What is the answer to solvefor y,(x^2)/3+(x^4)/6+x 8/12+16y=28 ?

The answer to solvefor y,(x^2)/3+(x^4)/6+x 8/12+16y=28 is y=(168+2(-x^2-2x)-x^4)/(96)

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solvefor y,(x^2)/3+(x^4)/6+x 8/12+16y=28

solve for

y, \frac{x ^{2}}{3} + \frac{x ^{4}}{6} + x \frac{8}{12} + 16 y = 28

y = \frac{168 + 2 ( - x ^{2} - 2 x ) - x ^{4}}{96}

Solution steps

\frac{x ^{2}}{3} + \frac{x ^{4}}{6} + x \frac{8}{12} + 16 y = 28

x \frac{8}{12} = \frac{2 x}{3}

\frac{x ^{2}}{3} + \frac{x ^{4}}{6} + \frac{2 x}{3} + 16 y = 28

Move

\frac{x ^{2}}{3}

to the right side

\frac{x ^{4}}{6} + \frac{2 x}{3} + 16 y = 28 - \frac{x ^{2}}{3}

Move

\frac{x ^{4}}{6}

to the right side

\frac{2 x}{3} + 16 y = 28 - \frac{x ^{2}}{3} - \frac{x ^{4}}{6}

Move

\frac{2 x}{3}

to the right side

16 y = 28 + \frac{- x ^{2} - 2 x}{3} - \frac{x ^{4}}{6}

Divide both sides by

16

y = \frac{168 + 2 ( - x ^{2} - 2 x ) - x ^{4}}{96}

6 y - 9 = 2 y + 15

\frac{( \frac{5}{13} )}{( 1 - ( \frac{5}{13} ) ^{2} )} = 3

x, 10 y - 3 = 5 x + 2

\frac{p - ( 3 p - 2 )}{3} = 2 + (p - 4)

3 x + \frac{1}{2} + \frac{a}{4} = 2 x

What is the answer to solvefor y,(x^2)/3+(x^4)/6+x 8/12+16y=28 ?
The answer to solvefor y,(x^2)/3+(x^4)/6+x 8/12+16y=28 is y=(168+2(-x^2-2x)-x^4)/(96)