9514 1404 393
Answer:
(x, y) = (4, 4)
Step-by-step explanation:
If you attempt to solve this by clearing fractions, you end up with an extraneous solution. Here, we'll solve the linear equations ...
7a +4b = 5/4
8b -14a = 3/2
where a = 1/(4x+3y) and b = 1/(4x-3y)
Dividing the second equation by 2 and adding the first, we have ...
(7a +4b) +1/2(8b -14a) = (5/4) +1/2(3/2)
8b = 8/4
b = 1/4
Substituting into the first equation gives ...
7a +4/4 = 5/4
a = 1/28
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Now, we can get back to solving for x and y.
4x +3y = 1/a = 28
4x -3y = 1/b = 4
8x = 32 . . . . . . . . . add the two equations
x = 4
3y = 28 -4x = 28 -4(4) = 12 . . . . . use x in the equation for 'a'; rearrange
y = 4 . . . . divide by 3
The solution to the system of equations is (x, y) = (4, 4).
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<em>Additional comment</em>
If you graph these equations, you find they describe hyperbolas that intersect at (0, 0) and (4, 4). The "solution" (0, 0) is extraneous, as both equations are undefined there.
and 
. Get everything on one side of the equals sign, set it equal to 0 and factor. When you do this you get (x-3)(x+27). The Zero Product Property rule tells us that either x-3 = 0 or x+27 = 0 and that x = 3 and -27. The only thing in math that will NEVER be negative besides time is distance/length, therefore, x cannot be 27 and has to be 3.