Question

1. add $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}m$.

Explanation:

Step1: Identify like terms

We have two expressions: $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}m$. The like terms are the $n$ terms and the constant terms (but there's only one constant term in the first expression, and the second has a $m$ term which is unique). So we add the $n$ terms and the constant terms (and keep the $m$ term as is).

First, add the $n$ terms: $\frac{1}{3}n + \frac{1}{6}n$. To add these, find a common denominator, which is 6. So $\frac{1}{3}n = \frac{2}{6}n$. Then $\frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$.

Step2: Add the constant terms

Now add the constant terms: $\frac{2}{3} + 0$ (wait, no, the second expression has no constant term, only the first has $\frac{2}{3}$? Wait, no, the second expression is $\frac{1}{6}n + \frac{1}{6}m$, so the constant terms: first expression has $\frac{2}{3}$, second has 0? Wait, no, maybe I misread. Wait the problem is to add $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}m$. So the terms are:

- $n$ terms: $\frac{1}{3}n$ and $\frac{1}{6}n$
- $m$ term: $\frac{1}{6}m$ (only in the second expression)
- constant term: $\frac{2}{3}$ (only in the first expression)

So adding the $n$ terms: $\frac{1}{3}n + \frac{1}{6}n = \frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$.

Adding the constant term: $\frac{2}{3}$ (since the second expression has no constant term, we just keep it? Wait, no, the second expression is $\frac{1}{6}n + \frac{1}{6}m$, so when we add the two expressions:

$(\frac{1}{3}n + \frac{2}{3}) + (\frac{1}{6}n + \frac{1}{6}m)$

= $\frac{1}{3}n + \frac{1}{6}n + \frac{2}{3} + \frac{1}{6}m$

As we did before, $\frac{1}{3}n = \frac{2}{6}n$, so $\frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$.

Then the constant term is $\frac{2}{3}$, and the $m$ term is $\frac{1}{6}m$. Wait, but maybe I made a mistake. Wait, the problem says "Add $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}m$". So combining like terms:

For $n$: $\frac{1}{3}n + \frac{1}{6}n = \frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$

For $m$: $\frac{1}{6}m$ (only in the second expression, so we keep it)

For constants: $\frac{2}{3}$ (only in the first expression, so we keep it)

Wait, but maybe the problem was written as $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}$? Wait, the image shows " $\frac{1}{6}n + \frac{1}{6}m$"? Wait, maybe I misread the second term. Let me check again. The user's image: "Add $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}m$". So that's correct.

Wait, but maybe the problem is to add $\frac{1}{3}n + \frac{2}{3}$ and $\frac{1}{6}n + \frac{1}{6}$ (maybe a typo, $m$ instead of a constant). But assuming the image is correct, with $\frac{1}{6}m$, then the result is $\frac{1}{2}n + \frac{1}{6}m + \frac{2}{3}$.

But maybe I made a mistake. Wait, let's re-express all terms with common denominators.

First expression: $\frac{1}{3}n + \frac{2}{3} = \frac{2}{6}n + \frac{4}{6}$

Second expression: $\frac{1}{6}n + \frac{1}{6}m = \frac{1}{6}n + \frac{1}{6}m + 0$

Now add them:

$\frac{2}{6}n + \frac{4}{6} + \frac{1}{6}n + \frac{1}{6}m$

Combine like terms:

$n$ terms: $\frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$

$m$ term: $\frac{1}{6}m$

constant term: $\frac{4}{6} = \frac{2}{3}$

So the sum is $\frac{1}{2}n + \frac{1}{6}m + \frac{2}{3}$.

Wait, but maybe the original problem had a typo, and the second term is $\frac{1}{6}n + \frac{1}{6}$ (without $m$). Let's check that case too, in case of a typo.

If the second term is $\frac{1}{6}n + \frac{1}{6}$, then:

First expression: $\frac{1}{3}n + \frac{2}{3}$

Second expression: $\frac{1}{6}n + \frac{1}{6}$

Add $n$ terms: $\frac{1}{3}n + \frac{1}{6}n = \frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$

Add constant terms: $\frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}$

So the sum would be $\frac{1}{2}n + \frac{5}{6}$.

But since the image shows $\frac{1}{6}m$, we have to go with that. But maybe the user made a typo. However, based on the image, the second term has $\frac{1}{6}m$. So the final answer is $\frac{1}{2}n + \frac{1}{6}m + \frac{2}{3}$.

But let's confirm the steps again.

Step 1: Write the two expressions to add: $(\frac{1}{3}n + \frac{2}{3}) + (\frac{1}{6}n + \frac{1}{6}m)$

Step 2: Remove parentheses: $\frac{1}{3}n + \frac{2}{3} + \frac{1}{6}n + \frac{1}{6}m$

Step 3: Combine like terms for $n$: $\frac{1}{3}n + \frac{1}{6}n = \frac{2}{6}n + \frac{1}{6}n = \frac{3}{6}n = \frac{1}{2}n$

Step 4: The $m$ term remains as $\frac{1}{6}m$ (since no other $m$ term)

Step 5: The constant term remains as $\frac{2}{3}$ (since no other constant term)

So the sum is $\frac{1}{2}n + \frac{1}{6}m + \frac{2}{3}$.

Answer:

$\frac{1}{2}n + \frac{1}{6}m + \frac{2}{3}$ (or if there was a typo and the second term is $\frac{1}{6}n + \frac{1}{6}$, then $\frac{1}{2}n + \frac{5}{6}$)