Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 21 By Herbert I. Gross and Richard A. Medeiros next.

Answer: x = 40, y = - 45 Solution for #1a: We may begin by eliminating y from… 6x + 5y = 15 5x + 4y = 20 next © 2007 Herbert I. Gross next To this end, we multiply both sides of the top equation by 4 and both sides of the bottom equation by 5 to obtain the equivalent system… 24x + 20y = 60 25x + 20y = 100

Solution for #1a: We may return to… 6x + 5y = 15 5x + 4y = 20 but this time eliminate x next © 2007 Herbert I. Gross next To this end, we multiply the top equation by 5 and the bottom equation by 6 to obtain the equivalent system… 30x + 25y = 75 30x + 24y = 120

Solution for #1a: If we subtract the bottom equation from the top equation, we obtain the result y = - 45. next © 2007 Herbert I. Gross next 30x + 25y = 75 30x + 24y = 120 30x + 25y = 75 30x + 24y = 120 – – – y = - 45

next Notes on #1a To make sure our answer is correct we have to check it. That is, in our system we must replace x by 40 and y by - 45 and see if 6x + 5y equals 15 and 5x + 4y equals 20. We see that… © 2007 Herbert I. Gross 6(40) + 5( - 45) = 240 – 225 = 15 next 6x + 5y = 15 5x + 4y = 20 5(40) + 4( - 45) = 200 – 180 = 20 next

Notes on #1a There are other methods we can use to solve our system. One way is that after we eliminated y and found that x = 40, we could have solved for y by replacing x by 40 in either the top or bottom equation of the system. © 2007 Herbert I. Gross 6(40) + 5y = 15 next 6x + 5y = 15 5x + 4y = 20 240 + 5y = 15 next For example, if we replace x by 40 in the equation 6x + 5y = 15; we see that… 5y = 15 – 240 = - 225y = - 225 ÷ 5 = - 45 next

Notes on #1a Because there is an emphasis placed on it, there is a tendency for students to want to rewrite every linear relationship in the form y = mx + b. This can be done in this exercise, but in our opinion, it is cumbersome and conducive toward making computational errors. © 2007 Herbert I. Gross 6x + 5y = 15 next 6x + 5y = 15 5x + 4y = 20 5y = - 6x + 15 next For example, given the equation 6x + 5y = 15 we can rewrite it as … y = 1 / 5 ( - 6x + 15) y = -6 / 5 x + 3 next

Notes on #1a In a similar way, we can rewrite the equation 5x + 4y = 20 as… © 2007 Herbert I. Gross 5x + 4y = 20 next 6x + 5y = 15 5x + 4y = 20 4y = - 5x + 20 next y = 1 / 4 ( - 5x + 20) y = -5 / 4 x + 5 next

Notes on #1a We may then equate the values of y in the equations y = -6 / 5 x + 3 and y = -5 / 4 x + 5 to obtain… © 2007 Herbert I. Gross next y = -5 / 4 x + 5 y = -6 / 5 x + 3 -5 / 4 x + 5 -6 / 5 x + 3 =

next Notes on #1a To eliminate fractions, we may then multiply both sides of the equation by 20 (that is, the least common multiple of 5 and 4) to obtain… © 2007 Herbert I. Gross next -6 / 5 x + 3 = -5 / 4 x + 5 20( -6 / 5 x + 3) = 20( -5 / 4 x + 5) - 24x + 60 = - 25x + 100 x + 60 = 100 and if we now add 25x to both sides of our equation we obtain… or x = 40

Answer: x = 40, y = - 45 Solution for #1b: We can proceed exactly in the same way as we did in Problem #1a. However, if we prefer to work with whole numbers rather than with decimal fractions, we can use our “rules of the game” to multiply both equations below… next © 2007 Herbert I. Gross next 0.06x + 0.05y = 0.15 0.05x + 0.04y = 0.20 by 100 to obtain the equivalent system…

Solution for #1b: next © 2007 Herbert I. Gross next 100(0.06x + 0.05y) = 100(0.15) 100(0.05x + 0.04y) = 100(0.20) or 6x + 5y = 15 5x + 4y = 20 …which is the same system we solved in Problem #1a. Hence, the solution is x = 40 and y = - 45.

next Notes on #1b Problem #1b illustrates the advantage we have in dealing with equations rather than with expressions. Namely by our “rules of the game”, we are allowed to do whatever we want to on one side of the equation provided that we do the same thing on the other side. © 2007 Herbert I. Gross next This idea will be illustrated again in our solution for Problem #1c.

