If one is added to both numenator ans denominator fraction become 3/4 however 3 is subtracted from both the fraction become 1/2.fnd the fraction.
1st case (x+1)/(y+1)= 3/4
2nd case. (x-3)/(y-3)= 1/2 not y-4
Well I have no idea how I got 4 there. I meant 3
To solve this problem, we need to represent the given situation algebraically and then solve the resulting equations.
Let's start by assigning variables to the numerator and denominator of the fraction. Let's call the numerator 'x' and the denominator 'y'.
According to the problem, when 1 is added to both the numerator and the denominator, the fraction becomes 3/4. This can be written as:
(x + 1) / (y + 1) = 3/4 -- equation 1
Similarly, when 3 is subtracted from both the numerator and the denominator, the fraction becomes 1/2. This can be written as:
(x - 3) / (y - 3) = 1/2 -- equation 2
Now, we have two equations with two variables. To solve these equations, we can use the method of substitution or elimination.
Let's solve using the method of substitution:
Rearrange equation 1 to solve for x:
(x + 1) / (y + 1) = 3/4
4(x + 1) = 3(y + 1)
4x + 4 = 3y + 3
4x = 3y - 1
x = (3y - 1)/4
Plug this value of x into equation 2:
[(3y - 1)/4 - 3)] / (y - 3) = 1/2
[(3y - 1 - 12)/4)] / (y - 3) = 1/2
[(3y - 13)/4)] / (y - 3) = 1/2
Now, cross-multiply:
2(3y - 13) = 4(y - 3)
6y - 26 = 4y - 12
6y - 4y = -12 + 26
2y = 14
y = 7
Now, substitute the value of y back into equation 1 to find x:
x = (3y - 1)/4
x = (3*7 - 1)/4
x = (21 - 1)/4
x = 20/4
x = 5
So, the resulting fraction is 5/7.
Let numerator be x and denominator be y
Fraction is x/y
1st case x+1/y+1 = 3/4
2nd case. X-3/y-4= 1/2
Solve for x and y