# Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 3 By Herbert I. Gross and Richard A. Medeiros next.

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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 3 By Herbert I. Gross and Richard A. Medeiros next

Problem #1 © 2007 Herbert I. Gross next L 1 is the set of points defined by L 1 = {(x,y): y = 2x + 3} and L 2 is the set of points defined by L 2 = {(x,y):y = 5x + 2} Is the point (1,7) a member of the set L 1 ? Answer: No next

Answer: No Solution for #1: The test for membership in L 1 is that the point (x,y) must satisfy the equality… y = 2x + 3. next © 2007 Herbert I. Gross Hence, (1,7) is on the line L 1 if and only if 7 = 2(1) + 3. next Since 7 = 2(1) + 3 = 5 is a false statement, we know that the point (1,7) is not on the line L 1.

Notes on #1 Notice that this problem doesn’t mention functions. However, we may think of L 1 as being the geometric graph of the function f, where f(x) = 2x + 3. © 2007 Herbert I. Gross In the language of functions, we have shown that when x = 1, f(x) ≠ 7. In fact, our equation tells us that if x = 1, the point (1,y) will be on L 1 only if y = 5. That is, if we replace x by 1 in the equation, we see that y = 2(1) + 3 = 5. next

Notes on #1 We know that L 1 has to be a straight line because the relationship between x and y as defined in the equation y = 2x + 3 is linear. If we wanted to draw the line, we could locate any two points in L 1 and connect them by a straight line. © 2007 Herbert I. Gross next

Notes on #1 © 2007 Herbert I. Gross For example, if we replace x by 0 in the equation y = 2x + 3, we see that y = 3, and hence, (0,3) is on L 1. next y x (0,3) (1,5) L1L1 Knowing that (1,5) and (0,3) are on this line, we then connect them by a straight line; and that straight line, of course, is L 1.

next Notes on #1 Recall that to make sure we haven’t made an error, it is a good practice to locate a third point. Namely, even if two points were incorrectly located, we could still pass a straight line through them. © 2007 Herbert I. Gross next However, if these three points do lie on a straight line, it is likely more than a sheer coincidence.

Notes on #1 Nevertheless, there is always room for error when we use inductive methods. For that reason, it’s better, whenever possible, to use the equation of the line to determine what points are on it. © 2007 Herbert I. Gross next For example, the form of the equation y = 2x + 3 immediately tells us two things… The point (0,3) is on L 1. Every time x increases by 1, y increases by 2.

next Notes on #1 So if we wanted to see what point on L 1 had its x-coordinate equal to 7, we would see that in going from (0,3) to the point in question, x increases by 7. © 2007 Herbert I. Gross next Therefore, y increases by 14, and consequently the point in question is (7,3+14) or (7,17).

Notes on #1 However, it is much more direct to use the equation y = 2x + 3 since we can then simply replace x by 7 in the equation… © 2007 Herbert I. Gross next y = 2x + 3 y = 2(7) + 3 y = 14 + 3 …to see immediately that y has to be 17.

Problem #2 © 2007 Herbert I. Gross next L 1 is the set of points defined by L 1 = {(x,y): y = 2x + 3}, and L 2 is the set of points defined by L 2 = {(x,y):y = 5x + 2} Is the point (1,7) a member of the set L 2 ? Answer: Yes next

Answer: Yes Solution for #2: The test for membership in L 2 is that the point (x,y) must satisfy the equality… y = 5x + 2. next © 2007 Herbert I. Gross Hence, (1,7) is on the line L 2 if and only if 7 = 5(1) + 2. next Since 7 = 5(1) + 2 is a true statement, we know that the point (1,7) is on the line L 2.

Notes on #2 The notes on Problem #1 apply as well to this problem. To reinforce the previous notes, notice that… © 2007 Herbert I. Gross next ► The problem doesn’t mention functions. However, we may think of L 2 as being the geometric graph of the function g, where g(x) = 5x + 2. In the language of functions, we have shown that when x = 1, g(x) = 7.

Notes on #2 ► Note that we use g rather than f here because later we will be talking about L 1 and L 2 in the same problem. This will let us keep track of the different functions that are defined by L 1 and L 2. © 2007 Herbert I. Gross next ► We know that L 2 has to be straight line because the relationship between x and y as defined he equation y = 5x + 2 is linear. If we wanted to draw the line, we could locate any two points in L 2 and connect them by a straight line.

Notes on #2 ► For example, if we replace x by 0 in the equation y = 5x + 2, we see that y = 2, and hence, (0,2) is on L 2. © 2007 Herbert I. Gross next (0,2) (1,7) L2L2 Knowing that (1,7) and (2,0) are on this line, we then connect them by a straight line, and that straight line, of course, is L 2.

next Notes on #2 © 2007 Herbert I. Gross ► As always, to make sure that we haven’t made an error, it is a good idea to locate a third point. Namely, recall that if we didn’t locate two points correctly, we could still pass a straight line through them. However, if all three points lie on a straight line, it is probably more than a coincidence.

next Notes on #2 ► Since there is always room for error when we use inductive methods, it’s always better to use the equation of the line to determine what points are on it. © 2007 Herbert I. Gross next ► For example, the form of the equation y = 5x + 2 immediately tells us two things… The point (0,2) is on L 1. Every time x increases by 1, y increases by 5.

