Let's denote the original number as $\frac{x}{x+6}$, where:

• The numerator is $x$,
• The denominator is 6 more than the numerator, so it's $x + 6$.

According to the problem, subtracting the number from $1$ gives $\frac{1}{4}$, so we can set up the equation:

$1 - \frac{x}{x+6} = \frac{1}{4}$

Step 1: Solve for $\frac{x}{x+6}$

Subtract $\frac{x}{x+6}$ from 1:

$\frac{(x+6) - x}{x+6} = \frac{1}{x+6}$

This simplifies to:

$\frac{1}{x+6}$

So, the equation becomes:

$\frac{1}{x+6} = \frac{1}{4}$

Step 2: Solve for $x$

Now, set the denominators equal to each other:

$x + 6 = 4$

Solve for $x$:

$x = 4 - 6 = -2$

Step 3: Verify the solution

The original number is $\frac{-2}{-2 + 6} = \frac{-2}{4} = -\frac{1}{2}$.

Let's check if the condition holds: $1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$ And: $\frac{3}{2} \neq \frac{1}{4}$

It appears there was an error, let me rework the steps and find the correct result.

Reattempt

Let the original number be $x$. Then we are told:

$1 - x = \frac{1}{4}$

So, subtracting $x$ from 1 gives $\frac{1}{4}$, which leads to:

$x = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$

Thus, the original number is $\frac{3}{4}$.

Would you like more details or clarifications on any step? Here are some related questions for further practice:

1. If the denominator is 8 more than the numerator, and the difference from 1 is $\frac{1}{5}$, what is the number?
2. Solve for $x$ if $\frac{x}{x+5} = \frac{2}{3}$.
3. What is the value of $x$ if the denominator is twice the numerator, and subtracting from 1 results in $\frac{1}{6}$?
4. Find a fraction where the denominator is 4 more than the numerator and the sum of the fraction and $\frac{1}{3}$ is $1$.
5. If subtracting a fraction from 1 gives $\frac{2}{7}$, and the denominator is 5 more than the numerator, what is the fraction?

Tip: Always ensure to verify solutions by substituting the values back into the original problem.