Question 92592: In a weight-lifting competition, the total weight of Joe's two lifts was 750 pounds. If twice the weight of is first lift was 300 pounds more than the weight of his second lift, what was the weight, in pounds, of his first lift?
This what I have so far: (2n (300+n)= ?
I would like some assistance in setting up the problem. Thanks.Th
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let's call F the weight of the first lift and S the weight of the second lift.
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The total weight of two lifts was 750 lbs. In equation form this becomes:
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F + S = 750 <=== first equation
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Twice the weight of the first lift is 2F. This is 300 pounds more than the weight of the
second lift. So if we take 300 lbs away from 2F the result will equal the second lift S.
In equation form, this is:
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2F - 300 = S <=== second equation
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From the second equation we know that S is equal to 2F - 300. Therefore, in the first equation
we can replace S with 2F - 300. Let's do that and the first equation becomes:
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F + (2F - 300) = 750
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The parentheses are preceded by a plus sign so they can be removed without changing
the signs of the terms inside. So the equation becomes:
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F + 2F - 300 = 750
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Add the two terms that contain F to get:
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3F - 300 = 750
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Get rid of the - 300 on the left side by adding 300 to both sides. On the left side this
addition cancels out the - 300 and the equation becomes:
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3F = 750 + 300 = 1050
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Now you can solve for F by dividing both sides by 3 to get:
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F = 1050/3 = 350
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Now we know that the first lift F was 350 lbs. Since the sum of the two lifts was 750
lbs, that means that with a first lift F of 350 lbs, the second lift S must be the remaining
400 lbs to make the total of both lifts 750 lbs.
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The two equations above can be solved in ways other than the substitution method shown above.
For example, you could do the problem by variable elimination or by graphing the two equations
(one axis representing F and the other S) and find the value of F and the corresponding
value of S at the point where the two graphs cross.
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Hope this helps you to see that problems of this type usually are worked by establishing
a set of two equations and solving this set simultaneously.
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