Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 12 By Herbert I. Gross and Richard A. Medeiros next

Problem #1a © 2007 Herbert I. Gross Answer: yes next Is π x + linear in x?√2

Answer: yes Solution for #1a: In order for an expression to be linear in x, it has to be written in the form y = mx + b (where m and b are constants). next © 2007 Herbert I. Gross next We may rewrite… into the equivalent form… x – √2 π x + - √2 π Then, the expression does have the form mx + b, where m = π and b =. - √2 Both π and are constants. Hence, x – √2 π is linear in x. next

π Notes on #1a next Although and are irrational numbers, they are constants. √2 Recall that π is the ratio between the circumference and the diameter of a circle. And since all circles have the same shape, this ratio is π f or all circles. Regarding : this is exactly that number which, when multiplied by itself, is equal to 2. There is only one such (positive) number. - √2 next © 2007 Herbert I. Gross

next So we may rewrite the given expression in the form… Notes on #1a next b What this means is that every time x increases by 1 unit, y increases by π units. © 2007 Herbert I. Gross x + - √2 π m

next More generally, if… Notes on 1a next …y increases by m every time x increases by 1. In fact, if we replace x by x + 1, our equation becomes… © 2007 Herbert I. Gross y = mx + b y new = m(x + 1) + b = mx + m + b = mx + b + m y

next Notes on #1a © 2007 Herbert I. Gross If the relationship between x and y can be written in the form y = mx + b, m represents the rate of change of y with respect to x. Key Point For example if m = 4, then y increases by 4 units whenever x increases by 1 unit. Or if m = - 4, then y decreases by 4 units whenever x increases by 1 unit.

Problem #1b © 2007 Herbert I. Gross Answer: no next Is 4x linear in x?

Answer: no Solution for # 1b: If y = 4x were linear in x, its rate of change with respect to x would be constant. However, the following chart shows that this is not the case… next © 2007 Herbert I. Gross next x x2x2x2x2 4x 2 4x = y Change in y with respect to x

Solution for #1b: The chart shows that as x increases, the rate of change of 4x with respect to x increases quite significantly. next © 2007 Herbert I. Gross next For example: when x = 100, x 2 = 10,000, 4x 2 = 40,000, and 4x = 40,005. However, when x = 101, x 2 = 10,201, 4x 2 = 40,804, and 4x = 40,809. Hence, as x changes from 100 to 101, 4x changes from 40,005 to 40,809, which is a difference of 804.

If 4x were linear in x, we would be able to write it in the form mx + b, where m and b are constants. However, the given expression has the form mx 2 + b, where m = 4 and b = 5. That is… Notes on #1b next © 2007 Herbert I. Gross mx 2 + b 4x 2 + 5

next While 4 and 5 are constants, the fact that x is not the same as x 2 means that the expression is not linear in x. Notes on #1b next © 2007 Herbert I. Gross mx 2 + b 4x What is true is that 4x is linear in x 2. That is, every time x 2 increases by 1, 4x increases by 4. In the above vein, we would say that if m and b are constants, mx 2 + b is linear in x 2. next

Problem #2a © 2007 Herbert I. Gross Answer: true next True or False ? When we double the length of the side of a square, we also double the perimeter of the square.

Answer: true Solution for #2a: The answer is easy to visualize. Namely… Figure 1 shows a square the length of whose side is 1 inch, and Figure 2 shows a square the length of whose side is 2 inches. next © 2007 Herbert I. Gross next 1 inch 2 inches Figure 1 Figure 2

next Solution for #2a: 1 inch 2 inches © 2007 Herbert I. Gross Notice that the perimeter of the 1 inch square is 4 inches while the perimeter of the 2 inch square is 8 inches. That is: doubling the length of its side also doubled the perimeter of the square.

next Solution for #2a: If the side of the square is s units, we see that… s units 2s units © 2007 Herbert I. Gross Notice that the perimeter of the “s unit” square is 4s units, while the perimeter of the “2s units” square is 8s units. Hence, doubling the length of a side also doubles the perimeter of the square.

Solution for #2a: In the spirit of our Game of Mathematics, we would also like to demonstrate the result of problem #2a by using our rules. To this end, if we let s denote the length of the side of the square, then the perimeter of the square, P, is given by the formula P = 4s. If we now double the length of each side of the square, the new length of each side will be 2s; and hence the new perimeter, P new, is given by the formula P new = 4(2s) next © 2007 Herbert I. Gross

Solution for #2a: P = 4s; P new = 4(2s) By the commutative and associative properties of multiplication, we may rewrite 4(2s) in the equivalent form 2(4s). Thus… next © 2007 Herbert I. Gross P new = 4(2s) = 2(4s) = 2P, from which we see that the new perimeter (P new ) is double the original perimeter (P). next

