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Use the elimination method to solve each system. …

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104

Problem 20 Medium Difficulty

Use the elimination method to solve each system.
$$
\left\{\begin{array}{l}
{-2 x+4 y=12} \\
{2 x+4 y=28}
\end{array}\right.
$$

Answer

$y=5$
$x=4$

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Chapter 4

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Section 3

Solving Systems of Equations by Elimination (Addition)

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Video Transcript

Let's help this system with the elimination method. So looking at the two equations notice that we have opposite ex terms. So if we add these equations together, the ex terms will be eliminated. And now we have four y plus four y. So that's eight. Why? And we have 12 plus 28 that's 40. So eight y equals 40. And now let's divide both sides of that equation by eight, and we have y equals five. Now, to find the value of X, we're going to take the value of Y equals five and substituted into one of the two equations I'll call the first equation Equation one and the 2nd 1 equation, too. So I'm going to use Equation one and I have negative two X plus four times five. Substituting the five in for y equals 12 so that gives us negative two X plus 20 equals 12 and then subtract 20 from both sides, and we have negative two X equals negative. Eight. Divide both sides by negative to an X equals four. Now let's go ahead and check that answer by substituting both numbers into equation, too. So that would be two times for plus four times five equals 28. So two times for eight plus four times 20 equals 28. That does check. And I will tell you this is actually the second time that I saw this problem because I made a small mistake the first time. And I didn't realize it until I did the check. So I really encourage you to do a check. So now that we know that that works, we can write her answer as an ordered pair X comma y. So we have four comma five.

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Alan S. Tussy, R. David Gustafson

Elementary and Intermediate Algebra

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