Answer: x = 40, y = - 45 Solution for #1c: Since working with fractions can be rather cumbersome, we eliminate the fractions as we did in Problem #1b. Namely, starting with… next © 2007 Herbert I. Gross next …we multiply the top equation by 30 (which is the least common multiple of the denominators 2, 5, and 6) to obtain… 1 / 5 x + 1 / 6 y = 1 / 2 1 / 4 x + 1 / 5 y = 1 30( 1 / 5 x + 1 / 6 y) = 30( 1 / 2 ) or 6x + 5y = 15

Solution for #1c: And we multiply both sides of the bottom equation in our system by 20 (which is the least common multiple of the denominators 2, 4, and 5). In this way we obtain… next © 2007 Herbert I. Gross next 5x + 4y = 20 20( 1 / 4 x + 1 / 5 y) = 20(1) or 1 / 5 x + 1 / 6 y = 1 / 2 1 / 4 x + 1 / 5 y = 1

Solution for #1c: is the same system we solved in Problem #1a. Hence, the solution to this problem is also x = 40 and y = - 45. 6x + 5y = 15 5x + 4y = 20 next © 2007 Herbert I. Gross As a check, we replace x by 40 and y by - 45 and see if we obtain true statements. In particular… 1 / 5 (40) + 1 / 6 ( - 45) = 8 – 7 1 / 2 = 1 / 2 1 / 4 (40) + 1 / 5 ( - 45) = 10 – 9 = 1 and…

Problem #1d © 2007 Herbert I. Gross next Answer: x = 40, y = - 45 next The equation of the line L 1 is 6x + 5y = 15, and the equation of the line L 2 is 5x + 4y = 20. At what point do these two lines intersect?

Answer: x = 40, y = - 45 Solution for #1d: The point (x,y) is on the line L 1 if and only if 6x + 5y = 15. It is on the line L 2 if and only if 5x + 4y = 20. Therefore, in order for (x,y) to be the point of intersection of the two lines it must be a point on both lines. In other words, the required point (x,y) must satisfy the system of equations… next © 2007 Herbert I. Gross next 6x + 5y = 15 5x + 4y = 20

Solution for #1d: is the same system we solved in the previous parts of Problem #1. 6x + 5y = 15 5x + 4y = 20 next © 2007 Herbert I. Gross Therefore, the point of intersection of L 1 and L 2 is (40, - 45).

next Notes on #1d This problem illustrates the advantage of our sometimes using algebra to solve geometric questions. © 2007 Herbert I. Gross next For example, we might have chosen to graph the two lines and visually locate the point at which they intersect. However, we might need a rather large sheet of paper (or a clever choice in making a scale) in order to have the point (40, - 45) be on the paper.

Notes on #1d By the way, we can draw the lines L 1 and L 2 without having to resort to rewriting their equations in the “y = mx + b” form. © 2007 Herbert I. Gross next For example, with respect to the equation 5x + 4y = 20, we can set x equal to 0 and solve the resulting equation to find that y = 5. In other words the point (0,5) is one point on the line.

Notes on #1d In a similar way, we can replace y by 0 in the equation 5x + 4y = 20 to see that (4,0) is another point on the line. © 2007 Herbert I. Gross next Since two points uniquely determine the line, we then simply draw the line that passes through (0,5) and (4,0) (which are referred to as the y-intercept and the x-intercept, respectively).

Notes on #1d Other problems that exist in constructing geometrical graphs are… © 2007 Herbert I. Gross next ► If the two lines intersect at a point whose coordinates are not integers we have to estimate the coordinates of the point of intersection. This is illustrated in the following slide.

Notes on #1d ► If the angle of intersection of the two lines is close to 0° (that is, if the two lines are “almost parallel” it is hard to read the exact point at which they intersect. © 2007 Herbert I. Gross next This is illustrated in the following slide.

Notes on #1d © 2007 Herbert I. Gross ► Of special importance, keep in mind that if the there are more than 3 variables, it is not possible to represent the graph in a traditional geometric form. In such cases, we have to rely on other computational techniques such as algebra.