The relationship between the perimeter of a square and the length of one of its sides is not just linear; it is a direct proportion. Notes on #2a next © 2007 Herbert I. Gross In other words, a direct proportion can always be written in the form y = mx A Generalization Definition: The linear relationship y = mx + b is called a direct proportion if and only if b = 0. next

If we divide both sides of y = mx by x, we obtain the equivalent equation… Notes on #2a next © 2007 Herbert I. Gross The equation tells us that if we divide y by the corresponding value of x, we always get the same result (namely, m). So if y = y 1 when x = x 1 and if y = y 2 when x = x 2 then… = m yxyx = y1x1y1x1 y2x2y2x2

next The above equation is the form in which we usually express proportions… Notes on #2a next © 2007 Herbert I. Gross With respect to our present exercise, P = 4s (with P the perimeter of a square, and s the length of a side): and thus, = y1x1y1x1 y2x2y2x2 Therefore, if P = P o when s = s o, and P = P 1 when s = s 1 then = P1s1P1s1 PosoPoso = 4 PsPs next

The above equation can be rewritten as… Notes on #2a next © 2007 Herbert I. Gross This equation tells us that the ratio between the perimeters of two squares is equal to the ratio of the lengths of the sides of the two squares. For example, if we quadruple the length of a side of a square, we also quadruple the perimeter of the square. = P1s1P1s1 PosoPoso = P1PoP1Po s1sos1so next

Notes on #2a © 2007 Herbert I. Gross The above discussion gives us a nice way to look at the number π. Namely, since by definition all circles have the same shape: the ratio between the circumference of a circle and its diameter is constant; and we name this constant π. That is, if C denotes the circumference of the circle and D denotes the diameter of the circle, then or equivalently, C = π D. In other words, the ratio between the circumference of a circle and its diameter is a constant that we call π. = π CDCD

Problem #2b © 2007 Herbert I. Gross Answer: False next True or False ? When we double the length of each side of a square, we also double the area of the square.

Answer: false Solution for #2b: Again the answer is easy to visualize. Namely... Figure 1 again shows a square the length of whose side is 1 inch and Figure 2 shows a square the length of whose side is 2 inches. next © 2007 Herbert I. Gross next 1 inch 2 inches Figure 1 Figure 2

next Solution: 1 inch 2 inches © 2007 Herbert I. Gross Looking at the diagrams, notice that there are four 1-inch squares in one 2-inch square. That is; when we doubled the length of each side, we quadrupled the area of the square.

next Notes on #2b © 2007 Herbert I. Gross In terms of direct proportions, the fact that A = s 2 tells us that the area of a circle is proportional to the square of the length of a side. So, if we multiply the length of a side by 3, the area of the square is multiplied by 3 2, not by 3. For example, if we start with A = s 2 and replace s by 3s, the new area, A n, is given by A n = (3s) 2 =9(s 2 ) = 9A. next

Notes on #2b © 2007 Herbert I. Gross Notice in the diagrams below that: when the length of a side of a square is 1 unit, its area is 1 square unit. next However, if we triple the length of the side, it becomes 3 units long; and the resulting square has an area of 9 square units. 1 unit 1 square unit 3 units 9 square units

Notes on #2b © 2007 Herbert I. Gross In other words, the area of a square is proportional to the square of the length of one of its sides. In terms of a chart… next s S 2 (=A) Change in A with respect to s P = 4s Change in P with respect to s , , next

Problem #3 © 2007 Herbert I. Gross Answer: No next There is a certain relationship between x and y, such that y = 3 when x = 1; y = 6 when x = 2; and y = 12 when x = 3. Is the relationship linear in x? next

Answer: no Solution for #3: As x changes from 1 to 2, y changes from 3 to 6. Hence, a change of 1 unit in x produced a change of 3 units in y. next © 2007 Herbert I. Gross next If the relationship were linear, the rate of change of y with respect to x would have to be constant. Therefore, every time x changed by 1 unit, y would have to change by 3 units.

Solution for #3: However, we see from the given information that as x changes from 2 to 3, y changes from 6 to 12. That is, this time a change of 1 unit in x produces a change of 6 units in y. next © 2007 Herbert I. Gross next Therefore, the relationship is not linear.

