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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