the sum of numerator and denominator is 3 less than twice the denominator.If the numerator and denominator is decreased by 1 , the numerator becomes half the denominator.Find the fraction?
n+d = 2d-3
n+1 = (d+1)/2
2/5
To solve this problem, let's denote the numerator as "x" and the denominator as "y."
According to the problem, the sum of the numerator and denominator (x + y) is 3 less than twice the denominator (2y - 3).
So, we can write the equation: x + y = 2y - 3
If the numerator and denominator are decreased by 1 (x - 1 and y - 1, respectively), the numerator becomes half the denominator (1/2 * y).
Therefore, we can write another equation: x - 1 = 1/2 * (y - 1)
Now, we can solve the system of equations to find the values of x and y.
Step 1: Solve the second equation for x in terms of y:
x = 1/2 * (y - 1) + 1
Step 2: Substitute this value for x in the first equation:
(1/2 * (y - 1) + 1) + y = 2y - 3
Step 3: Simplify and solve for y:
1/2 * y - 1/2 + 1 + y = 2y - 3
3/2 * y + 1/2 = 2y - 3
2y - 3 - 3/2 * y - 1/2 = 0
4y - 6 - 3y - 1 = 0
y - 7 = 0
y = 7
Step 4: Substitute y = 7 into the second equation to find x:
x - 1 = 1/2 * (7 - 1)
x - 1 = 1/2 * 6
x - 1 = 3
x = 4
Therefore, the fraction is x/y, which is 4/7.