Answer: x = 24, y = 21 Solution for #2a: Using the method that was stressed in this lesson we may begin by eliminating either x or y from the system… 2x + 3y = 111 5x + 4y = 204 next © 2007 Herbert I. Gross 5(2x + 3y) = 5(111) 2(5x + 4y) = 2(204) If we elect to eliminate x, we may multiply the top equation by 5 and the bottom equation by 2 to obtain the equivalent system… 10x + 15y = 555 10x + 8y = 408 or next

Solution for #2a: If we now subtract the bottom equation from the top equation we see that… next © 2007 Herbert I. Gross And if we divide both sides by 7, we obtain… 10x + 15y = 555 10x + 8y = 408 7y = 147 next 7 y = 21

Solution for #2a: For further practice in using this method we can then solve for x by eliminating y from both equations … 2x + 3y = 111 5x + 4y = 204 next © 2007 Herbert I. Gross 4(2x + 3y) = 4(111) 3(5x + 4y) = 3(204) To do this we may multiply the top equation by 4 and the bottom equation by 3 to obtain… 8x + 12y = 444 15x + 12y = 612 or next

Solution for #2a: If we now subtract the top equation from the bottom equation we see that… next © 2007 Herbert I. Gross And if we divide both sides by 7, we obtain… 7x = 168 next 7 x = 24 8x + 12y = 444 15x + 12y = 612

Solution for #2a: To complete the solution, we then replace x by 21 and y by 24 in the system… next © 2007 Herbert I. Gross 2(24) + 3(21) = 48 + 63 = 111 2x + 3y = 111 5x + 4y = 204 5(24) + 4(21) = 120 + 84 = 204 and verify that…

next Notes on #2a In vertical format, it is often confusing trying to subtract a greater number from a lesser number. If this is true in your case, simply observe that the solution to a system of equation does not depend on the order in which the equations are listed. © 2007 Herbert I. Gross

next © 2007 Herbert I. Gross More specifically, rather than subtracting the top equation from the bottom equation, we may first rewrite… next 8x + 12y = 444 15x + 12y = 612 Notes on #2a in the equivalent form.. …and we then subtract the bottom equation from the top equation. —

next Notes on #2a If you prefer to add rather than to subtract, you may use the “add the opposite” rule. Namely, you change the sign of each term in either the top equation or the bottom equation (changing the sign of each term in an equation is equivalent to multiplying each term in the equation by - 1). © 2007 Herbert I. Gross

next Notes on #2a For example, with respect to our system… © 2007 Herbert I. Gross next …if we multiply both sides of the bottom equation by - 1, we obtain the equivalent system… 8x + 12y = 444 15x + 12y = 612 --- 8x + 12y = 444 15x + 12y = 612

Problem #2b © 2007 Herbert I. Gross next Answer: 21¢ next Two (equally priced) apples and three (equally priced) pears cost $1.11 while five apples and four pears cost $2.04. What is the cost of each pear ?

Answer: 21¢ Solution for #2b: If we let A represent the cost, in cents, of each apple and P represent the cost, in cents, of each pear; we obtain the system of equations… 2A + 3P= 111 5A + 4P = 204 next © 2007 Herbert I. Gross

Solution for #2b: If we were to replace A by x and P by y, we would have obtained… 2 A + 3 P = 111 5 A + 4 P = 204 next © 2007 Herbert I. Gross …which is the same system we solved in Problem #2a. In that case, we saw that y = 21. Since y now represents the cost of each pear (P), we see that each pear costs 21 cents. x xy y next

Notes on #2b In Lessons 23 and 24, we will study the solution of word problems in greater detail. © 2007 Herbert I. Gross For now, we simply want to illustrate how the solutions of systems of equations can be used to solve “real world” problems. next

Notes on #2b Another “real world” application is with respect to graphs. © 2007 Herbert I. Gross For example, what our solution also tells us is that the lines whose equations are 2x + 3y = 111 and 5x + 4y = 204 intersect at the point (24,21). next

Notes on #2b Notice the advantage in using A rather than x to denote the cost of each apple and P rather than y to denote the cost of each pear. Namely, if we use x and y it is easy to forget whether x denotes the number of apples or whether it denotes the number of pears. © 2007 Herbert I. Gross next However, it’s rather easy to remember that A represents the cost of each apple and P represents the cost of each pear.

