How do I figure out the following: If xy=546 and x+y=26, then what are x and y?



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samhouston's profile pic

Posted on (Answer #1)

The easiest way is to graph both equations and locate the point(s) of intersection.  In this case, there are no points of intersection.  Therefore, this has no solution.

gs01's profile pic

Posted on (Answer #2)

Since x+y = 26 ...........(1),

 and xy = 546.............(2)

x = 26-y

therefore substituting the value of x in equation 2

(26-y)y = 546

26Y - y^2 = 546  Taking all the numbers on the right side

Y^2 -26Y +546 = 0

Since it is a quadratic or 2nd degree eauation we will have two values for Y using equation for ax^2+ bx+c =


(b+and -(Sqrt(b^2-4ac)/2a)

-26+Sqrt(26^2-4*1*546)/2*1) and -26-Sqrt(26^2-4*1*546)/2*1)

 Y= 13 + 19.41648783i or Y= 13 - 19.41648783i

i = sqrt (-1) is called the imaginary part.

and x = 26-Y


callayne's profile pic

Posted on (Answer #3)

Thanks for the help, people.  The 2nd answer is more advanced than where we are right now.  We're supposed to be using the ac method or, if that doesn't work evenly, then we're supposed to use a factoring method.

atyourservice's profile pic

Posted on (Answer #4)

I graphed it since i kept getting confused at my math and there are no real solutions

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