Notes on #2 ► So if we wanted to see what point on L 2 had its x-coordinate equal to 7, we would see that in going from (0,2) to the point in question, x increases by 7. Therefore, y increases by 35, and consequently the point in question is (7,2+35) or (7,37). © 2007 Herbert I. Gross next ► However, using the equation y = 5x + 2 is much more direct in the sense that we would simply replace x by 7 in the our equation and see immediately that y has to be 37.

Problem #3 © 2007 Herbert I. Gross next L 1 is the set of points defined by L 1 = {(x,y): y = 2x + 3} and L 2 is the set of points defined by L 2 = {(x,y):y = 5x + 2} What point on L 1 has its x-coordinate equal to 7.5? Answer: (7.5, 18) next

Answer: (7.5, 18) Solution for #3: As we saw in Problem #1, the test for membership in L 1 is that the point (x,y) must satisfy the equality… y = 2x + 3 next © 2007 Herbert I. Gross Hence, (7.5, y) is on the line L 1 if and only if y = 2( 7.5 ) + 3 next y = 15 + 3 = 18 next Therefore, the desired point is (7.5,18).

Notes on #3 This problem introduces a troublesome feature that exists whenever we rely on a drawing. Namely, in theory a point has no thickness, but the dot we draw to represent the point does have thickness. © 2007 Herbert I. Gross next Thus, for example, if there are no markings on the x-axis between x = 7 and x = 8, we would have to estimate the location of the point (7.5,0).

Notes on #3 However, when we use the algebraic relationship between the coordinates, we can always find the exact location of points without resorting to estimation. © 2007 Herbert I. Gross

Problem #4 © 2007 Herbert I. Gross next L 1 is the set of points defined by L 1 = {(x,y): y = 2x + 3} and L 2 is the set of points defined by L 2 = {(x,y):y = 5x + 2} What point on L 2 has its y-coordinate equal to 20? Answer: ( 18 / 5,20) next

Answer: ( 18 / 5,20) Solution for #4: This problem is similar to what we did in Problem #2. The test for membership in L 2 is that the point (x,y) must satisfy the equality… y = 5x + 2 next © 2007 Herbert I. Gross Hence, if the point (x, 20) is on the line L 2, x must satisfy the equation… 20 = 5(x) + 2 next Solving our equation, we see that x = 18 / 5 Hence, the point is ( 18 / 5, 20).

Notes on #4 This problem is designed to emphasize the advantage the equation of the line has versus drawing the graph geometrically. Namely, if we were to draw the line L 2 as suggested in one of our notes in the solution of Problem #3, we would have to estimate the x-coordinate of the point. © 2007 Herbert I. Gross More specifically, we would see from the drawing that the x-coordinate of the desired point was between 3 and 4, but we would have to estimate what the exact value was. next

Notes on #4 A reasonable estimate based on the drawing might be (3.5, 20) which is very close to the exact point (which is (3.6, 20)). However, while 3.5 and 3.6 are close in value; if the noun they modify is “billions of dollars”, it is the difference between \$3,500,000,000 and \$3,600,000,000. While the percentage of error is not great between these two amounts; it does represent a difference of \$100,000,000 which is far from a trivial amount. © 2007 Herbert I. Gross

Problem #5 © 2007 Herbert I. Gross next L 1 is the set of points defined by L 1 = {(x,y): y = 2x + 3} and L 2 is the set of points defined by L 2 = {(x,y):y = 5x + 2} What point(s) belongs to both lines L 1 and L 2 ? Answer: ( 1 / 3, 11 / 3 ) next

Answer: ( 1 / 3, 11 / 3 ) Solution for #5: In order to be on the line L 1, the point (x,y) has to satisfy the equation… y = 2x + 3 next © 2007 Herbert I. Gross And to be on the line L 2 the point (x,y) has to satisfy the equation… y = 5x + 2 next Hence, to be on both lines, the point (x,y) has to satisfy both of the equations.

Solution for #5: One way to determine the desired point is to replace y in the equation y = 2x + 3 by its value in the equation y = 5x + 2. Doing this we obtain… next © 2007 Herbert I. Gross y = 2x + 3 next We may then subtract 2x from both sides of the equation 5x + 2 = 2x + 3 to obtain… 5x + 2 3x + 2 = 3 y = next

Solution for #5: And if we now subtract 2 from both sides of the equation and then divide by 3, we see that x = 1 / 3. next © 2007 Herbert I. Gross 3x = 1 next If we then replace x by 1 / 3 in the equation y = 2x + 3, we see that y = 2( 1 / 3 ) + 3, which we may write as y = 11 / 3. Hence, the only point that that can possibly belong to both lines is ( 1 / 3, 11 / 3 ). x = 1 / 3 3x + 2 = 3

Solution for #5: What we have to do next is replace x by 1/3 in the equation y = 5x + 2 and verify that y = 11/3. next © 2007 Herbert I. Gross next Replacing x by 1 / 3 in the equation y = 5x + 2, we see that y = 5( 1 / 3 ) + 2, or y = 11 / 3. Hence, the only point that belongs to both lines is ( 1 / 3, 11 / 3 ).

Notes on #5 The point, ( 1 / 3, 11 / 3 ) belongs to both lines and is the intersection of L 1 and L 2. © 2007 Herbert I. Gross next (0,2) L2L2 (1,7) (1,5) (0,3) ( 1 / 3, 11 / 3 ) L1L1

next Note on #5 More generally, while two curves can share several points in common, two nonparallel straight lines can meet in one and only one point. © 2007 Herbert I. Gross

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