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