Finding the Integer Value x for Equal Fractions

The problem asks for a positive integer $x$ such that when it's added to the numerator and denominator of $\frac{3}{4}$, and subtracted from the numerator and denominator of $\frac{15}{17}$, the resulting fractions are equal.

Setting Up the Equations

The first fraction $\frac{3}{4}$ is modified to $\frac{3+x}{4+x}$.

The second fraction $\frac{15}{17}$ is modified to $\frac{15-x}{17-x}$.

According to the problem statement, these two new fractions are equal:

$ \frac{3+x}{4+x} = \frac{15-x}{17-x} $

Solving for x

To solve for $x$, we cross-multiply:

$ (3+x)(17-x) = (15-x)(4+x) $

Expand both sides of the equation:

Left side: $ (3 \times 17) + (3 \times -x) + (x \times 17) + (x \times -x) = 51 - 3x + 17x - x^2 = 51 + 14x - x^2 $

Right side: $ (15 \times 4) + (15 \times x) + (-x \times 4) + (-x \times x) = 60 + 15x - 4x - x^2 = 60 + 11x - x^2 $

Now, set the expanded forms equal:

$ 51 + 14x - x^2 = 60 + 11x - x^2 $

The $ -x^2 $ terms cancel out on both sides:

$ 51 + 14x = 60 + 11x $

Rearrange the terms to isolate $x$. Subtract $ 11x $ from both sides:

$ 51 + 14x - 11x = 60 $

$ 51 + 3x = 60 $

Subtract 51 from both sides:

$ 3x = 60 - 51 $

$ 3x = 9 $

Divide by 3:

$ x = \frac{9}{3} $

$ x = 3 $

Verification

The value found is $ x = 3 $. This is a positive integer, satisfying the condition.

Let's check the fractions:

• First fraction: $ \frac{3+3}{4+3} = \frac{6}{7} $
• Second fraction: $ \frac{15-3}{17-3} = \frac{12}{14} = \frac{6}{7} $

Since both fractions simplify to $\frac{6}{7}$, the value $ x=3 $ is correct.