Question

how you solve for y? lol
x=(2y+3)/(3-2y)

Answer 1

@helder_edwin

Answer 2

well u have
\[ \large x=\frac{2y+3}{3-2y} \]
what is dividing on one side changes sides to multiply so
\[ \large x(3-2y)=2y+3 \]

Answer 3

now distribute
\[ \large 3x-2xy=2y+3 \]
\[ \large 3x-3=2y+2xy=y(2+2x) \]
therefore
\[ \large y=\frac{3x-3}{2+2x} \]

Answer 4

nice but in 2y+2xy can't you factor 2y? why did you factor the y only :D

Answer 5

because u r solving for y. u could factor 3 in the numerator and 2 in the denominator but u gain nothing.

Answer 6

oooh thank you very much :d