Notes on #3 © 2007 Herbert I. Gross Again it might be helpful to see the solution in terms of a chart. Namely, from the given information we see that… Hence, the change in y with respect to x is not constant. next xy Change in y with respect to x next

Notes on #3 © 2007 Herbert I. Gross Don’t confuse the fact that, in the above chart, y doubled each time x increased by 1 with the fact that the rate of change is constant. In this example the rate of change is not constant. xy Change in y with respect to x

next Notes on #3 © 2007 Herbert I. Gross This is how compound interest works. (Recall that compound interest is when you receive interest on the interest.) Suppose you invest money in such a way that your investment doubles every ten years. In this case, 1 unit of time is 10 years. So if you start with $20,000, then in 10 years you’ll have $40,000. Then during the next 10 years, it is the $40,000 that is doubling. Hence, at the end of 20 years, you’ll have $80,000.

next Notes on #3 © 2007 Herbert I. Gross Had the process been linear (as is the case with simple interest), if you had earned $20,000 in simple interest for the first 10 years, then you would also have earned $20,000 in interest for the next 10 years. Thus, you would have had your original $20,000 plus an additional $40,000 in interest, or a total of $60,000; rather than the $80,000 you got by having the interest compounded.

Problem #4a © 2007 Herbert I. Gross Answer: y = 23 next In a certain linear relationship, y = 7 when x = 0 and y = 15 when x = 2. What is the value of y when x = 4? next

Answer: y = 23 Solution for #4a: When x increases by 2 (that is, x increases from 0 to 2), then y increases by 8 (that is, from 7 to 15). next © 2007 Herbert I. Gross next Since the relationship is linear (that is, the rate of change of y with respect to x is constant) it means that every time x increases by 2, y has to increase by 8. Therefore, as x increases from 2 to 4, y must increase from 15 to (23).

next Notes on #4a © 2007 Herbert I. Gross Notice in the chart below that, in going from x = 2 to x = 4, the value of x doubled. But, y only increased from 15 to 23, not from 15 to 30. The reason is: y is not directly proportional to x. In other words, being linear in x tells us only that the change in y with respect to x is constant. next x Change in x Change in xy Change in y Change in y

next Notes on #4a © 2007 Herbert I. Gross Notice that even though every entry in the “change in y” column is 8, the rate of change of y with respect to x (m) is 4. That is, we find the value of m by dividing the change in y by the corresponding change in x. Hence, in this case since y changes by 8 whenever x changes by 2, and the value of m is given by 8 ÷ 2 or 4.

next Notes on #4a © 2007 Herbert I. Gross The fact that the relationship is linear means that the relationship between x and y can be expressed in the form… …and since when x = 0, y = 7 we may rewrite the above equation as… next y = mx + b 7 = m0 + b Since m0 = 0, we see that 7 = b. next

Notes on #4a © 2007 Herbert I. Gross y = mx + b Replacing b by 7, we see that… …and since when x = 2, y = 15 we see from the above equation that… next y = mx = 2x + 7 Subtracting 7 from both sides of this equation, we see that 2x = 8; and then dividing both sides by 2, we see that m = 4; and if we now replace m by 4 in the formula y = mx + 7, we obtain … next y = 4x + 7

next Notes on #4a © 2007 Herbert I. Gross One of the nice things about the formula y = 4x + 7 is that it gives us a convenient way to find the value of y for any given value of x. next y = 4 x + 7 (4) For example, we may use formula y = 4x + 7 to verify our previous answer. Namely, we see that when x = 4 then… = = 23y

next Notes on #4a © 2007 Herbert I. Gross More generally, the formula y = 4x + 7 tells us that, for any given value of x, we find the corresponding value of y by multiplying the x-value by 4 and then adding 7. For example, if we replace x by 3.57 in our formula, we see that… y = 4(3.57) + 7 = = 21.28y next

Notes on #4a © 2007 Herbert I. Gross And this shows us one advantage the formula y = 4x + 7 has over making a chart. With the formula, we can find the corresponding y value for any given value of x; whereas no matter how many rows we put into the chart, it is impossible to list all of the possibilities.

next Notes on #4a © 2007 Herbert I. Gross In this exercise, the change happened to involve an increase in y with respect to an increase in x. However, the change could just have easily involved a decrease in y as x increased. In this case, the theory would be exactly the same. The only difference would be that m would have been negative. This is illustrated in Problem #4b.

Problem #4b © 2007 Herbert I. Gross Answer: y = 3 next In a certain linear relationship, y = 7 when x = 0 and y = 5 when x = 2. What is the value of y when x = 4? next

Answer: y = 3 Solution for #4b: The logic in this part is the same as it was in part (a). The only difference this time is that when x increases by 2 (from 0 to 2), y decreases by 2 (from 7 to 5). next © 2007 Herbert I. Gross next In terms of signed numbers, we can also say that every time x increases by 2, y increases by - 2. In any case, since the relationship is linear we now know that every time x increases by 2, y will decrease by 2. Hence, when x changes from 2 to 4 (an increase of 2), y changes from 5 to 3 (a decrease of 2).

