2007 Herbert I. Gross next 2x + y = 0 4x + 2y = 0 In other words, one solution of the above system is x = 3 and y = - 6.

next Notes on #2 From a geometric point of view, the two equations in the system represent the same line. © 2007 Herbert I. Gross next More specifically, if we solve for y in the equation 2x + y = 0, we see that y = - 2x, and if we solve for y in the equation 4x + 2y = 0 we see that 2y = - 4x or y = - 2x. 2x + y = 0 4x + 2y = 0

next Notes on #2 In other words, each of the equations in the system represents the line that passes through (0,0) and whose slope is - 2. © 2007 Herbert I. Gross next In still other words, the point (x,y) is on this line if and only if y = - 2x; that is, we may choose x to be any value we wish, whereupon (x, - 2x) will be a point on this line. 2x + y = 0 4x + 2y = 0

Problem #3 © 2007 Herbert I. Gross next Answer: n = 41 next What is the value of n if x and y satisfy the system of equations… x + y = 17 x – y = 13 3x – 2y = n

Answer: n = 41 Solution for # 3: If we add the top two equations in the system we see that 2x = 30 and hence, that x = 15. next © 2007 Herbert I. Gross next x + y = 17 x – y = 13 3x – 2y = n Knowing that x = 15, we see from the top equation in the system that y = 2. A quick check verifies that x = 15 and y = 2 also satisfies the middle equation of the system.

Solution for # 3: If we now replace x by 15 and y by 2 in the bottom equation of the system we see that 3x – 2y = n becomes… next © 2007 Herbert I. Gross next 3 x – 2 y = n x + y = 17 x – y = 13 3x – 2y = n (15)(2) 45 – 4 = n n = 41

next Notes on #3 is an example where we have more constraints than we have variables. © 2007 Herbert I. Gross next x + y = 17 x – y = 13 3x – 2y = n More specifically, we have only the 2 variables x and y, but there are 3 constraints placed on these 2 variables.

next Notes on #3 Whenever we have more constraints than we have variables, one of two situations has to apply… © 2007 Herbert I. Gross next (1) The constraints are contradictory; in which case the system has no solutions. (2) Enough of the constraints are redundant and/or non-contradictory; in which case the system will have one or more solutions.

Notes on #3 In terms of this problem, if we replace n by 41, the bottom equation in the system is redundant. That is, once we have the solution for the top 2 equations, the bottom equation has to be a true statement. © 2007 Herbert I. Gross next If, on the other hand, we choose the value of n to be any number other than 41, the constraints will be contradictory. That is, when we replace x and y in the bottom equation by the values that satisfy the other two equations we will obtain a false statement. x + y = 17 x – y = 13 3x – 2y = n

next Notes on #3 Notice that we could have imposed many other redundant constraints. © 2007 Herbert I. Gross next For example, since x = 15 and y = 2 satisfy such equations as… 4x + 3y = 4(15) + 3(2) = = 66 next 2x – 3y = 2(15) – 3(2) = 30 – 6 = 24… and…

next Notes on #3 … t he following system of 5 equations with only 2 variables will also have as its solution x = 15 and y = 2. © 2007 Herbert I. Gross next x + y = 17 x – y = 13 3x – 2y = 41 4x + 3y = 66 2x – 3y = 24

Problem #4 © 2007 Herbert I. Gross next Answer: b = 2 next For what value of b will there be no solution for the system… 2(1 – x) + 3(x + 2) + 2(y –6) = 13 x + by = 30

Answer: b = 2 Solution for # 4: The sequence of steps in the following slide allows us to replace the top equation in the system below… next © 2007 Herbert I. Gross next 2(1 – x) + 3(x + 2) + 2(y –6) = 13 x + by = 30 by the equivalent system … x + 2y = 17 x + by = 30

Solution for # 4: More specifically… next © 2007 Herbert I. Gross next 2(1 – x) + 3(x + 2) + 2(y – 6) = 13 x + by = 30 x + 2y = 17 2(1 + - x) + 3(x + 2) + 2(y + - 6) = x + 3x y = 13 ( - 2x + 3x) + 2y = 13 x + 2y + ( ) = 13 x + 2y – 4 = 13 2(1) + 2( - x) + 3(x) + 3(2) + 2(y) + 2( - 6) = 13 next x + 2y = 17 2(1 – x) + 3(x + 2) + 2(y – 6) = 13 next

