Find the value of :x^2+2xy+y^2-3x-3y-1 have: x^2-y=y^2-x (x
1 answer:
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Answer:
(x, y) = (1/2, -1)
Step-by-step explanation:
Subtracting twice the first equation from the second gives ...
(2/x +1/y) -2(1/x -5/y) = (3) -2(7)
11/y = -11 . . . . simplify
y = -1 . . . . . . . multiply by y/-11
Using the second equation, we can find x:
2/x +1/-1 = 3
2/x = 4 . . . . . . . add 1
x = 1/2 . . . . . . . multiply by x/4
The solution is (x, y) = (1/2, -1).
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<em>Additional comment</em>
If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.
A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).
