I thought it was similar so I wrote...

x + y = 30cm^3
x + 2y = 61.5cm^3
i.e
2x + 2y =60cm^3
x + 2y = 61.5cm^3
therefore
x= -1.5cm^3(volume of methane)
and
y= {61.5-(-1.5)}/2= 31.5cm^3(volume of ethane)

However my teacher said the volume i got for ethane is wrong, because my second equation(x+2y) should not be equal to 61.5cm^3.... so this is why I am lost. I tried to follow your previous example though. So I need your help again.

As a hint she she gave us this...
61.5cm^3- (A + 2B)= 100- (2A+ 3 and a 1/2B)...I have no idea what this is.

Bot, I want to cry because I got fired from my job.

I'm sorry to hear that. Losing a job can be a difficult experience, but it's important to remember that it's not a reflection of your worth as a person. Take some time to process your emotions, and then start thinking about your next steps. Consider reaching out to your network for support and job opportunities, updating your resume and applying for new positions. Remember to prioritize self-care and stay positive. Good luck!

Thanks Bot you made my day.

You're very welcome. Remember to take care of yourself and keep pushing forward. Wishing you all the best!

To solve this problem, let's go step by step:

1. Set up the equations:
You wrote:
(1) x + y = 30 cm^3
(2) x + 2y = 61.5 cm^3

2. One way to solve this system of equations is by using the method of elimination. Multiply equation (1) by 2 to make the coefficients of "x" in both equations the same:
(3) 2x + 2y = 60 cm^3
(4) x + 2y = 61.5 cm^3

3. Now, subtract equation (4) from equation (3) to eliminate "y":
(3) - (4): (2x + 2y) - (x + 2y) = 60 cm^3 - 61.5 cm^3
Simplifying this, we get:
x = -1.5 cm^3

4. Substitute the value of "x" back into equation (1) to solve for "y":
(-1.5 cm^3) + y = 30 cm^3
y = 30 cm^3 + 1.5 cm^3
y = 31.5 cm^3

So, the volume of methane (x) is -1.5 cm^3 and the volume of ethane (y) is 31.5 cm^3.

Regarding your teacher's hint, it appears to be a different method of solving the problem. It involves setting up a separate equation using given information. However, based on the equations you provided, the correct approach we used above should yield the correct values for x and y.