Solution for # 4: In summary, we may replace the system… next © 2007 Herbert I. Gross next 2(1 – x) + 3(x + 2) + 2(y –6) = 13 x + by = 30 by the equivalent system … x + 2y = 17 x + by = 30

Solution for # 4: If we subtract the top equation from the bottom equation we obtain… next © 2007 Herbert I. Gross next or… x + 2y = 17 x + by = 30 by – 2y = 13y(b – 2) = 13 …which becomes a false statement when b = 2. That is, when b = 2, the equation becomes the false statement 0 = 13. Thus, when b = 2, there are no values for x and y that satisfy the given system of equations.

next Notes on #4 This problem is almost the same as the previous problems in this set. The only difference is that we have to use our “rules of the game” to paraphrase the top equation in the system above into the more traditional form x + 2y = 17 before we can apply the methods described in the lesson. © 2007 Herbert I. Gross 2(1 – x) + 3(x + 2) + 2 (y – 6) = 13 x + by = 30

next Notes on #4 If b = 2 the equations x + 2y = 17 and x + 2y = 30 are contradictory. When we replace b by 2, the system becomes… © 2007 Herbert I. Gross next …and if we then subtract the top equation above from the bottom equation, we obtain the false statement 0 = 13. x + 2y = 17 x + by = 30 2y

next Notes on #4 Since 0 = 13 is a false statement, it means when b = 2 there are no values for x and y that can satisfy both of the equations in the given system… © 2007 Herbert I. Gross next And since both systems are equivalent, x + 2y = 17 x + 2y = 30 x + 2y = 17 0 = 13 it means that there are no values for x and y that can satisfy both of the equations in the given system. = next

Notes on #4 From a geometric point of view the two equations in the system above represent a pair of parallel lines. More specifically, we may rewrite… © 2007 Herbert I. Gross next in the form… x + 2y = 17 x + 2y = 30 x + 2y = 17 2y = - x + 17 or… y = -x-x

next Notes on #4 In a similar way we may rewrite… © 2007 Herbert I. Gross next in the form… x + 2y = 17 x + 2y = 30 2y = - x + 30 or… y = -x-x

next Notes on #4 From the standard y = mx + b form we see that the two lines have the same slope, © 2007 Herbert I. Gross next y = -x-x y = -x-x but different y-intercepts… next

Problem #5a © 2007 Herbert I. Gross next Answer: b + 2c = 5x + 7y +11z next Express b + 2c as a linear expression in x, y and z if… b = x + y + z and c = 2x + 3y + 5z

Answer: b + 2c = 5x + 7y +11z Solution for # 5a: Knowing that c = 2x + 3y + 5z, we then multiply both sides of the equation by 2 to obtain 2c = 4x + 6y + 10z. next © 2007 Herbert I. Gross next We are also told that b = x + y + z. Therefore… b + 2c = (x + y + z) + (4x + 6y + 10z) = 5x + 7y + 11z (x + 4x) + (y + 6y) + (z + 10z) =

Problem #5b © 2007 Herbert I. Gross next Answer: n = 31 next For what value of n will there be solutions for the system… x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = n

Answer: n =31 Solution for # 5a: By virtue of what we showed in part (a), we know that… next © 2007 Herbert I. Gross next x + y + z = b 2x + 3y + 5z = c 5x + 7y + 11z = b + 2c

Solution for # 5a: If we now replace b by 7 and c by 12 in the system below we obtain… next © 2007 Herbert I. Gross next x + y + z = b 2x + 3y + 5z = c 5x + 7y + 11z = b + 2 c ( ) Comparing both systems, we see that n = 31. next = 3131

