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anonymous

  • one year ago

how do i solve x = (2y-3)/(y+1) for y?

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  1. anonymous
    • one year ago
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    @campbell_st @mathmate @mathway @aaronq @Teddyiswatshecallsme

  2. anonymous
    • one year ago
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    @rishavraj @Nnesha @Owlcoffee @LazyBoy @Shalante

  3. anonymous
    • one year ago
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    i multiplied y+1 on both sides is that right

  4. anonymous
    • one year ago
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    not sure how to isolate y from there tho

  5. campbell_st
    • one year ago
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    if you multiply by the denominator you get \[x(y + 1) = 2y - 3\]

  6. anonymous
    • one year ago
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    i got that far @campbell_st

  7. campbell_st
    • one year ago
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    then distribute \[xy + x = 2y - 3\]

  8. campbell_st
    • one year ago
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    subtract 2y from both sides \[xy - 2y + x = -3\] factor and you get \[y(x - 2) + x = -3 \]

  9. campbell_st
    • one year ago
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    next subtract x \[y(x -2) = -x -3\] lastly divide both sides by (x - 2)

  10. anonymous
    • one year ago
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    so y = (-x-3) over (x-2)? @campbell_st

  11. campbell_st
    • one year ago
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    that's what I got

  12. Owlcoffee
    • one year ago
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    Whenever we are told to "solve for a variable" we mean doing the necessary operations to isolate it on any side of the "=" sign. So, on the equation: \[x = \frac{(2y-3) }{ (y+1) }\] We are asked to solve for "y" so we will do what is necessary to leave "y" on any side of the equality. It is also important to note that this equation is solved for "x", since "x" is isolated on the left side of the equation. "x" itself represents a fraction with denominator "1" so therefore, we can treat this equation as a proportion: \[\frac{ x }{ 1 }= \frac{(2y-3) }{ (y+1) }\] So we will cross multiply: \[x(y+1)=2y-3\] Applying distributive property: \[xy+x=2y-3\] This is the key step, we will transfer all the "y"s to the left side of the equation and obtain: \[xy-2y=-x-3\] Now, we will take common factor "y", since it is repeated in both terms of the left side: \[y(x-2)=-x-3\] If we divide both sides by (\(x+2\)) we will obtain: \[y=\frac{ -x-3 }{ x-2 }\]

  13. Owlcoffee
    • one year ago
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    (x-2) *

  14. anonymous
    • one year ago
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    thank you! i get it now @Owlcoffee

  15. Owlcoffee
    • one year ago
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    No problem.

  16. mathmate
    • one year ago
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    @Owlcoffee Whenever we divide an expression by another ( x-2, in this case), it is necessary to specify the condition that x-2\(\ne\)0 to be mathematically correct. This is because the results are invalid if we divide by zero.

  17. Owlcoffee
    • one year ago
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    That would be true if the relationship between the variables were functional, since we don't know that, we cannot specify any conditions.

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