The numerator of a fraction is two less than its denominator. When both numerator and denominator are increased by 3, the fraction is increased by 3/20. Find the original fraction. Explain your solution

Let the original fraction be x/(x+2)

(x+3)/(x+5) = x/(x+2) + 0.15

Solve for x. Try x = 3,

6/8 = 3/5 + 0.15 It works!

So the original fraction is 3/5.

thanks

To solve this problem, let's assume the original fraction is x/y, where x is the numerator and y is the denominator.

According to the given information, the numerator of the fraction is two less than its denominator. So we can write the first equation as:

x = y - 2

The second piece of information states that when both the numerator and denominator are increased by 3, the fraction is increased by 3/20. So the new fraction can be expressed as:

(x + 3)/(y + 3) = x/y + 3/20

To solve this equation, we need to eliminate the fractions, so let's multiply both sides of the equation by 20y:

20y(x + 3) = 20xy + 3y(y + 3)

Now we can expand and simplify the equation:

20xy + 60y = 20xy + 3y^2 + 9y

Rearranging the equation:

3y^2 + 9y - 60y = 0

Simplifying further:

3y^2 - 51y = 0

Factoring out y:

y(3y - 51) = 0

From this equation, we see that either y = 0 (which is not possible in this context) or 3y - 51 = 0. Solving for y:

3y - 51 = 0
3y = 51
y = 17

Now substitute the value of y back into the first equation (x = y - 2) to find the value of x:

x = 17 - 2
x = 15

Therefore, the original fraction is 15/17.