I think I would eliminate variables.
multiply the first equation by 2/3
multiply the second equation by 4
add the equations.
solve for abs(2-5x), you will have two solutions.
Then, put each of those solutions into the first equation, and solve for y.
Get a large blank pad on which to do this problem, and check it twice. You can finally check each solution by substitution into the second equation.
That's quite nasty for a test.
I would try the following, along the lines suggested by bobpursley
multiply the first by 2
18/|2-5x| - 8/(y^2-3y) = 10
multiply the second by 9
18/|2-5x| + 6/(y^2-3y) = 3
-14/(y^2-3y) = 7
7y^2 - 21y + 14 = 0
y^2 - 3y + 2 = 0
(y-2)(y-1) = 0
y = 2 or y = 1
I will do the y=2
If y = 2 , then
9/|2-5x| - 4/-2 = 5
9/|2-5x| = 3
3/|2-5x| = 1
|2-5x| = 3
-5x = 1
x = -1/5
-2 + 5x = 3
5x = 5
x = 1
so two pairs from that part:
(1,2) and (-1/5, 2)
you do the other two solutions, using y = 1
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