The number of a fraction is 6x+1 and the denomination is 7-4x x can have any value between - 2 and 2, both included. The value of x for which the numerator is greater than the denominator, are

Options

A. `3/5 lt x le 2`
B. `3/5 le x le 2`
C. `0 lt x le 2`
D. `-2 le x lt 2`

Answer

To solve the problem step by step, we need to find the values of \( x \) for which the numerator \( 6x + 1 \) is greater than the denominator \( 7 - 4x \).

### Step 1: Set up the inequality
We start with the inequality:
\[ 6x + 1 > 7 - 4x \]

### Step 2: Rearrange the inequality
To solve for \( x \), we first move all terms involving \( x \) to one side and constant terms to the other side. We can do this by adding \( 4x \) to both sides and subtracting \( 1 \) from both sides:
\[ 6x + 4x > 7 - 1 \]
This simplifies to:
\[ 10x > 6 \]

### Step 3: Solve for \( x \)
Next, we divide both sides of the inequality by \( 10 \):
\[ x > \frac{6}{10} \]
This simplifies to:
\[ x > \frac{3}{5} \]

### Step 4: Consider the range of \( x \)
We are given that \( x \) can take any value between \(-2\) and \(2\), inclusive. Therefore, we need to combine this condition with the result from the inequality:
\[ \frac{3}{5} < x \leq 2 \]

### Step 5: Write the final answer
The values of \( x \) for which the numerator is greater than the denominator are:
\[ \frac{3}{5} < x \leq 2 \]

### Conclusion
Thus, the final answer is that \( x \) can take any value in the range:
\[ \frac{3}{5} < x \leq 2 \]