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Question 114217This question is from textbook
: I need some help....please!
The sum of two numbers is 40. The sum of twice the larger and 4 times the smaller is 108. Find the numbers.
This question is from textbook
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let L represent the larger number and S represent the smaller number.
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The problem tells you that the sum of these two numbers equals 40. You can write this in
equation form as:
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L + S = 40
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Next the problem tells you that if you find twice the larger and add it to four times the
smaller, the sum is 108. Twice the larger is 2L and 4 times the smaller is 4S. Add these two
products and the result equals 108. In equation form this is:
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2L + 4S = 108
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If you go back to the first equation ( L + S = 40 ) and subtract S from both sides you
get:
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L = 40 - S
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Now you can go to the second equation (2L + 4S = 108 ) and substitute 40 - S for L because
L is equal to 40 - S. When you do that substitution you get:
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2(40 - S) + 4S = 108
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Multiply out the left side by taking 2 times each of the terms in the parentheses to
get:
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80 - 2S + 4S = 108
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On the left side combine the -2S and the +4S to get +2S and this makes the equation become:
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80 + 2S = 108
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Get rid of the 80 on the left side by subtracting 80 from both sides to reduce the
equation to:
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2S = 28
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Finally, solve for S by dividing both sides by 2 to get:
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S = 28/2 = 14
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Now you know that S = 14. But the problem tells you that the sum of L and S equals 40.
This means that L has to be 26 because 40 - 14 = 26.
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So the answer to this problem is that the larger number is 26 and the smaller number is 14.
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Check ... do the two numbers add up to 40? Yes, they do.
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and does twice the larger (2*26 = 52) and 4 times the smaller (4*14 = 56) add up to be 108?
Yes, because 52 + 56 = 108.
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The answer checks.
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Hope this helps you to see your way through the problem and understand what you can do to
solve it.
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