Question 80684: The sun of two numbers is 51. twice the first plus 4 times the second is 128. what are the numbers? Found 2 solutions by bucky, praseenakos@yahoo.com:Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! Call the first number F and the second number S
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The sum of the first number and the second number equals 51. In equation form this is:
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F + S = 51
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Then the problem says that twice the first (2*F) plus four times the second (4*S) equals 128.
In equation form this is:
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2F + 4S = 128
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So we now have a set of two equations:
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F + S = 51
2F + 4S = 128
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We can solve this set of equations by the process of variable elimination. One way is
to multiply the to equation (all terms on both sides) by -2 to get:
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-2F - 2S = -102
2F + 4S = 128
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Now add the two equations vertically. When you do, the -2F in the top equation cancels the
2F in the bottom equation. Continuing with the vertical addition you get:
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2S = 26
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Dividing both sides of this equation by 2 results in:
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S = 13
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Now return to the original first equation which said that the sum of the first and second
numbers is 51 ...
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F + S = 51
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Substitute 13 for S and this equation becomes:
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F + 13 = 51
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Solve for F by subtracting 13 from both sides:
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F = 38
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In summary, the two numbers are 13 and 38
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We already know that 13 plus 38 equals 51 as required by the problem.
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Then 2 times the first is 2 * 38 and that equals 76. Add to that 4 times 13 which is 52.
The result of 76 + 52 is 128 as was also required by the problem. The answer checks.
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Hope this helps you to understand the problem.
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