Can't figure out a linear equation

I am sorry, this is my first time here. I don't know if I put the question correctly.

Anyways, I have a simple question on me, middle school. Well, I am 14 meh. I am not being able to solve it. I would like it if someone could help me in it. Using Algebra and Linear Equations.. No AP or anything, we haven't studied it yet.

$\sqrt{x/y}$ + $\sqrt{y/x} = 10/3$ Find xy

• Not a linear equation. Because there is a square root sign that you want to go away, try squaring both sides as a first step. What happens? – T.J. Gaffney Mar 6 '14 at 5:17
• I tried. Couldn't get the answer though – Bone Mar 6 '14 at 5:21
• (x^2+y^2)9/82=xy is what u get after manipulation. there must be one more equation, coz there are two unknowns – ketan Mar 6 '14 at 5:31
• Try multiplying through by $\sqrt{x/y}$ first. (Then square both sides.) – Eric Towers Mar 6 '14 at 5:38
• We cannot determine $xy$, because if we multiply each of $x$ and $y$ by $k\ne \pm 1$, we don't change your square-rooted expressions but we change $xy$. – André Nicolas Mar 6 '14 at 5:39

If you square both sides you get $$\frac{x}{y}+2\sqrt{\frac{x}{y}}\sqrt{\frac{y}{x}}+\frac{y}{x}=\frac{100}{9}\ ,$$ that is, $$\frac{x}{y}+2+\frac{y}{x}=\frac{100}{9}$$ or $$\frac{x}{y}-\frac{82}{9}+\frac{y}{x}=0\ .$$ Multiply both sides by $x/y$ to get $$\Bigl(\frac{x}{y}\Bigr)^2-\frac{82}{9}\Bigl(\frac{x}{y}\Bigr)+1=0\ .$$ If you have studied quadratics you can now treat $x/y$ as a single variable to find two possible values $$\frac{x}{y}=\cdots$$ Can you finish this?

However this will not give you a value for $x$ and $y$ separately, or for $xy$. I don't think it is actually possible to answer the question you have been asked.

• Thank you. I guess it's a wrong question then – Bone Mar 6 '14 at 5:50
• Yes, but maybe only a little bit wrong. Have another look at the source and see if you were given any more information about $x$ and $y$. – David Mar 6 '14 at 5:51
• nope i wasnt... – Bone Mar 6 '14 at 5:52
• When you simplified $\sqrt{\dfrac xy}\sqrt{\dfrac yx}$ to $1$ you have made the assumption that $x/y$ and $y/x$ can't be both negative. But there's nothing in the énoncé that says that $x/y$ and $y/x$ aren't both negative AFAIK. – Hakim Mar 9 '14 at 11:26
• As it's a middle school problem I think we can reasonably assume that complex numbers are not involved. – David Mar 9 '14 at 22:12

This is not a linear equation, rather, it is a radical equation. To solve it, first square both sides and simplify. $$\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=\dfrac{10}{3}$$ $$\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right)^2=\left(\dfrac{10}{3}\right)^2$$ $$\dfrac{x}{y}+2\sqrt{\dfrac{x}{y}}\sqrt{\dfrac{y}{x}}+\dfrac{y}{x}=\dfrac{100}{9}$$ $$\dfrac{x}{y}+2\sqrt{\dfrac{xy}{yx}}+\dfrac{y}{x}=\dfrac{100}{9}$$ $$\dfrac{x}{y}+2+\dfrac{y}{x}=\dfrac{100}{9}$$ $$\dfrac{x}{y}+\dfrac{y}{x}-\dfrac{100}{9}+2=0$$ $$\dfrac{x}{y}+\dfrac{y}{x}-\dfrac{100}{9}+\dfrac{18}{9}=0$$ $$\dfrac{x}{y}+\dfrac{y}{x}-\dfrac{82}{9}=0$$ Multiply both sides by $\dfrac{x}{y}$ and simplify. We will get a quadratic equation (an equation of the form $ax^2+bx+c=0$, where $a$, $b$, and $c$ are constants), which we can solve for. $$\dfrac{x}{y}\left(\dfrac{x}{y}+\dfrac{y}{x}-\dfrac{82}{9}\right)=\dfrac{x}{y}\left(0\right)$$ $$\left(\dfrac{x}{y}\right)^2-\dfrac{82}{9}\cdot\dfrac{x}{y}+1=0$$ Let $\dfrac{x}{y}=a$. This allows for easier calculation. $$a^2-\dfrac{82}{9}a+1=0$$ Multiply by $9$ on both sides to get rid of the denominator in $\dfrac{82}{9}x$. $$9a^2-82a+9=0$$ Use the quadratic formula. If you did not know, the quadratic formula states:

For any equation $ax^2+bx + c=0$ ($a$, $b$, and $c$ are constants), its roots are: $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$a=\dfrac{-(-82)\pm\sqrt{(-82)^2-4(9)(9)}}{2(9)}$$ $$a=\dfrac{82\pm\sqrt{6724-324}}{18}$$ $$a=\dfrac{82\pm\sqrt{6400}}{18}$$ $$a=\dfrac{82\pm80}{18}$$ $$a=\dfrac{41\pm40}{9}$$ $$a=\dfrac{1}{9}, \ 9$$ $$\dfrac{x}{y}=\dfrac{1}{9}, \ 9$$ Now, we will check for extraneous roots. Extraneous roots can be brought into the solution by squaring the square root. These answers are incorrect. Let's start with $\dfrac{x}{y}=\dfrac{1}{9}$ $$\sqrt{\dfrac{1}{9}}+\sqrt{9}=\dfrac{10}{3}$$ $$\dfrac{1}{3}+3=\dfrac{10}{3}$$ $$\dfrac{10}{3}=\dfrac{10}{3}$$ $$\text{True.}$$ Now for $x=9$: $$\sqrt{9}+\sqrt{\dfrac{1}{9}}=\dfrac{10}{3}$$ $$3+\dfrac{1}{3}=\dfrac{10}{3}$$ $$\dfrac{10}{3}=\dfrac{10}{3}$$ $$\text{True.}$$ So: $$\displaystyle \boxed{\dfrac{x}{y}=\dfrac{1}{9}, \ 9}$$ Everything from here on out are educated guesses.

I do not think you can solve for $xy$. My logic may be wrong, but know that I am just trying to help.

We know that: $$\dfrac{x}{y}=\dfrac{1}{9}, \ 9$$ This can be rewritten as: $$\dfrac{x}{y}=\dfrac{1}{9}, \ \dfrac{9}{1}$$ This splits the equality into two cases.

Case 1: $\dfrac{x}{y}=\dfrac{1}{9}$ $$\dfrac{x}{y}=\dfrac{1}{9}$$ $$x=1$$ $$y=9$$ Case 2: $\dfrac{x}{y}=\dfrac{9}{1}$ $$\dfrac{x}{y}=\dfrac{9}{1}$$ $$x=9$$ $$y=1$$ You can see that in each of the solutions $x=1, \ y=9$ and $x=9, \ y=1$, $\boxed{xy=9}$. But we must not forget that there are the sets where $x=2, \ y=18$, $x=18, \ y=2$, $x=3, \ y=27$, $x=27, \ y=3$, etc. And that is just for whole numbers. What about decimals? There are an infinite number of solutions. So finding $xy$ is impossible.

Hopefully you were looking for $\dfrac{x}{y}$. Then, the solution is: $$\displaystyle \boxed{\dfrac{x}{y}=\dfrac{1}{9}, \ 9}$$ But if you really need to find the value of $xy$, then there are an infinite number of solutions.

• From what I had solved it, x/y came to be equal to 9. Following David's Answer. How is your answer 1/9 – Bone Mar 7 '14 at 12:49
• @Bone $\dfrac{41\pm40}{9}=\dfrac{41+40}{9}, \ \dfrac{41-40}{9}$. Hopefully you knew that... – TrueDefault Mar 8 '14 at 3:54
• meh... oh well.. – Bone Mar 8 '14 at 4:37
• @Bone I added how you cannot solve for $xy$. – TrueDefault Mar 8 '14 at 7:04
• When you simplified $\sqrt{\dfrac xy}\sqrt{\dfrac yx}$ to $1$ you have made the explicit assumption that $x/y$ and $y/x$ can't be both negative. But there's nothing in the énoncé that says that $x/y$ and $y/x$ aren't both negative AFAIK. – Hakim Mar 9 '14 at 11:27