# how to create barcode in c#.net Review Questions and Answers in Software Encoding QR Code 2d barcode in Software Review Questions and Answers

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That s in the form we want! Again from Answer 15-4, the slope-intercept form of the equation for line QR is y = (5/2)x 1 When we multiply each side by 2, we get 2y = 5x 2 Subtracting 5x from each side, we obtain 5x + 2y = 2 That s in the form we want! Once again referring to Answer 15-4, the slope-intercept form of the equation for line PR is y=x+2 Subtracting x from each side gives us x + y = 2 That s in the form we want!
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Question 16-1
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In Chap 16, we learned how a two-by-two linear system in variables x and y can be solved by the following process: Morph both equations into SI form with y all by itself on the left side of the equals sign Mix the two equations to get a first-degree equation in x Solve the first-degree equation for x Substitute that solution back into one of the SI equations to solve for y How can we solve such a system by morphing and mixing alone, without substituting either variable for the other
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We can go through the morph-and-mix process twice, first for one variable and then for the other We proceed like this: Morph both equations into SI form with y all by itself on the left side of the equals sign Mix the two equations to get a first-degree equation in x
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Part Two 327
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Solve the first-degree equation for x Morph both equations into SI form with x all by itself on the left side of the equals sign Mix the two equations to get a first-degree equation in y Solve the first-degree equation for y
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Question 16-2
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How can we put the following two-by-two linear system into a pair of SI equations with y all by itself on the left side of the equals sign 2x y + 8 = 0 and x 3y + 9 = 0
By now, we re good enough at equation manipulation to write down the steps one after another, without having to justify everything For the first original equation, we can do this: 2x y + 8 = 0 y + 8 = 2x y = 2x 8 y = 2x + 8 and for the second original equation, we can do this: x 3y + 9 = 0 3y + 9 = x 3y = x 9 3y = x + 9 y = (1/3)x + 3
Question 16-3
How can we combine the two equations from Answer 16-2 to get a first-degree equation and solve the original system for x
We can mix the right sides of the two SI equations together and then solve the resulting firstdegree equation in x by manipulation Here it goes, one step at a time:
2x + 8 = (1/3)x + 3 6x + 24 = x + 9 6x + 15 = x 5x + 15 = 0 5x = 15 x = 3
Question 16-4
How can we put the two-by-two linear system from Question 16-2 into a pair of SI equations with x all by itself on the left sides of the equals signs
For the first original equation, we can do this: 2x y + 8 = 0 2x + 8 = y 2x = y 8 x = (1/2)y 4 and for the second original equation, we can do this: x 3y + 9 = 0 x + 9 = 3y x = 3y 9
Question 16-5
How can we combine the two equations from Answer 16-4 to get a first-degree equation and solve the original system for y
We can mix the right sides of the two SI equations together and then solve the resulting firstdegree equation in y by manipulation, as follows: (1/2)y 4 = 3y 9 y 8 = 6y 18 y + 10 = 6y 10 = 5y y=2
Question 16-6
How can we be sure the solution we obtained in Answers 16-3 and 16-5 is in fact the correct solution to the original two-by-two linear system