What integer $x$ satisfies $\frac{1}{4}<\frac{x}{7}<\frac{1}{3}$?
One way to do this is to find the common denominator of the three fractions and compare the numerators. The least common multiple of 4, 7, and 3 is 84, so I can multiply each fraction by $\frac{84}{84}$ to get equivalent fractions with the same denominator. This gives me $\frac{21}{84}<\frac{12x}{84}<\frac{28}{84}$. Now I can see that I need to find an integer $x$ that makes $21<12x<28$. To simplify this inequality, I can divide everything by 12, which gives me $\frac{21}{12}<x<\frac{28}{12}$. I can reduce these fractions to lowest terms by dividing the numerator and denominator by their greatest common factor. For $\frac{21}{12}$, the GCF is 3, so I get $\frac{7}{4}$. For $\frac{28}{12}$, the GCF is 4, so I get $\frac{7}{3}$. Now the inequality is $\frac{7}{4}<x<\frac{7}{3}$. I can see that the only integer that satisfies this inequality is 2, since it is the only integer between $\frac{7}{4}$ and $\frac{7}{3}$. To check, I can plug in 2 for $x$ and see if the original inequality is true. This gives me $\frac{1}{4}<\frac{2}{7}<\frac{1}{3}$, which is true. Therefore, the integer $x$ that satisfies the given inequality is 2. 