D'Nalor 10 Posted September 3, 2008 Share Posted September 3, 2008 I had the following question in a maths exam the other day [math]xy = 10[/math] and [math]x^2 + y^2 = 61[/math] HOW DO I DO THIS, IT'S DRIVING ME MAD BECAUSE I CAN'T FIGURE IT OUT! I tried just the primitive guess and check method, but IT DIDN'T WORK! can you please give me some help? Link to post Share on other sites

iNow 5907 Posted September 3, 2008 Share Posted September 3, 2008 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. Link to post Share on other sites

Gilded 180 Posted September 3, 2008 Share Posted September 3, 2008 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. ) Link to post Share on other sites

ajb 1567 Posted September 3, 2008 Share Posted September 3, 2008 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. Link to post Share on other sites

Gilded 180 Posted September 3, 2008 Share Posted September 3, 2008 Whoops, haven't really done much mathematics during the last year or so. Nevermind my previous post. Link to post Share on other sites

chitrangda 42 Posted September 6, 2008 Share Posted September 6, 2008 do it by substituion method. you'll get equation 100=60y^2 after finding y solve for x by substituion he value of y. its same as given in post no 2 by inow. Link to post Share on other sites

Bignose 946 Posted September 6, 2008 Share Posted September 6, 2008 (edited) 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 September 6, 2008 by Bignose Link to post Share on other sites

chitrangda 42 Posted September 6, 2008 Share Posted September 6, 2008 sorry missed that. the equaion which comes is 100=-y^2(y^2-61) which when can be solved via quadratic equation Link to post Share on other sites

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