the numerator of a certain fraction is four less then the denominator, if the numerator is doubled and the denominator is diminished by two, the sum of the original fraction and the new one is three. find the original fraction
old fraction: (x-4)/x
new fraction: 2(x-4)/(x-2)
and then it said:
(x-4)/x + 2(x-4)/(x-2) = 3
multiply each term by x(x-2)
(x-4)(x-2) + 2(x)(x-4) = 3x(x-2)
x^2 - 6x + 8 + 2x^2 - 8x = 3x^2 - 6x
-8x = -8
x = 1
the fraction was -3/1 ???
new fraction = -6/-1 = 6
The original fraction was -3/1 ?????
Check :
old fraction + new fraction
= -3 + 6 = 3 , it works
(poorly designed problem)
To solve this problem, let's start by defining the fraction.
Let x be the denominator of the fraction. According to the problem, the numerator is four less than the denominator, so the numerator can be expressed as (x - 4).
The original fraction can then be written as (x - 4) / x.
Now, let's consider the second part of the problem. The numerator is doubled, so it becomes 2(x - 4). The denominator is diminished by two, so it becomes (x - 2).
The new fraction can be expressed as 2(x - 4) / (x - 2).
According to the problem, the sum of the original fraction and the new fraction is three:
(x - 4) / x + 2(x - 4) / (x - 2) = 3.
To solve this equation, we can first eliminate the denominators by finding a common denominator. In this case, the common denominator is x * (x - 2).
(x - 4)(x - 2) / (x * (x - 2)) + 2(x - 4)(x) / (x * (x - 2)) = 3.
Simplifying the equation:
(x - 4)(x - 2) + 2(x - 4)(x) = 3(x * (x - 2)).
Expanding and rearranging the equation:
x^2 - 6x + 8 + 2x^2 - 8x = 3(x^2 - 2x).
Combining like terms:
3x^2 - 14x + 8 = 3x^2 - 6x.
Canceling out the common terms (3x^2):
-14x + 8 = -6x.
Moving all x terms to one side of the equation:
-14x + 6x = -8.
Simplifying:
-8x = -8.
Dividing both sides by -8:
x = 1.
Therefore, the original denominator of the fraction is 1.
Substituting this value back into the original fraction equation:
(x - 4) / x = (1 - 4) / 1 = -3.
So, the original fraction is -3/1 or simply -3.