Notes on #5b In more advanced courses, one would show that if b, c, and d were any constants, the system of equations… © 2007 Herbert I. Gross next …would only have solutions if b + 2c = d. x + y + z = b 2x + 3y + 5z = c 5x + 7y + 11z = d

next Notes on #5b However, since this isn’t such a course, we used part (a) of this problem to establish the above result. © 2007 Herbert I. Gross next In the present problem, b = 7 and c = 12. Therefore, the system will have solutions only if b + 2c = 7 + 2(12) = 31. In other words, if n ≠ 31 there are no values for x, y, and z for which… x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = n

next Notes on #5b On the other hand, if n = 31, the bottom equation in the given system gives us no additional information. In other words, the system… © 2007 Herbert I. Gross next has the same solution set as the system… x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = 31 x + y + z = 7 2x + 3y + 5z = 12

next Notes on #5b In the system below, there are three variables but only two constraints. Hence, we have 1 degree of freedom. That is, we can pick a value for one of the variables and our choice will determine the value of the other 2 variables. © 2007 Herbert I. Gross x + y + z = 7 2x + 3y + 5z = 12

next Notes on #5b For example, suppose we choose to let x = 4. If we replace x by 4 in the system below, we obtain the following linear system of 2 equations in 2 unknowns… © 2007 Herbert I. Gross next x + y + z = 7 2x + 3y + 5z = 12 or in simpler form… 4 8 y + z = 3 3y + 5z = 4

next Notes on #5b If we multiply the top equation in the system below by - 3 we obtain the equivalent system… © 2007 Herbert I. Gross next And if we now replace the bottom equation in the above system by the sum of the two equations, we obtain… y + z = 3 3y + 5z = 4 - 3y - 3z y + - 3z = - 9 2z = - 5 and since 2z = - 5, z = next

next Notes on #5b If we now replace z by © 2007 Herbert I. Gross next or… y + z = 3 3y + 5z = next in the top equation in the system above we see that… y + z = y = = 11 2

next Notes on #5b The values for y and z were obtained under the condition that x = 4. © 2007 Herbert I. Gross next Therefore; x = 4, y = 11 2 next x + y + z = 7 2x + 3y + 5z = 12, and z = is a solution of… and, hence also of the system… x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = 31

next Notes on #5b © 2007 Herbert I. Gross next and… next As a check we see that… x + y + z = = = 7 2x + 3y + 5z = ) + 5( ) = (4) + 2( = = 8282 x + y + z = 7 2x + 3y + 5z = 12

next Notes on #5b There’s no need to check whether x = 4, y = © 2007 Herbert I. Gross and z = 11 2 is a solution of the equation 5x + 7y + 11z = 31 because we have already shown that it is an inescapable conclusion that follows from the first two equations of the system above x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = 31

next Notes on #5b © 2007 Herbert I. Gross next However, if we wish to check this we see that… 5x + 7y + 11z = ) + 11( ) = (4) + 7( = = x + 7y + 11z = 31

next Notes on #5b If we hadn’t presented part (a) of this exercise first, we could still have solved part (b) by using the general method that has been described in this lesson. © 2007 Herbert I. Gross …we could have eliminated x from the bottom 2 equations. Namely… next x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = n That is, starting with…

next Notes on #5b We could multiply both sides of the top equation by 10; © 2007 Herbert I. Gross next next x + y + z = 7 2x + 3y + 5z = 12 5x + 7y + 11z = n 10x + 10y + 10z = x y z = x y z = - 2n both sides of the middle equation by - 5; and both sides of the bottom equation by - 2 to obtain the equivalent system

next Notes on #5b If we replace the middle equation in the system below by the sum of the middle equation and the top equation, we obtain… © 2007 Herbert I. Gross next and if we replace the bottom equation by the sum of the bottom equation and the top equation, we obtain the equivalent system… 10x + 10y + 10z = x y z = x y z = - 2n - 5y z = y z = 70 – 2n next

Notes on #5b If we then divide both sides of the middle equation in the system by - 5, we obtain… © 2007 Herbert I. Gross next and if we divide both sides of the bottom equation by - 2 we obtain the equivalent system… 10x + 10y + 10z = y z = y z = 70 – 2n y + 3z = - 2 2y + 6z = n next

Notes on #5b If we next multiply the middle equation by - 2, we obtain the equivalent system… © 2007 Herbert I. Gross next 10x + 10y + 10z = 70 y + 3z = - 2 2y + 6z = n - 2y + - 6z = 4 Finally, we replace the bottom equation by the sum of the bottom equation and the middle equation to obtain the equivalent system… …and from the bottom equation, we see that n = 31. next 0 = − 31 + n next