The denominator of a fraction is three more than twice the numerator. If both numerator and denominator are decreased by seven, the simplified result is 7 divided by19. Find the original fraction. (Do NOT simplify.)
In original fraction:
numerator ---- x
denominator ---- 2x+3
so the fraction was x/(2x+3)
after change:
fraction is (x-7)/(2x+3 - 7) = 7/19
(x-7)/(2x-4) = 7/19
19x - 133 = 14x - 28
5x = 105
x = 21
the original fraction is 21/45
original fraction: n/(n+3)
(n-7)/(n+3-7) = 7/19
So solve for n
my bad. I missed the "twice"
To solve this problem, let's start by setting up the equation based on the given information.
Let's say the numerator of the fraction is "x". Then, the denominator of the fraction would be "3 + 2x", as it is three more than twice the numerator.
So, the original fraction can be written as "x / (3 + 2x)".
According to the problem, both the numerator and denominator are decreased by 7, resulting in a simplified fraction of 7/19.
So, the new fraction can be written as "(x - 7) / ((3 + 2x) - 7)", which simplifies to "(x - 7) / (2x - 4)".
Now we can set up the equation:
"(x - 7) / (2x - 4) = 7/19"
To solve this equation, we can cross-multiply:
19(x - 7) = 7(2x - 4)
19x - 133 = 14x - 28
5x = 105
x = 21
Now that we have the value of x, we can substitute it back into the original fraction to find the original fraction:
Original Fraction = x / (3 + 2x)
Original Fraction = 21 / (3 + 2(21))
Original Fraction = 21 / (3 + 42)
Original Fraction = 21 / 45
Therefore, the original fraction is 21/45.