System of equations

[AvgStudent]

Problem:
Find the value of $x+y$, given:
$x\sqrt{y}+y\sqrt{x}=182\\ x\sqrt{x}+y\sqrt{y}=183\\$

My attempt:
Adding equations 1 and 2, collect like terms and factor give:
$$x\sqrt{y}+y\sqrt{x}+x\sqrt{x}+y\sqrt{y}=365\\ x\sqrt{y}+x\sqrt{x}+y\sqrt{x}+y\sqrt{y}=365\\ x(\sqrt{y}+\sqrt{x})+y(\sqrt{x}+\sqrt{y})=365\\ (x+y)(\sqrt{x}+\sqrt{y})=365$$
I'm not sure how to continue from here.

 

[JeffM]

Notice that you are not asked to find x and y individually.

$$\text {Let } u = \sqrt{x} \text { and } v = \sqrt{y} \implies \\ u^2v + uv^2 = 182 \text { and } u^3 + v^3 = 183.$$
What does that suggest to you?

 

[AvgStudent]

Notice that you are not asked to find x and y individually.

$$\text {Let } u = \sqrt{x} \text { and } v = \sqrt{y} \implies \\ u^2v + uv^2 = 182 \text { and } u^3 + v^3 = 183.$$
What does that suggest to you?

Thank Jeff, that was very helpful.
$$(u+v)^3=u^3+v^3+3u^2v+3v^2u=183+3(182)=729\implies u+v=9\\$$Also,
$$u^2v+uv^2=uv(u+v)=uv(9)=182\implies uv=\frac{182}{9}\\$$Lastly,
$$u^3+v^3=(u+v)(u^2+v^2-uv)\\ 183=(9)\left(u^2+v^2-\frac{182}{9}\right)\\ u^2+v^2=x+y=\frac{183}{9}+\frac{182}{9}=\boxed{\frac{365}{9}}$$
Hope I didn't make any mistakes.

 

[JeffM]

Looks good to me.

As a follow up to my previous post, it frequently happens that knowing what topic you are currently studying helps us give you the most intuitive answer. Telling us context is a good idea. Here I am guessing that you are studying cubics.

Finding x and y individually is one way to try, but it is fraught with difficulties. I suspect it ends up with needing to solve a quintic. Not all quintics are solvable by algebra.

 

[AvgStudent]

Looks good to me.

As a follow up to my previous post, it frequently happens that knowing what topic you are currently studying helps us give you the most intuitive answer. Telling us context is a good idea. Here I am guessing that you are studying cubics.

Finding x and y individually is one way to try, but it is fraught with difficulties. I suspect it ends up with needing to solve a quintic. Not all quintics are solvable by algebra.

Hi Jeff,
I'm not "learning" any topic in particular. I'm reviewing and practising algebra in general for my entrance exam. The format of the exam is multiple-choice, so there isn't any restriction to the methodology.

 

[JeffM]

Just because I am curious.

$$(u + v)^3 = 3 * 182 + 183 = 546 + 183 = 729 \implies \\ u + v = 9 \implies u = 9 - v \implies u^2 = 81 - 18v + v^2 \implies\\ (81 - 18v + v^2)v + v^2(9 - v) = 182 \implies\\ 81v - 18v^2 + v^3 + 9v^2 - v^3 = 182 \implies\\ 9v^2 - 81v + 182 = 0 \implies\\ v = \dfrac{81 \pm \sqrt { 6561 - 4(9)(182)}}{18} = \dfrac{81 \pm \sqrt{6561 - 6552}}{18} \implies\\ v = \dfrac{39}{9} \implies u = \dfrac{42}{9} \text { or }\\ v = \dfrac{42}{9} \implies u = \dfrac{39}{9} \implies \\ x = \dfrac{1764}{81} \text { and } y = \dfrac{1521}{81} \text { or } x = \dfrac{1521}{81} \text { and } y = \dfrac{1764}{81}.$$
Because they are symmetric, we need not check both answers.

$$x\sqrt{y} + y\sqrt{x} = \dfrac{1764}{81} * \dfrac{39}{9} + \dfrac{1521}{81} * \dfrac{42}{9} =\\ \dfrac{68796 + 63882}{729} = \dfrac{132678}{729} = 182. \ \checkmark\\ x\sqrt{x} + y\sqrt{y} = \dfrac{1764}{81} * \dfrac{42}{9} + \dfrac{1521}{81} *\dfrac{39}{9} =\\ \dfrac{133407}{729} = 183. \ \checkmark$$.

It checks.

$$\dfrac{1764}{81} + \dfrac{1521}{81} = \dfrac{3285}{81} = \dfrac{365}{9}.$$
AvgStudent is correct.