Simplify $\frac{\sqrt{507}}{\sqrt{48}}-\frac{\sqrt{175}}{\sqrt{112}}$.
We have:

$\frac{\sqrt{507}}{\sqrt{48}}-\frac{\sqrt{175}}{\sqrt{112}}=\frac{13\sqrt3}{4\sqrt3}-\frac{5\sqrt7}{4\sqrt7}=\frac{13}{4}-\frac54=\frac84=\boxed{2}$.
2
Level 5
Prealgebra
Problem: Simplify $\frac{\sqrt{507}}{\sqrt{48}}-\frac{\sqrt{175}}{\sqrt{112}}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first, we can rewrite the fractions using the property of square roots: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Thus, we have:

1. Rewrite the first term: $\frac{\sqrt{507}}{\sqrt{48}} = \sqrt{\frac{507}{48}}$.
2. Rewrite the second term: $\frac{\sqrt{175}}{\sqrt{112}} = \sqrt{\frac{175}{112}}$.

Next, we need to simplify the fractions inside the square roots:

3. For $\frac{507}{48}$, we can factor both numbers. The prime factorization of 507 is $3 \times 169 = 3 \times 13^2$. The prime factorization of 48 is $16 \times 3 = 4^2 \times 3$. Thus, $\frac{507}{48} = \frac{3 \times 13^2}{4^2 \times 3} = \frac{13^2}{16} = \left(\frac{13}{4}\right)^2$.
4. Therefore, $\sqrt{\frac{507}{48}} = \frac{13}{4}$.

5. For $\frac{175}{112}$, the prime factorization of 175 is $25 \times 7 = 5^2 \times 7$. The prime factorization of 112 is $16 \times 7 = 4^2 \times 7$. Thus, $\frac{175}{112} = \frac{5^2 \times 7}{4^2 \times 7} = \frac{25}{16} = \left(\frac{5}{4}\right)^2$.
6. Therefore, $\sqrt{\frac{175}{112}} = \frac{5}{4}$.

Now we can substitute these simplified terms back into the original expression:

7. The expression becomes $\frac{13}{4} - \frac{5}{4}$.
8. Since both fractions have a common denominator, we can combine them: $\frac{13 - 5}{4} = \frac{8}{4}$.

Finally, simplify the result:

9. $\frac{8}{4} = 2$.

Thus, the final answer is $2$.<|END_OF_LATENT|> The final answer is $2$.