# Thread: How do I simplify this?

1. ## How do I simplify this?

Hi folks!

I am not sure of what to do next here:

Numerator: $% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaIXaaaba GaaG
% OmaaaacaWG4bWaaWbaaSqabeaacqGHsislcaaIXaGaai4laiaa ikda
% aaGccaWG5bWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccq GHsi
% slcaGGBbGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG 4bWa
% aWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccaWG5bWaaWbaaS qabe
% eyOeI0IaamiEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaa Gcba
% GaamyEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaaaaaOGa ayjk
% aiaawMcaaiaac2faaaa!4E3E!
${ - \frac{1} {2}x^{ - 1/2} y^{1/2} - [ - \frac{1} {2}x^{1/2} y^{ - 1/2} \left( {\frac{{ - x^{1/2} }} {{y^{1/2} }}} \right)]}$
$

All over this denominator: $% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaWaaeWaaeaacaWG5bWaaWbaaSqabe aaca
% ikdaaaaaaa!33E3!
${\left( {y^{1/2} } \right)^2 }$
$

(too much latex for the post) How should I approach this? Is there something I should be checking for first? Here's the answer, but I can't figure out how to get from the above to the answer:

$% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHsislcaaIXaaaba GaaG
% baaaaaaa!338F!
$\frac{{ - 1}} {{2y^2 \sqrt x }}$
$

2. Well, for simplicity note a= $x^\frac{1}{2}$ and b= $y^\frac{1}{2}$. So you have: $\frac{-\frac{1}{2}a^{-1}b-\frac{1}{2}ab^{-1}\cdot \frac{a}{b}}{b^2}$

Use that $a^{-1}=\frac{1}{a}$. Hope you can solve it now.

Edit: You sure wrote that correctly/ completely? Because $a^3+b^3$ should be equal with 1 which is not necessary true.

3. Originally Posted by kungfumonkey
Hi folks!

I am not sure of what to do next here:

Numerator: $% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaIXaaaba GaaG
% OmaaaacaWG4bWaaWbaaSqabeaacqGHsislcaaIXaGaai4laiaa ikda
% aaGccaWG5bWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccq GHsi
% slcaGGBbGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG 4bWa
% aWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccaWG5bWaaWbaaS qabe
% eyOeI0IaamiEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaa Gcba
% GaamyEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaaaaaOGa ayjk
% aiaawMcaaiaac2faaaa!4E3E!
${ - \frac{1} {2}x^{ - 1/2} y^{1/2} - [ - \frac{1} {2}x^{1/2} y^{ - 1/2} \left( {\frac{{ - x^{1/2} }} {{y^{1/2} }}} \right)]}$
$

All over this denominator: $% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaWaaeWaaeaacaWG5bWaaWbaaSqabe aaca
% ikdaaaaaaa!33E3!
${\left( {y^{1/2} } \right)^2 }$
$

(too much latex for the post) How should I approach this? Is there something I should be checking for first? Here's the answer, but I can't figure out how to get from the above to the answer:

$% MathType!MTEF!2!1!+-
% faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% aiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHsislcaaIXaaaba GaaG
% baaaaaaa!338F!
$\frac{{ - 1}} {{2y^2 \sqrt x }}$
$
is this what you wrote there ?

$\displaystyle \frac{-\frac {\sqrt{y}}{2\sqrt{x}} - \big{[} -\frac {\sqrt{x}}{2\sqrt{y}}\cdot (\frac{-\sqrt{x}}{\sqrt{y}})\big{]}}{y}$

4. Hello, kungfumonkey!

$\displaystyle \text{Simplify: }\;\frac{-\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}} - \bigg[-\frac{1}{2}x^{\frac{1}{2}}y^{-\frac{1}{2}}\left(-\frac{x^{\frac{1}{2}}}{y^{\frac{1}{2}}}\right)\big g]}{\left(y^{\frac{1}{2}}\right)^2}$

$\text{Answer: }\:\dfrac{ - 1}{2y^2 \sqrt x }$ . How?

$\displaystyle \text{We have: }\;\frac{-\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}}- \frac{1}{2}xy^{-1}}{y} \;\;=\;\;-\tfrac{1}{2}x^{-\frac{1}{2}}y^{-\frac{1}{2}} - \tfrac{1}{2}xy^{-2}$

. . . . $\displaystyle =\;-\tfrac{1}{2}\left(\frac{1}{x^{\frac{1}{2}}y^{\frac {1}{2}}} + \frac{x}{y^2}\right) \;\;=\;\;-\tfrac{1}{2}\left(\frac{1}{x^{\frac{1}{2}}y^{\frac {1}{2}}}\cdot\frac{y^{\frac{3}{2}}}{y^{\frac{3}{2} }} + \frac{x}{y^2}\cdot\frac{x^{\frac{1}{2}}}{x^{\frac{ 1}{2}}}\right)$

. . . . $\displaystyle =\;\;-\tfrac{1}{2}\left(\frac{y^{\frac{3}{2}}}{x^{\frac{ 1}{2}}y^2} + \frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}y^2}\right) \;\;=\;\;-\tfrac{1}{2}\left(\frac{x^{\frac{3}{2}} + y^{\frac{3}{2}}}{x^{\frac{1}{2}}y^2}\right)$

. . . . $\displaystyle =\;\;-\frac{x\sqrt{x} + y\sqrt{y}}{2y^2\sqrt{x}}$

As veileen pointsd out, their answer suggests that
. . the numerator equals 1, which is not true.

The numerator factors: . $x\sqrt{x} + y\sqrt{y} \;=\;(\sqrt{x} + \sqrt{y})(x - \sqrt{xy} + y)$
. . which is of absolutely no help.

5. Thanks for the help all! The solution provided is what my solution manual gives but it has been incorrect before, even on testing. I don't have any examples that match this format so I can't compare my work to anything in my course book.

I'll follow up Monday with what my professor says.

6. My professor said there was a mistake in the book. The correct solution was what Soroban posted. Thanks again gurus!