Notes on #4b © 2007 Herbert I. Gross Again we could have started with the “mx + b” form and written… Since y = 7 when x = 0, we may replace x by 0 and y by 7 to obtain… next y = mx + b 7 = m0 + b Since m0 = 0, we see that 7 = b. next

Notes on #4b © 2007 Herbert I. Gross Since b is a constant and it is equal to 7 when x = 0, it is always 7 in this example. Hence, we may replace b by 7 in the formula y = mx + b to obtain… We know that when x = 2, y = 5. Therefore, we may replace x by 2 and y by 5 in the above formula to conclude that… y = mx = m2 + 7 next = 2m + 7

next Notes on #4b © 2007 Herbert I. Gross If we now subtract 7 from both sides of the equation below… And if we then divide both sides by 2, we will see that m = - 1… 5 = 2m + 7 next – 7 – = 2m = m

next Notes on #4b © 2007 Herbert I. Gross We can now replace m by - 1 in formula y = mx + 7 to obtain the formula… In the “y = 7 – x” form, we see that: given any value for x, we find the corresponding value of y simply by subtracting the value of x from 7. In this sense, we could have found the answer to Problem #4b simply by subtracting 4 from 7. y = - 1x + 7 next = x = 7 – x

next Notes on #4b © 2007 Herbert I. Gross Yet another way to view the “y = 7 – x” result is, to “pretend” that the relationship between x and y is a direct proportion. In that case, since m would still equal - 1, the relationship would then have been y = - 1x. Thus, when x = 0, y would also equal 0. However, we know that y is 7 when x = 0. To convert y = 0 to y = 7, we have to add 7. This tells us that in the “y = mx + b” form, that b = 7. next

Notes on #4b © 2007 Herbert I. Gross However, if m is positive, it means that y increases as x increases; while if m is negative, it means that y decreases as x increases. Problem #4 emphasizes that the form y = mx + b represents a linear relationship between x and y, regardless of the sign of m. next

Problem #5a © 2007 Herbert I. Gross Answer: 2 next If the relationship between x and y is y = 5(x – 3) – 3(x + 4), what is the rate of change of y with respect to x? next

Answer: 2 Solution for #5a: Our approach is to rewrite 5(x – 3) – 3(x – 4) in the “mx + b” form. To this end, we see that… next © 2007 Herbert I. Gross next In the form y = 2x , we see that m = 2, and this means that the rate of change of y with respect to x is 2. y=5(x – 3) – 3 (x+4)=5(x + - 3) + - 3(x + 4)=5x + 5( - 3) + - 3x + - 3(4)=5(x) + - 3(x) + 5( - 3) + - 3(4)=[5x + - 3x] + [ ]=2x =2x – 27y=5(x – 3) – 3 (x+4)=2x =2x – 27 next

Notes on #5a © 2007 Herbert I. Gross The method illustrated the value of paraphrasing before we begin to answer the question. However, we could also have constructed a chart… xx – 35x – 3x + 43x + 45(x – 3) – 3(x + 4) = ychange – 72 = – 75 = – 78 = – 81 = – 84 = – 87 = – 90 = – 93 = – 96 = 292 next

Notes on #5a © 2007 Herbert I. Gross We started our chart with x = 20 in order to avoid having to deal with negative numbers. Notice that the chart suggests that every time x increases by 1, y will increase by 2. xx – 35x – 3x + 43x + 45(x – 3) – 3(x + 4) = ychange – 72 = – 75 = – 78 = – 81 = – 84 = – 87 = – 90 = – 93 = – 96 = 292

Problem #5b © 2007 Herbert I. Gross Answer: 1 / 2 next If the relationship between x and y is y = 5(x – 3) – 3(x + 4), what is the rate of change of x with respect to y? next

Answer: 1 / 2 Solution for #5b: In doing part (a) we found that… next © 2007 Herbert I. Gross next Change in y Change in x = 2= 1 / 2 Change in x Change in y This is equivalent to…

next Aside on #5b © 2007 Herbert I. Gross Our adjective/noun theme sheds some light on the relationship between 2 and 1/2 (or more generally, between m and 1/m). Clearly: as numbers (adjectives), 2 ≠ 1/2. next However, 2 pounds per 1 dollar is the same rate as 1/2 dollar for 1 pound. In other words, while 2 ≠ 1/2, 2 pounds per dollar = 1/2 dollars per pound

Notes on #5b © 2007 Herbert I. Gross More formally, we already know from part (a) that… next y = 2x If we add 17 to both sides of the equation we obtain… y + 17 = 2x If we now multiply both sides of the latter equation by 1 / 2, we see that… or2x = y + 17 x = 1 / 2 y + 17 / 2 …and we again see that the rate of change of x with respect to y is 1 / 2. next

Notes on #5b © 2007 Herbert I. Gross To generalize this result, starting with the generic form y = mx + b, we may add - b to both sides of the formula to obtain… next y + - b = mx and then multiply both sides of the equation by 1/m, to obtain 1/m y + 1/m ( - b) = x … from which we see that the rate of change of x with respect to y is 1/m.