Answer: -1 < x < 3
Explanation:
![(5)/((x+2)(4-x))<1](https://img.qammunity.org/2021/formulas/mathematics/high-school/qp6mpmyg2e0ilapook04ympffklnuqmydx.png)
Step 1 The denominator cannot equal zero:
x + 2 ≠ 0 and 4 - x ≠ 0
x ≠ -2 4 ≠ x
Place these restrictive values on the number line with an OPEN dot:
<----------o-------------------o--------->
-2 4
Step 2 Find the zeros (subtract 1 from both sides and set equal to zero):
![(5)/((x+2)(4-x))-1=0\\\\\\(5)/((x+2)(4-x))-((x+2)(4-x))/((x+2)(4-x))=0\\\\\\(5-(-x^2+2x+8))/((x+2)(4-x))=0\\\\\\(5+x^2-2x-8)/((x+2)(4-x))=0\\\\\\(x^2-2x-3)/((x+2)(4-x))=0\\\\\\\text{Multiply both sides by (x+2)(4-x) to eliminate the denominator:}\\x^2-2x-3=0\\(x-3)(x+1)=0\\x-3=0\quad x+1=0\\x=3\quad x=-1](https://img.qammunity.org/2021/formulas/mathematics/high-school/tesgq446rc5f44jx6z9t0o6g2qy2re3dm6.png)
Add the zeros to the number line with an OPEN dot (since it is <):
<----------o-----o----------o----o--------->
-2 -1 3 4
Step 3 Choose test points to the left, between, and to the right of the points plotted on the graph. Plug those values into (x - 3)(x + 1) to determine its sign (+ or -):
Left of -2: Test point x = -3: (-3 - 3)(-3 + 1) = Positive
Between -2 and -1: Test point x = -1.5: (-1.5 - 3)(-1.5 + 1) = Positive
Between -1 and 3: Test point x = 0: (0 - 3)(0 + 1) = Negative
Between 3 and 4: Test point x = 3.5: (3.5 - 3)(3.5 + 1) = Positive
Right of 4: Test point x = 5: (5 - 3)(5 + 1) = Positive
+ + - + +
<----------o-----o----------o----o--------->
-2 -1 3 4
Step 4 Determine the solution(s) based on the inequality symbol. Since the original inequality was LESS THAN, we want the solutions that are NEGATIVE.
Negative values only occur between -1 and 3
So the solution is: -1 < x < 3