Hi. So we had a long test with this question.. I was wondering how to solve it? Coz our teacher didn't discuss the answers.. and most probably we'll have the same type of question during midterms..

Find the Solution Set

9/|2-5x| - 4/y^2-3y = 5
2/|2-5x| + 2/ 3(y^2-3y) = 1/3

thanks!

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
subtract:
-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

case1: 2-5x=3
-5x = 1
x = -1/5

case2:
-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

To find the solution set for the given equations, we need to solve the system of equations simultaneously. Here's how you can approach it:

1. Start with the first equation:

9/|2-5x| - 4/(y^2-3y) = 5

2. Multiply both sides of the equation by |2-5x| and (y^2-3y) to eliminate the absolute value and denominators:

9(y^2-3y) - 4|2-5x|(|2-5x|) = 5|2-5x|(y^2-3y)

Note: When multiplying both sides by |2-5x|, the absolute value sign disappears because we consider both the positive and negative cases.

3. Simplify the equation:

9y^2 - 27y - 4(2-5x)^2 = 5(y^2-3y)(2-5x)

4. Expand the squared term and simplify further:

9y^2 - 27y - 4(4-20x+25x^2) = 5(y^2-3y)(2-5x)

9y^2 - 27y - 16 + 80x - 100x^2 = 5(2y-3y^2)(2-5x)

9y^2 - 27y - 16 + 80x - 100x^2 = 10y(2-5x) - 15y^2(2-5x)

9y^2 - 27y - 16 + 80x - 100x^2 = 20y - 50xy - 30y^2 + 75xy^2

5. Combine like terms and set the equation equal to zero:

30y^2 - 29y + 16 + 75xy^2 - 50xy - 20y + 100x^2 - 80x = 0

6. Now, let's move on to the second equation:

2/|2-5x| + 2/(3(y^2-3y)) = 1/3

7. Multiply both sides of the equation by |2-5x| and (3(y^2-3y)):

2(3(y^2-3y)) + 2|2-5x||2-5x| = (1/3)(3(y^2-3y))(|2-5x|)

8(y^2-3y) + 2(2-5x)(2-5x) = (y^2-3y)(|2-5x|)

8y^2 - 24y + 4 - 20x + 25x^2 = y^2-3y(2-5x)

8y^2 - 24y + 4 - 20x + 25x^2 = y^2-6y+15xy

7y^2 - 18y + 4 + 20x - 15xy + 25x^2 = 0

9. Now, you have a system of two polynomial equations:

30y^2 - 29y + 16 + 75xy^2 - 50xy - 20y + 100x^2 - 80x = 0
7y^2 - 18y + 4 + 20x - 15xy + 25x^2 = 0

To solve this system of equations, you can use methods such as substitution, elimination, or graphically. You can also use numerical methods if necessary.

Once you solve the system of equations, you will obtain the solution set for the given equations.