Asked by Jen
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!
Find the Solution Set
9/|2-5x| - 4/y^2-3y = 5
2/|2-5x| + 2/ 3(y^2-3y) = 1/3
thanks!
Answers
Answered by
bobpursley
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.
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.
Answered by
Reiny
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
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
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