Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 3...

Algebra Problems…Solutions

Algebra Problems…Solutions

© 2007 Herbert I. Gross

Set 17 part 3By Herbert I. Gross and Richard A. Medeiros

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Problem #1


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L1 is the set of points defined byL1 = {(x,y): y = 2x + 3}

and L2 is the set of points defined byL2 = {(x,y):y = 5x + 2}

Is the point (1,7) a member of the set L1?

Answer: No

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Answer: NoSolution for #1:

The test for membership in L1 is that the point (x,y) must satisfy the equality…

y = 2x + 3.

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Hence, (1,7) is on the line L1 if and only if7 = 2(1) + 3.

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Since 7 = 2(1) + 3 = 5 is a false statement, we know that the point (1,7)

is not on the line L1.

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Notes on #1

Notice that this problem doesn’t mention functions. However, we may think of L1 as being the geometric graph of the function f,where f(x) = 2x + 3.


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 L1 only if y = 5. That is, if we replace x by 1 in the equation, we see that

y = 2(1) + 3 = 5.

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Notes on #1

We know that L1 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 L1 and connect them by a straight line.


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Notes on #1


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 L1.

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y

x

(0,3)

(1,5)

L1

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 L1.

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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.


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However, if these three points do lie on a straight line, it is likely more than a

sheer coincidence.

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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.


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For example, the form of the equation y = 2x + 3 immediately tells us two things…

The point (0,3) is on L1.

Every time x increases by 1, y increases by 2.

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Notes on #1

So if we wanted to see what point on L1 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.


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Therefore, y increases by 14, and consequently the point in question is

(7,3+14) or (7,17).

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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…


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y = 2x + 3

y = 2(7) + 3

y = 14 + 3

…to see immediately that y has to be 17.

Problem #2


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L1 is the set of points defined byL1 = {(x,y): y = 2x + 3},


Is the point (1,7) a member of the set L2?

Answer: Yes

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Answer: YesSolution for #2:

The test for membership in L2 is that the point (x,y) must satisfy the equality…

y = 5x + 2.

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Hence, (1,7) is on the line L2 if and only if7 = 5(1) + 2.

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Since 7 = 5(1) + 2 is a true statement, we know that the point (1,7)

is on the line L2.

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Notes on #2

The notes on Problem #1 apply as well to this problem. To reinforce the previous notes, notice that…


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► The problem doesn’t mention functions. However, we may think of L2 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.

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Notes on #2

► Note that we use g rather than f here because later we will be talking about

L1 and L2 in the same problem. This will let us keep track of the different functions that are defined by L1 and L2.


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► We know that L2 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 L2 and connect them by a

straight line.

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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 L2.


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y

x

(0,2)

(1,7)

L2

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 L2.

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Notes on #2


► 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.

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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.


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► For example, the form of the equation y = 5x + 2 immediately tells us two things…

The point (0,2) is on L1.

Every time x increases by 1, y increases by 5.

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Notes on #2

► So if we wanted to see what point on L2 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).


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► 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


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What point on L1 has its x-coordinate equal to 7.5?

Answer: (7.5, 18)

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Answer: (7.5, 18) Solution for #3:

As we saw in Problem #1, the test for membership in L1 is that the point (x,y)

must satisfy the equality…y = 2x + 3

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Hence, (7.5, y) is on the line L1 if and only if

y = 2( 7.5 ) + 3

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y = 15 + 3 = 18

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Therefore, the desired point is (7.5,18).

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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.


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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).

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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.


Problem #4


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What point on L2 has its y-coordinate equal to 20?

Answer: (18/5 ,20)

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Answer: (18/5 ,20)Solution for #4:This problem is similar to what we did in Problem #2. The test for membership in L2 is that the point (x,y) must satisfy the equality…

y = 5x + 2

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Hence, if the point (x, 20) is on the line L2, x must satisfy the equation…

20 = 5(x) + 2

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Solving our equation, we see that x = 18/5

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Hence, the point is (18/5 , 20).

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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 L2 as

suggested in one of our notes in the solution of Problem #3, we would have to estimate

the x-coordinate of the point.


More specifically, we would see from the drawing that the x-coordinate of the desired

point was between 3 and 4, but we wouldhave to estimate what the exact value was.

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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


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What point(s) belongs to both lines L1 and L2?

Answer: (1/3 , 11/3)

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Answer: (1/3 ,11/3)Solution for #5:In order to be on the line L1, the point (x,y) has to satisfy the equation…

y = 2x + 3

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And to be on the line L2 the point (x,y) has to satisfy the equation…

y = 5x + 2

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Hence, to be on both lines, the point (x,y) has to satisfy both of the equations.

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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…

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y = 2x + 3

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We may then subtract 2x from both sides of the equation 5x + 2 = 2x + 3 to obtain…

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5x + 2

3x + 2 = 3

y =

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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.

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3x = 1

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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).

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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.

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Replacing x by 1/3 in the equation y = 5x + 2,

we see that y = 5(1/3) + 2, or y = 11/3.

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Hence, the only point that belongs to both lines is (1/3,11/3).

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Notes on #5

The point, (1/3,11/3) belongs to both lines

and is the intersection of L1 and L2 .


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y

x

(0,2)

L2

(1,7)

(1,5)

(0,3) (1/3,11/3)

L1

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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.


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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 3...

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