Re: rearranging an equation

Solution :- Given y(S^4 + 3S^2 -12) = (104/s)+3 -24 +12S^3 - 3S^2 + 25S -15

s(S^4+3S^2-12)Y=104+s(12S^3-3S^2+25S-36)

because s(S^4+35^2-12)?0

Y=(12sS^3-3sS^2+25sS-36s+104)/s(S^4+3S^2-12)

Re: rearranging an equation

Please clarify the right hand side of yuor equation. What you wrote is this:

$\displaystyle \frac {104} {S} + 3 -24 +12S^3-3S^2 + 25S - 15$

But I suspect what you meant is this:

$\displaystyle \frac {104} {S + 3} -24 +12S^3-3S^2 + 25S - 15$

Why didn't you combine the -24 and -15 terms?

One option is to multiply both sides through by (S+3), then divide both sides by the Y coeffeicient:

$\displaystyle \small y(S^4 + 3S^2 -12)(S+3) = (\frac {104}{s+3} -24 +12S^3 - 3S^2 + 25S -15)(s+3)$

$\displaystyle \small y(S^5+3S^4+3S^3+9S^2-12S-36) = 104 -39(S+3)+12S^3(S+3)-3S^2(S+3)+25S(S+3)$

$\displaystyle \small y= \frac {(12S^4+33S^3+16S^2+10S+36S+179)}{(S^5+3S^4+3S^3+9 S^2-12S-36)}$