Answer: x = 20, y = 30 Solution for #3a: To avoid having to work with decimal fractions we can multiply both sides of the bottom equation in… next © 2007 Herbert I. Gross by 2 to obtain the equivalent system of equations… x + y = 50 7.5x + 5y = 300 x + y = 50 15x + 10y = 600

Solution for #3a: To eliminate y from the system… next © 2007 Herbert I. Gross we may first multiply the top equation in our system by 10 to obtain the equivalent system… x + y = 50 15x + 10y = 600 10x + 10y = 500 15x + 10y = 600

Solution for #3a: If we now subtract the top equation from the bottom equation, we see that 5x = 100, or x = 20. next © 2007 Herbert I. Gross We may then replace x by 20 in the equation x + y = 50 to conclude that y = 30. 10x + 10y = 500 15x + 10y = 600 To check our result we replace x by 20 and y by 30 in the equation 7.5x + 5y = 300 and see that… = 150 + 150 = 3007.5 x + 5 y (20) (30) next

Notes on #3a Like in any other game, strategy often depends on the given situation. © 2007 Herbert I. Gross next For example, in this case the equation x + y = 50 gives us a very efficient way to find the value of y, once we know the value of x is 20.

Problem #3b next Answer: 20 pounds @$7.50 and 30 pounds @$5 next Coffee beans worth $7.50 per pound are blended with coffee beans worth $5 per pound to make a mixture of 50 pounds worth $6 per pound. How many pounds of each mixture are there in the blend? © 2007 Herbert I. Gross

Answer: 20 lbs.@ $7.50, 30 lbs.@ $5.00 Solution for #3b: If we let x represent the number of pounds of the $7.50 per pound coffee, and y represent the number of pounds of the $5 per pound coffee, then the given information tells us that… next © 2007 Herbert I. Gross x + y = 50 If the 50 pound mixture costs $6 per pound, then the total value of the mixture is 50 × $6 or $300.

Solution for #3b: At $7.50 per pound, x pounds cost 7.50x. next © 2007 Herbert I. Gross And at $5 per pound, y pounds cost 5y. The total value of the mixture, in dollars (300) is equal to the sum of the values of the $7.50 per pound portion (7.5x) and the $5 per pound portion (5y). In terms of an equation, this means that… 7.5x + 5y = 300 next

Solution for #3b: And in order for the above equations to be true at the same time, x and y have to be the solution of the system… next © 2007 Herbert I. Gross …which is exactly the same system we solved in Problem #3a. 7.5x + 5y = 300 next x + y = 50 7.5x + 5y = 300

Solution for #3b: Since x represents the number of pounds of the coffee that costs $7.50 per pound we know that there must be 20 pounds of this coffee in the mixture; and since the mixture has a total weight of 50 pounds, the remaining 30 pounds must represent the number of pounds of the $5 per pound coffee is in the mixture. next © 2007 Herbert I. Gross

next Notes on #3b In many textbooks, Exercise 3b is treated as a linear equation in x. Namely, the approach is usually something like the following… © 2007 Herbert I. Gross next Let x denote the number of pounds of the $7.50 per pound coffee in the mixture. Since the mixture weighs 50 pounds, the number of pounds of the $5 per pound coffee is 50 – x.

next Notes on #3b In essence, this is equivalent to subtracting x from both sides of the equation x + y = 50 and obtaining that y = 50 – x. © 2007 Herbert I. Gross next Especially for beginning students we prefer to begin with x + y = 50 because we believe it’s helpful not having to keep track of the steps in your head.

next Notes on #3b In other words, many textbooks prefer to use a different strategy in solving the system of equations… © 2007 Herbert I. Gross next Namely, they solve the top equation for y to obtain y = 50 – x, and they then replace y by 50 – x in the bottom equation. x + y = 50 7.5x + 5y = 300

Answer: x = 2, y = 8, z = 17 Solution for #4: If we had wished, we could have used the method that was discussed in the appendix of this lesson (and we’ll show how in the notes that follow this solution). next © 2007 Herbert I. Gross However, it is not too difficult to paraphrase the system below into an equivalent system of two linear equations. x + 3y + 5z = 111 y = 3x + 2 z = 2y + 1

Solution for #4: For example, we may replace z in the top equation of the system below by its value in the bottom equation. next © 2007 Herbert I. Gross x + 3y + 5z = 111 y = 3x + 2 z = 2y + 1 x + 3y + 5 z = 111 ( ) 2y + 1 In other words, we may rewrite the equation as… next

Solution for #4: In simplifying the equation below, we obtain… x + 3y + 5(2y + 1) = 111 next © 2007 Herbert I. Gross We may now combine the above equation with the middle equation of the original system (y = 3x + 2) to obtain the system… x + 3y + 10y + 5 = 111 x + 13y = 106 y = 3x + 2

Solution for #4: To solve the above system we may replace y in the top equation by its value in the bottom equation to obtain… next © 2007 Herbert I. Gross x + 13(3x + 2) = 106 x + 13y = 106 y = 3x + 2 x + 39x + 26 = 106 40x + 26 = 106 40x = 80 x = 2 Replacing x by 2 in the equation y = 3x + 2 we obtain the equation… y = 3(2) + 2y = 8 next

