Question

add.
\\(\frac{7y + 1}{4y} + \frac{9y - 5}{4y}\\)
simplify your answer as much as possible.

Explanation:

Step1: Add the numerators (common denominator)

Since the denominators are the same ($4y$), we add the numerators: $(7y + 1)+(9y - 5)$.

Step2: Combine like terms in numerator

Simplify the numerator: $7y+9y + 1-5=16y - 4$.

Step3: Simplify the fraction

Now we have $\frac{16y - 4}{4y}$. Factor out 4 from the numerator: $\frac{4(4y - 1)}{4y}$. Cancel the common factor of 4: $\frac{4y - 1}{y}$. We can also split this into $\frac{4y}{y}-\frac{1}{y}=4-\frac{1}{y}$, but $\frac{4y - 1}{y}$ is also simplified. Alternatively, further simplifying $\frac{16y - 4}{4y}=\frac{4(4y - 1)}{4y}=\frac{4y - 1}{y}=4-\frac{1}{y}$, but the simplified form can also be written as $\frac{16y - 4}{4y}=\frac{4(4y - 1)}{4y}=\frac{4y - 1}{y}$ or $4-\frac{1}{y}$. However, let's check the addition again. Wait, $(7y + 1)+(9y - 5)=16y - 4$, then divide by $4y$: $\frac{16y - 4}{4y}=\frac{4(4y - 1)}{4y}=\frac{4y - 1}{y}$. Alternatively, we can factor numerator and denominator: $\frac{16y - 4}{4y}=\frac{4(4y - 1)}{4y}=\frac{4y - 1}{y}$. Another way: $\frac{16y}{4y}-\frac{4}{4y}=4-\frac{1}{y}$. Both are correct, but let's see which is simpler. Let's do the addition of fractions properly:

$\frac{7y + 1}{4y}+\frac{9y - 5}{4y}=\frac{(7y + 1)+(9y - 5)}{4y}=\frac{16y - 4}{4y}$. Now, factor numerator and denominator: numerator $16y - 4 = 4(4y - 1)$, denominator $4y = 4\times y$. Cancel 4: $\frac{4y - 1}{y}$. Which can be written as $4-\frac{1}{y}$ or $\frac{4y - 1}{y}$.

Answer:

$\frac{4y - 1}{y}$ (or $4 - \frac{1}{y}$)