Asked by Catie
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.
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.
Answers
Answered by
Catie
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
-.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
Answered by
DrBob222
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
http://www.jiskha.com/display.cgi?id=1201639160.1201669433
Answered by
Catie
thanks!
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