Solution for #4: We then replace x by 2 and y by 8 in the equation x + 3y + 5z = 111 to obtain the equation… next © 2007 Herbert I. Gross x + 3y + 5z = 111 2 + 3(8) + 5z = 111 2 + 24 + 5z = 111 26 + 5z = 111 5z = 85 next z = 17 To complete our proof, we replace x by 2, y by 8, and z by 17 in the expression x + 3y + 5z and see that… 2 + 3(8) + 5(17) = 111 2 + 24 + 85 = 111

next Notes on #4 An interesting thing about most mathematics problems is that there is only one correct answer but hardly ever only one way to find the correct answer. © 2007 Herbert I. Gross next Another approach would have been to express y and z in terms of x, and thus reduce the top equation in the system to a linear equation in x.

The middle equation in the system, namely, y = 3x + 2 tells us that whenever we wish, we may replace y by 3x + 2. © 2007 Herbert I. Gross x + 3y + 5z = 111 y = 3x + 2 z = 2y + 1 Hence, we may replace y by 3x + 2 in the bottom equation of the system to obtain the equivalent equation… Notes on #4 z = 2y + 1 z = 2(3x + 2) + 1 z = 6x + 4 + 1 z = 6x + 5

Going now to the top equation, we may replace y by its value in the second equation, and z by its value in the third equation to obtain… next © 2007 Herbert I. Gross x + 3y + 5z = 111 y = 3x + 2 z = 6x + 5 Notes on #4 x + 3y + 5z = 111 x + 3(3x + 2) + 5(6x + 5) = 111 x + 9x + 6 + 30x + 25 = 111 40x + 31 = 111 40x = 80 x = 2 next

Once we know that x = 2, we may replace x by 2 in the equations y = 3x + 2 and z = 6x + 5 to conclude that… © 2007 Herbert I. Gross x + 3y + 5z = 111 y = 3x + 2 z = 6x + 5 Notes on #4 y = 3(2) + 2 = 6 + 2 = 8 z = 6(2) + 5 = 12 + 5 = 17 y 8 z 17

In this particular exercise the equations were of such a nature that it would have been more tedious if we had used the method that was highlighted in this Lesson. However, it could have been done. next © 2007 Herbert I. Gross x + 3y + 5z = 111 y = 3x + 2 z = 2y + 1 Notes on #4 For example, with just a little algebraic manipulation, we could have rewritten… in the form… x + 3y + 5z = 111 - 3x + y = 2 - 2y + 1z = 1

To eliminate x from all but the top equation in the system above, we may first multiply both sides of the top equation in the system by 3 to obtain… next © 2007 Herbert I. Gross Notes on #4 x + 3y + 5z = 111 - 3x + y = 2 - 2y + 1z = 1 x + 3y + 5z = 111 - 3x + y = 2 - 2y + 1z = 1 3x + 9y + 15z = 333

after which we can replace the middle row in the system above by the sum of the middle row and the top row to obtain… next © 2007 Herbert I. Gross Notes on #4 3x + 9y + 15z = 333 - 3x + y = 2 - 2y + 1z = 1 3x + 9y + 15z = 333 - 3x + y = 2 - 2y + 1z = 1 10y + 15z = 335

To eliminate y from the bottom equation in the system above, we may first multiply both sides of the bottom equation by 5 to obtain… next © 2007 Herbert I. Gross Notes on #4 3x + 9y + 15z = 333 10y + 15z = 2 - 2y + 1z = 1 3x + 9y + 15z = 333 10y + 15z = 335 - 2y + 1z = 1 - 10y + 5z = 5

We may then replace the bottom row in the system above by the sum of the bottom row and the middle row to obtain… next © 2007 Herbert I. Gross Notes on #4 3x + 9y + 15z = 333 10y + 15z = 335 - 10y + 5z = 5 3x + 9y + 15z = 333 10y + 15z = 335 - 10y + 5z = 5 20z = 340

We can simplify the system above by dividing both sides of the top equation by 3, next © 2007 Herbert I. Gross Notes on #4 3x + 9y + 15z = 333 10y + 15z = 335 20z = 340 3x + 9y + 15z = 333 10y + 15z = 335 20z = 340 x + 3y + 5z = 111 2y + 3z = 67 both sides of the middle equation by 5, and both sides of the bottom equation by 20 to obtain… z = 17 next

