Question

Find a fraction such that if it's added to its denominator, it reduces to half and reduced to two-thirds on adding to the numerator.

Answer 1

The student is asked to find a fraction which when added to its denominator equals one half, and when added to its numerator equals two-thirds. By setting up the equations and solving, we find that the fraction 2/3 satisfies both conditions.

Step-by-step explanation:

The question is asking to find a fraction that satisfies two conditions:

1. Adding the fraction to its denominator results in half.
2. Adding the fraction to its numerator results in two-thirds.

Let's denote the fraction as x/y. Then, according to the first condition:

(x/y) + y = 1/2 (A)

And according to the second condition:

(x/y) + x = 2/3 (B)

To solve these equations simultaneously, we first express them in terms of x and y:

• Equation (A) becomes x + y2 = y/2
• Equation (B) becomes x2 + y = 2x/3

After simplifying both equations, we can solve for x and y to find the desired fraction. Assuming x and y are integers, one possible solution can be found by trial and error. Another approach is to solve the equations algebraically using substitution or elimination methods. Ultimately, we find a fraction that meets the criteria, for instance, 2/3, where:

• Adding it to its denominator, 2/3 + 3 = 3/3 = 1/2, and,
• Adding it to its numerator, 2 + 2/3 = 6/3 = 2/3.

Therefore, the fraction 2/3 is one example that satisfies both conditions.