Asked by Laura
Solve the inequality algebraically:
5/x-3 + 3/2-x > 1
(greater then or equal to one)
I know how to do it in theory but I keep getting the answer of just y=2 and I know that can't be right.
Help is really appreciated! Thank you
5/x-3 + 3/2-x > 1
(greater then or equal to one)
I know how to do it in theory but I keep getting the answer of just y=2 and I know that can't be right.
Help is really appreciated! Thank you
Answers
Answered by
Reiny
I am sure you mean:
5/(x-3) + 3/(2-x) ≥ 1
5/(x-3) + 3/(2-x) - 1 ≥ 0
(5(2-x) + 3(x-3) - (x-3)(2-x) ≥ 0 , skipping common denominator for easier typing
10 - 2x + 3x - 9 + (x-3)(x-2) ≥ 0
1 + x + x^2 - 5x + 6 ≥ 0
(x^2 -7x + 7)/((x-3)(2-x)) ≥ 0 , brought back the LCD
The quadratic does not factor, but I found x = 1.2 and x = 5.8 to be approximate solutions.
then checking for the different intervals, we have
1.2 ≤ x < 2 OR 3 < x ≤ 5.8
5/(x-3) + 3/(2-x) ≥ 1
5/(x-3) + 3/(2-x) - 1 ≥ 0
(5(2-x) + 3(x-3) - (x-3)(2-x) ≥ 0 , skipping common denominator for easier typing
10 - 2x + 3x - 9 + (x-3)(x-2) ≥ 0
1 + x + x^2 - 5x + 6 ≥ 0
(x^2 -7x + 7)/((x-3)(2-x)) ≥ 0 , brought back the LCD
The quadratic does not factor, but I found x = 1.2 and x = 5.8 to be approximate solutions.
then checking for the different intervals, we have
1.2 ≤ x < 2 OR 3 < x ≤ 5.8
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