Next, we would replace z by 17 in the middle equation in the system above to see that… © 2007 Herbert I. Gross Notes on #4 x + 3y + 5z = 111 2y + 3z = 67 z = 17 2y + 3 z = 67(17) next 2y + 51 = 67 2y = 16 y = 8 next

Finally, we would replace z by 17 and y by 8 in the top equation of the system to obtain… © 2007 Herbert I. Gross Notes on #4 x + 3y + 5z = 111 y = 8 z = 17 x + 3 y + 5 z = 111(17) next x + 24 + 85 = 111 x + 109 = 111 x = 2 next (8)(8)

Notes on #4 In this case, our method is quite cumbersome. However, notice that it yields the solution in a very logical step-by-step way. © 2007 Herbert I. Gross next As we shall illustrate in the next problem, this step by step method becomes more and more valuable as the system of equations becomes more and more complicated (especially as the number of variables increases).

Answer: x = 1, y = - 2, z = 3 Solution for #5: This system is more “complicated” than the system in Problem 4, at least in the sense that each of its three equations contains all of the variables x, y and z. x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1 next © 2007 Herbert I. Gross Therefore, we will use the systematic approach described in this Lesson.

Solution for #5: More specifically we will multiply both sides of the top equation by 6; next © 2007 Herbert I. Gross both sides of the middle equation by - 3, and both sides of the bottom equation by - 2 to obtain the equivalent system of equations… x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1 6x + 6y + 6z = 12 - 6x + - 9y + - 9z = - 15 - 6x + - 8y + - 4z = - 2 next x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1

Solution for #5: Next, we replace the middle equation in the system above by the sum of the middle equation and the top equation; next © 2007 Herbert I. Gross 6x + 6y + 6z = 12 - 6x + - 9y + - 9z = - 15 - 6x + - 8y + - 4z = - 2 - 3y + - 3z = - 3 - 2y + 2z = 10 and we replace the bottom equation by the sum of the bottom equation and the top equation to obtain… 6x + 6y + 6z = 12 - 6x + - 9y + - 9z = - 15 - 6x + - 8y + - 4z = - 2

Solution for #5: We can make the system a bit simpler by dividing both sides of the top equation in the system by 6; next © 2007 Herbert I. Gross both sides of the middle equation by - 3, and both sides of the bottom equation by 2 to obtain the equivalent system of equations… 6x + 6y + 6z = 12 - 3y + - 3z = - 3 - 2y + 2z = 10 next 6x + 6y + 6z = 12 - 3y + - 3z = - 3 - 2y + 2z = 10 x + y + z = 2 y + z = 1 - y + z = 5

Solution for #5: We can then eliminate y from the bottom equation in the system above by replacing the bottom equation by the sum of the bottom equation and the middle equation to obtain… next © 2007 Herbert I. Gross x + y + z = 2 y + z = 1 - y + z = 5 2z = 6 … and from the bottom equation we conclude that z = 3. x + y + z = 2 y + z = 1 - y + z = 5

Solution for #5: Finally, to complete our proof we return to the system… x + y + z = 2 next © 2007 Herbert I. Gross and replace x by 1, y by - 2 and z by 3 and then verify that the following three equations are all true statements… 2 x + 3 y + 3 z = 5 3 x + 4 y + 2 z = 1 -2-2 ( - 2) 1 (1) ( - 2) 3 (3) next x = 1 y = - 2 z = 3

next Concluding Note © 2007 Herbert I. Gross next As a preface to this note, consider a question such as, “ What was the color of Paul Revere’s white horse?” This is a legitimate question but one which already contains the answer. Surprising as it might seem, the same analysis applies to the system of equations. x = 1 y = - 2 z = 3

next Concluding Note © 2007 Herbert I. Gross next While this might sound like a “dumb question”, it has a profound consequence. More specifically, in the form of a question, it is asking us… “For what values of x, y, and z is it true that x = 1, y = - 2, and z = 3?”

Concluding Note © 2007 Herbert I. Gross next Namely, we have just shown that the system of equations… x = 1 y = - 2 z = 3 x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1 …is equivalent to… …and therefore must have the same solution set.

next Concluding Note © 2007 Herbert I. Gross …we were checking that this was indeed the case. So in the last step of our solution, when we replaced x by 1, y by - 2, and x by 3 in the system… x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 21 By Herbert I. Gross and Richard A. Medeiros next.

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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 21 By Herbert I. Gross and Richard A. Medeiros next.

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