Figuring out three variables based on two givens

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I've got two sets of data



(x * 80000) / y = 103800


and



(z * 80000) / y = 60700


the same x, y, and z values would need to work for



(x * 27000) / y = 28000

(z * 27000) / y = 16350


The values on the right hand side are approximates, but I'm looking to figure out how I would solve for x, y and z to satisfy these conditions




calculus




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  • $x=y=z=0$. All of your equations are linear functions through the origins.
    – Karn Watcharasupat
    12 hours ago






  • 1




    The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
    – lulu
    11 hours ago














up vote
0
down vote

favorite
1












I've got two sets of data



(x * 80000) / y = 103800


and



(z * 80000) / y = 60700


the same x, y, and z values would need to work for



(x * 27000) / y = 28000

(z * 27000) / y = 16350


The values on the right hand side are approximates, but I'm looking to figure out how I would solve for x, y and z to satisfy these conditions




calculus




share|cite|improve this question



















  • $x=y=z=0$. All of your equations are linear functions through the origins.
    – Karn Watcharasupat
    12 hours ago






  • 1




    The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
    – lulu
    11 hours ago












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I've got two sets of data



(x * 80000) / y = 103800


and



(z * 80000) / y = 60700


the same x, y, and z values would need to work for



(x * 27000) / y = 28000

(z * 27000) / y = 16350


The values on the right hand side are approximates, but I'm looking to figure out how I would solve for x, y and z to satisfy these conditions




calculus




share|cite|improve this question











I've got two sets of data



(x * 80000) / y = 103800


and



(z * 80000) / y = 60700


the same x, y, and z values would need to work for



(x * 27000) / y = 28000

(z * 27000) / y = 16350


The values on the right hand side are approximates, but I'm looking to figure out how I would solve for x, y and z to satisfy these conditions





calculus





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 12 hours ago









adam3039

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  • $x=y=z=0$. All of your equations are linear functions through the origins.
    – Karn Watcharasupat
    12 hours ago






  • 1




    The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
    – lulu
    11 hours ago
















  • $x=y=z=0$. All of your equations are linear functions through the origins.
    – Karn Watcharasupat
    12 hours ago






  • 1




    The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
    – lulu
    11 hours ago















$x=y=z=0$. All of your equations are linear functions through the origins.
– Karn Watcharasupat
12 hours ago




$x=y=z=0$. All of your equations are linear functions through the origins.
– Karn Watcharasupat
12 hours ago




1




1




The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
– lulu
11 hours ago




The first equation in $x,y$ tells you that $x=frac 10380080000y=frac 519400yapprox 1.2975 y$. The second tells you that $x=frac 2800027000=frac 2827yapprox 1.037y$. Those are not really compatible.
– lulu
11 hours ago















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