Question

Solve the rational equation. Express numbers as integers or simplified fractions.
4/3y^2-12 -6/y+2+1/y-2=0

Answer 1

i guess it is something like this ' cause its not clear in the way u have type it ..
\[\frac{ 4 }{ 3y^{2} - 12 } - \frac{ 6 }{ y + 2} + \frac{ 1 }{ y - 2 }\]
now u can write
\[\frac{ 4 }{ 3y^{2} - 12 } = \frac{ 4 }{ 3( y^{2} - 4) }\]
now using the difference of 2 squares
\[\frac{ 4 }{ 3( y^{2} - 4) } = \frac{ 4 }{ 3( y - 2) ( y+ 2) }\]
so... u can write ur question in this way now
\[\frac{ 4 }{ 3( y- 2)(2 + 2) } - \frac{ 6 }{ y + 2 } + \frac{ 1 }{ y - 2 }\]
now take a common denominator... i h think u know how to find it...
here it's
\[3(y - 2)( y +2)\]
now write the whole question as a single fraction
\[\frac{ 4 - 6[3(y-2)] + 1[3(y +2) }{ 3(y+2)(y-2) }\]
which is
\[\frac{ 4 + 18(y-2) + 3(y+2) }{ 3(y-2)(y+2) } = 0\]
now u can move the denominator to the other side. since the other side is 0 the whole term will become zero , giving u
\[4 + 18(y - 2) + 3(y +2) = 0\]

now .. solve this for "y" .. i think u can do that :)

Answer 2

Would it be 4+18y-36+3y+6=0?

Answer 3

yep :)

Answer 4

Would it be -26=21y?

Answer 5

yep :)