I see you took my advice literally about carrying extra places. It isn't necessary to use ALL of those numbers that show up on your calculator. I think 1.5377X + 1.8447Y = 0.944 will be quite sufficient and I suspect even one number fewer still will work ok. As for making the coefficients equal, I think that is the hard way of doing it although it can be done. I would solve equation 1 for either X or Y and substitute into the other one. For example,

Question

I see you took my advice literally about carrying extra places. It isn't necessary to use ALL of those numbers that show up on your calculator. I think 1.5377X + 1.8447Y = 0.944 will be quite sufficient and I suspect even one number fewer still will work ok. As for making the coefficients equal, I think that is the hard way of doing it although it can be done. I would solve equation 1 for either X or Y and substitute into the other one. For example,

1.5377X + 1.8447Y = 0.944
X = (0.944-1.8447Y)/1.5377
Now substitute this value of X into equation 2 and solve. That will get a Y value which can be substituted into equation 1 to obtain X.

sorry, that was stupid for me to write all those places out. i just wasn't sure how far to take the decimal places out in order for it to be accurate. i'm so used to 3 sig figs.

this is what i have but i think its wrong because i got a negative value for x.

Answer 1

1.5748(0.944-1.8447Y/1.5377) + 1.90293Y = 0.953

-.4026 Y + 1.90293 Y = 0.953
1.50033 Y = 0.953
Y=.6352

1.5377 X +1.8447(.6352) = 0.944
1.5377 X +1.1717 = 0.944
1.5377 X = -.2277
X=-.148078

Answer 2

I forgot some of the question and had to go back to the original and find the problem. I posted a response at the initial question but here is a link, I think, which will find it for you. I just didn't want to type the response again.

http://www.jiskha.com/display.cgi?id=1201639160.1201669433

Answer 3

It's totally fine! It's always good to be thorough when working with calculations. Regarding your concern about the number of decimal places, it depends on the level of accuracy you require for your problem. In this case, you can keep as many decimal places as needed to maintain accuracy.

Now, let's address your equation and the negative value you obtained for X. You have the system of equations:

1.5377X + 1.8447Y = 0.944
X = (0.944 - 1.8447Y)/1.5377

To solve this system, you can substitute the expression for X from the second equation into the first equation:

1.5377((0.944 - 1.8447Y)/1.5377) + 1.8447Y = 0.944

Simplifying, you get:

0.944 - 1.8447Y + 1.8447Y = 0.944

The Y term cancels out, leaving you with:

0.944 = 0.944

This equation is true, which means that Y can have any value. There is no unique solution for X and Y in this case; it is an equation for a line, and any pair of X and Y that satisfies the equation will be a solution. Therefore, you did not make any mistake in obtaining a negative value for X.

In summary, you correctly solved the system of equations, and finding a negative value for X is completely valid in this case.

1.5748(0.944-1.8447Y/1.5377) + 1.90293Y = 0.953

I forgot some of the question and had to go back to the original and find the problem. I posted a response at the initial question but here is a link, I think, which will find it for you. I just didn't want to type the response again.

thanks!

It's totally fine! It's always good to be thorough when working with calculations. Regarding your concern about the number of decimal places, it depends on the level of accuracy you require for your problem. In this case, you can keep as many decimal places as needed to maintain accuracy.