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an odd exam queston

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I had the following question in a maths exam the other day

[math]xy = 10[/math] and [math]x^2 + y^2 = 61[/math]


I tried just the primitive guess and check method, but IT DIDN'T WORK!

can you please give me some help?

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Have you tried substitution? For example, you can set the first equation to [math]x = \frac{10}{y}[/math] , then put/substitute that [math] \frac{10}{y}[/math] in place of x in the second equation like this [math](\frac{10}{y})^2 + y^2 = 61[/math] and solve for y.


Once you've solved for y, you can substitute your solution for y (whatever number you get, put that where the Y is) and solve for x.

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One method is taking the square root of the second equation and moving the x to the other side you'll find that [math]y = \sqrt{61} - x[/math].


Thus the first equation can be expressed as [math]-x^2 + x(\sqrt{61}) - 10 = 0[/math]


...which is a quadratic equation and is easily solved using the quadratic formula.


(Since the [math]b[/math] in the formula's "[math]b^2[/math]" is conveniently [math]\sqrt{61}[/math] it might imply that the question was formed with the quadratic formula in mind. ;) )

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Gilded You mean [math] y = \sqrt{61-x^{2}}[/math].



To get an idea of what the answer should be you could always plot the function [math]g(x) = 10/x[/math] and and see where is touches the circle defined by your second equation. That is where the root 61 comes into play.


Then from there solve it algebraically using substitution etc.

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you'll get equation 100=60y^2


How the heck do you get this? You've missed some steps/made some errors here...


You forgot to multiply the [math]y^2[/math] term by the [math]y^2[/math] term that was in the denominator of [math](\frac{10}{y})^2[/math] to get [math]y^4[/math]. The equation is [math]y^4 - 61y^2 + 100 = 0[/math]. Which can then be solved via the quadratic equation.

Edited by Bignose
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