Question

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

Answer 1

@campbell_st @mathmate @mathway @aaronq @Teddyiswatshecallsme

Answer 2

@rishavraj @Nnesha @Owlcoffee @LazyBoy @Shalante

Answer 3

i multiplied y+1 on both sides is that right

Answer 4

not sure how to isolate y from there tho

Answer 5

if you multiply by the denominator you get

\[x(y + 1) = 2y - 3\]

Answer 6

i got that far @campbell_st

Answer 7

then distribute

\[xy + x = 2y - 3\]

Answer 8

subtract 2y from both sides

\[xy - 2y + x = -3\]

factor and you get

\[y(x - 2) + x = -3 \]

Answer 9

next subtract x

\[y(x -2) = -x -3\]

lastly divide both sides by (x - 2)

Answer 10

so y = (-x-3) over (x-2)?
@campbell_st

Answer 11

that's what I got

Answer 12

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:
\[y=\frac{ -x-3 }{ x-2 }\]

Answer 13

(x-2) *

Answer 14

thank you! i get it now @Owlcoffee

Answer 15

No problem.

Answer 16

@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.

Answer 17

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