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330 Review Questions and Answers
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To solve this, we add 15 to each side and then divide through by 5, as follows: 5x 15 = 0 5x = 15 x = 3
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Question 16-9
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How can we add multiples of the two original equations stated in Question 16-2 to solve the linear system for y
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We can multiply the second equation through by 2 and then add it to the first equation, getting the sum 2x y + 8 = 0 2x + 6y 18 = 0 5y 10 = 0 To solve this, we add 10 to each side and then divide through by 5, like this: 5y 10 = 0 5y = 10 y=2
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How can we solve the two-by-two linear system stated in Question 16-2 by the rename-andreplace (substitution) method Here are the original equations again, for reference: 2x y + 8 = 0 and x 3y + 9 = 0
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We start by converting either of the two equations to SI form, so one of the variables appears all alone on the left side of the equals sign Let s use the first equation and isolate y on the left side To manipulate the equation, we proceed just as we did in Answer 16-2: 2x y + 8 = 0 y + 8 = 2x y = 2x 8 y = 2x + 8
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Part Two 331
Next, we substitute the quantity (2x + 8) for y in the second equation and solve the result for x, as follows: x 3y + 9 = 0 x 3(2x + 8) + 9 = 0 x 6x 24 + 9 = 0 5x 15 = 0 5x = 15 x = 3 Now that we know the value of x, we can plug it into either equation and solve for y Let s use the first equation Then we proceed as follows: 2x y + 8 = 0 2 ( 3) y + 8 = 0 6 y + 8 = 0 6 + 8 = y y=2
17
Question 17-1
Let s consider again the two-by-two system we saw in Question 16-2 How can we graph this system in Cartesian coordinates, with x as the independent variable and y as the dependent variable Here are the original equations: 2x y + 8 = 0 and x 3y + 9 = 0
Answer 17-1
We can use the SI forms of the equations to find their y-intercepts, and the solution of the system to find a third point that lies on both lines We re lucky here, because the intersection point is fairly far away from the y axis The SI forms of the equations were derived in Answer 16-2 Respectively, they are: y = 2x + 8
332 Review Questions and Answers
and y = (1/3)x + 3 We know from these equations that the y-intercepts are 8 and 3, so the point (0, 8) is on the first line and the point (0, 3) is on the second line The solution to the system, as we ve already determined, is ( 3, 2) That point lies on both lines Figure 20-6 shows plots of these three points, along with plots of the lines connecting the points and representing the equations
Question 17-2
How can we determine the x-intercept of the line representing the equation y = 2x + 8 on the basis of the known slopes and the point data in Fig 20-6 (We can set y = 0 and then calculate x by arithmetic, but the purpose of this exercise is to demonstrate the geometric principles of slope and intercept)
Answer 17-2
The slope of the line is 2 Therefore, if we start from any point on the line and move in the positive x direction by n units, we must move in the positive y direction by 2n units to stay on the line In the opposite sense, if we start from any point on the line and move in the negative y direction by p units, we must move in the negative x direction by p /2 units to stay on the line Let s start at the point (0, 8), which is the y-intercept If we move in the negative y direction
(0,8)
y = (1/3)x + 3 Solution = ( 3,2) (0,3)
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