anonymous
  • anonymous
how do i solve x = (2y-3)/(y+1) for y?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@campbell_st @mathmate @mathway @aaronq @Teddyiswatshecallsme
anonymous
  • anonymous
@rishavraj @Nnesha @Owlcoffee @LazyBoy @Shalante
anonymous
  • anonymous
i multiplied y+1 on both sides is that right

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
not sure how to isolate y from there tho
campbell_st
  • campbell_st
if you multiply by the denominator you get \[x(y + 1) = 2y - 3\]
anonymous
  • anonymous
i got that far @campbell_st
campbell_st
  • campbell_st
then distribute \[xy + x = 2y - 3\]
campbell_st
  • campbell_st
subtract 2y from both sides \[xy - 2y + x = -3\] factor and you get \[y(x - 2) + x = -3 \]
campbell_st
  • campbell_st
next subtract x \[y(x -2) = -x -3\] lastly divide both sides by (x - 2)
anonymous
  • anonymous
so y = (-x-3) over (x-2)? @campbell_st
campbell_st
  • campbell_st
that's what I got
Owlcoffee
  • Owlcoffee
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: \[y=\frac{ -x-3 }{ x-2 }\]
Owlcoffee
  • Owlcoffee
(x-2) *
anonymous
  • anonymous
thank you! i get it now @Owlcoffee
Owlcoffee
  • Owlcoffee
No problem.
mathmate
  • mathmate
@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.
Owlcoffee
  • Owlcoffee
That would be true if the relationship between the variables were functional, since we don't know that, we cannot specify any conditions.

Looking for something else?

Not the answer you are looking for? Search for more explanations.