+0

# Inequality

0
31
2

Find all x that satisfy the inequality (2x+10)(x+3)<(3x+9)(x-8) . Express your answer in interval notation.

Jul 30, 2022

#1
+262
+2

Answer: $$x\in(-\infty, -3) \cup (34, \infty)$$

Solution:

Set the equations equal to find critical points:

(2x+10)(x+3)=3(x+3)(x-8)

x=-3 is a critical point, as is x=34.

If this inequality were to be graphed on a number line, it would result in a striped pattern. You can test the middle region (I will test -3

Testing for 0:

(0+10)(0+3)=3(0+3)(0-8)

The regions around it (x<-3 and x>34) are.

Therefore, the answer in interval notation is $$x\in(-\infty, -3) \cup (34, \infty)$$.

Jul 30, 2022
#2
+124524
+1

Expanding we have

2x^2 + 16x + 30 < 3x^2 -15x - 72       rearrange as

x^2 - 31x  - 102 >  0     factor as

( x - 34) ( x + 3)  >   0

This will be true   when  x < -3   and x > 34

So   the intervals that make this true are

(  -inf, -3)  U  ( 34, inf)

Jul 30, 2022