SOLUTION: the sum of the reciprocals of two integers is 3/4. if the larger number exceeds the smaller number by 2,what is the larger integers?

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Question 662666: the sum of the reciprocals of two integers is 3/4. if the larger number exceeds the smaller number by 2,what is the larger integers?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The easiest way to solve this problem is the 5th grader way.
However, your teacher expects you to write a quadratic equation and solve that equation.

USING A QUADRATIC EQUATION:
Let the smaller integer be x.
Since the larger number exceeds the smaller number by 2,
the larger number is x%2B2.
The sum of the reciprocals of the two integers is 3/4 translates as
1%2Fx%2B1%2F%28x%2B2%29=3%2F4
We use a common denominator to add those rational function reciprocals
1%2Fx%2B1%2F%28x%2B2%29=3%2F4-->%28x%2B2%29%2F%28x%28x%2B2%29%29%2Bx%2F%28x%28x%2B2%29%29=3%2F4-->%28%28x%2B2%29%2Bx%29%2F%28x%28x%2B2%29%29=3%2F4-->%282x%2B2%29%2F%28x%5E2%2B2x%29=3%2F4
Next, to get rid of denominators, we multiply both sides times the denominators.
(Some refer to that as solving a proportion by equating the cross-products).
%282x%2B2%29%2F%28x%5E2%2B2x%29=3%2F4 --> 3%28x%5E2%2B2x%29=4%282x%2B2%29+
We need to work further to get the standard form of the quadratic equation.
3%28x%5E2%2B2x%29=4%282x%2B2%29 --> 3x%5E2%2B6x=8x%2B8 --> 3x%5E2%2B6x-8x-8=0 --> 3x%5E2-2x-8=0
Now that we have the quadratic equation in the form ax%5E2%2Bbx%2Bc=0,
we solve using the quadratic formula, x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+:
x=%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%2A3%2A%28-8%29%29%29%2F%282%2A3%29-->x=2+%2B-+sqrt%284%2B96%29%29%2F6 -->x=2+%2B-+sqrt%28100%29%29%2F6-->x=2+%2B-+10%29%2F6
The equation solution x=2-10%29%2F6=-8%2F6=-4%2F3 is a solution of 1%2Fx%2B1%2F%28x%2B2%29=3%2F4,
but it is not a solution for the problem because -4%2F3 is not an integer.
The other solution to the equation,
x=2%2B10%29%2F6=12%2F6=2 gives us the smaller integer.
The larger integer, x%2B2, is 2%2B2=highlight%284%29
NOTE:
I could also have started by making the larger integer y,
and the smaller integer y-2,
which would lead to the equation 3y%5E2-14y%2B8=0,
and the integer solution y=4 (the larger integer).

THE FIFTH GRADER WAY:
The fifth grader knows that fifth graders are smarter than older people,
and knows that this problem will be easy.
The fifth grader knows that integers are number like 1, 2, 3, 4, and so on.
The fifth grader hopefully knows that the reciprocal of a number is 1 divided by the number,
so the reciprocal of 2 would be written as 1%2F2 or as 0.5,
and the reciprocal of 8 would be written as 1%2F8 or as 0.125.
The fifth grader figures out that larger numbers give small reciprocals,
and small number give large reciprocals,
so he/she figures that the solution integers must be smaller than 8 and 10,
because 1%2F8%2B1%2F10=0.125%2B0.1=0.225 which is much smaller than 3%2F4=0.75.
The fifth grader starts trying small numbers:
Could the two integers be 1 and 3?
1%2F1%2B1%2F3=1%2B1%2F3=1%261%2F3 and that is larger than 3%2F4
Could the two integers be 2 and 4?
1%2F2%2B1%2F4=2%2F4%2B1%2F4=3%2F4 so the two integers are 2 and 4,
and the larger of the two integers is highlight%284%29.