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5. determine which of the the right-hand side of a、\\(\frac{2}{3} = \fr…

Question

  1. determine which of the the right-hand side of a、\\(\frac{2}{3} = \frac{4}{9}\\) c \\(\frac{3}{5} = \frac{6}{15}\\)

Explanation:

Response

To determine if the fractions are equal, we can use cross - multiplication. The rule is: for two fractions $\frac{a}{b}$ and $\frac{c}{d}$, if $a\times d = b\times c$, then the fractions are equal.

For option a: $\frac{2}{3}=\frac{4}{9}$

Step 1: Cross - multiply

We calculate the product of the numerator of the first fraction and the denominator of the second fraction, and the product of the denominator of the first fraction and the numerator of the second fraction.
For $\frac{2}{3}$ and $\frac{4}{9}$, we have $2\times9 = 18$ and $3\times4=12$.
Since $18
eq12$, the two fractions $\frac{2}{3}$ and $\frac{4}{9}$ are not equal.

For option c: $\frac{3}{5}=\frac{6}{15}$

Step 1: Cross - multiply

For the fractions $\frac{3}{5}$ and $\frac{6}{15}$, we calculate $3\times15$ and $5\times6$.
$3\times15 = 45$ and $5\times6 = 30$? Wait, no, wait. Wait, $\frac{6}{15}$ can be simplified. $\frac{6\div3}{15\div3}=\frac{2}{5}$, and $\frac{3}{5}
eq\frac{2}{5}$. Wait, maybe I made a mistake in cross - multiplication. Wait, $3\times15 = 45$, $5\times6=30$, $45
eq30$, so $\frac{3}{5}
eq\frac{6}{15}$.

Wait, maybe the original problem is to find which pair of fractions is equal? But according to the calculation, neither a nor c is equal. But maybe there are other options not shown. If we assume that we made a mistake in option c: Wait, $\frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15}$, and $\frac{6}{15}$ is different from $\frac{9}{15}$. So both a and c are not equal. But if we re - check option a: $\frac{2}{3}=\frac{6}{9}$, and $\frac{4}{9}$ is less than $\frac{6}{9}$, so $\frac{2}{3}>\frac{4}{9}$. For option c: $\frac{3}{5}=\frac{9}{15}$, $\frac{6}{15}$ is less than $\frac{9}{15}$, so $\frac{3}{5}>\frac{6}{15}$.

If the question is to find which of the given fractions (a or c) is equal, then neither is equal. But maybe there is a typo. If we consider $\frac{2}{3}=\frac{6}{9}$ (if the second fraction in a was $\frac{6}{9}$ instead of $\frac{4}{9}$), but based on the given problem, for the given a ($\frac{2}{3}=\frac{4}{9}$) and c ($\frac{3}{5}=\frac{6}{15}$), both are not equal.

But if we have to choose between a and c, and maybe there is a mistake in my calculation. Let's re - calculate for c: $\frac{3}{5}$ and $\frac{6}{15}$. $\frac{6}{15}$ simplifies to $\frac{2}{5}$, and $\frac{3}{5}
eq\frac{2}{5}$. For a: $\frac{2}{3}\approx0.6667$, $\frac{4}{9}\approx0.4444$, so they are not equal.

If the question is to determine which of the fractions on the left - hand side is greater than the one on the right - hand side:

For a: $\frac{2}{3}$ vs $\frac{4}{9}$. We can find a common denominator, which is 9. $\frac{2}{3}=\frac{2\times3}{3\times3}=\frac{6}{9}$. Since $\frac{6}{9}>\frac{4}{9}$, so $\frac{2}{3}>\frac{4}{9}$.

For c: $\frac{3}{5}$ vs $\frac{6}{15}$. We can find a common denominator, which is 15. $\frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15}$. Since $\frac{9}{15}>\frac{6}{15}$, so $\frac{3}{5}>\frac{6}{15}$.

If we assume the question is to find which of the two (a or c) has the left - hand side greater than the right - hand side, both a and c satisfy this. But if we have to choose one, maybe a or c. But based on the calculation, for a: $\frac{2}{3}>\frac{4}{9}$ and for c: $\frac{3}{5}>\frac{6}{15}$.

If we take the first one (a) as an example of showing the process:

Step 1: Find a common denominator

To compare $\frac{2}{3}$ and $\frac{4}{9}$, we find a common denominator. The least common denominator of 3 and 9 is 9.

Step 2: Rewrite the fractions

Rewrite $\frac{2}{3}$ with denominator 9: $\frac{2\times3}{3\times3}=\frac{6}{9}$.

Step 3: Compare the fractions

Now we compare $\frac{6}{9}$ and $\frac{4}{9}$. Since the denominators are the same, we compare the numerators. $6 > 4$, so $\frac{6}{9}>\frac{4}{9}$, which means $\frac{2}{3}>\frac{4}{9}$.

For the fraction in option c:

Step 1: Find a common denominator

To compare $\frac{3}{5}$ and $\frac{6}{15}$, we find a common denominator. The least common denominator of 5 and 15 is 15.

Step 2: Rewrite the fractions

Rewrite $\frac{3}{5}$ with denominator 15: $\frac{3\times3}{5\times3}=\frac{9}{15}$.

Step 3: Compare the fractions

Now we compare $\frac{9}{15}$ and $\frac{6}{15}$. Since the denominators are the same, we compare the numerators. $9>6$, so $\frac{9}{15}>\frac{6}{15}$, which means $\frac{3}{5}>\frac{6}{15}$.

Answer:

Step 1: Find a common denominator

To compare $\frac{3}{5}$ and $\frac{6}{15}$, we find a common denominator. The least common denominator of 5 and 15 is 15.

Step 2: Rewrite the fractions

Rewrite $\frac{3}{5}$ with denominator 15: $\frac{3\times3}{5\times3}=\frac{9}{15}$.

Step 3: Compare the fractions

Now we compare $\frac{9}{15}$ and $\frac{6}{15}$. Since the denominators are the same, we compare the numerators. $9>6$, so $\frac{9}{15}>\frac{6}{15}$, which means $\frac{3}{5}>\frac{6}{15}$.