# Difference between revisions of "1958 AHSME Problems/Problem 34"

## Problem

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

$\textbf{(A)}\ \frac{3}{5} < x \le 2\qquad \textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad \textbf{(C)}\ 0 < x \le 2\qquad \\ \textbf{(D)}\ 0 \le x \le 2\qquad \textbf{(E)}\ -2 \le x \le 2$

## Solution

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