NotThatSmart

The wording of this problem is weird, but I think it's basically asking to find the total amount of students. 

Because there are 3 students in math club, science club has double that with 6!

Ok, so now we know there are 3 kids in math club, 3 kids in both, and 6 kids in science. Are we done? 

The answer is ofcourse not. We have to subtract the duplicates, which is the kids in both. We subtract 3 from both clubs so that we now have kids ONLY in math and ONLY in science. The kids ONLY in math are 0, and the kids ONLY in science is 3. 

$3+3+0=6$

So there are 6 students in total!

I hoped I answered your question! 

Thanks! :) :) :)

Ahhh, I see. I didn't think about using the two equations we make and combining them in the end. 

I was more focued on setting a variable, which I see is not needed. 

Thanks CPhill, I understand how to solve this problem now!

For this problem, you would have to set the amount of money as a variable. 

They key is to essentially write a solvable equation to figure it out!

In fact, my parents used to give me a lot of problems similar to this one :)

So let's let that be x. 

From the problem, we can get the equation $\frac{2}{5}x+\frac{1}{7}(x-360)+360=x$. 

We have $\frac{2}{5}x + \frac{1}{7}x - \frac{360}{7}+360 = x$. Combining like terms and isolating x, we have $\frac{2160}{7}=\frac{16}{35}x$. 

We multiply both sides by 35/16, and we have $\frac{35}{16} \cdot \frac{2160}{7} = x$. 

So we have $x = 675$

$675!

Thanks!

I don't think it is possible to achieve a real number from this function. 

This is because $\sqrt{2}$ is greater than 1, so in order for it to be real, we would need $\sqrt{3+\sqrt{x}}$ to be negative, which obviously is not possible...

I think the domain is essentially just $[0, \infty)$

Thanks!

Haha, this is a very funny problem and shows the world of fractions. 

a) When x = 10, we have $\frac{2(10)+3}{2(10)+3} = \frac{23}{23} = 1$. So yes, when x = 10, we have the first expression is equal to the second expression. 

b) Ok, so then why doesn't $-\frac{3}{2}$ work? It looks like the numerator and denominator are equal, so it's always 1 right? 

Well, let's simplify and go from there. We have $\frac{2(-2/3)+3}{2(-2/3)+3}=\frac{-3+3}{-3 + 3}=0/0$. WAIT a minute, the denominator can't be 0! That would make the value undefined, not 1!

c) Ok, so the reaon why the answer is not 1 in part (b) is because the denominator would become 0. 

However, as long as it doesn't equal 0, the numerator and denominator will always be equal, meaning the value of the fraction will always be 1!

I hope I answered your question!

Thanks!