How do you solve #(4x-5)/(x+3)0#?

Question

How do you solve #(4x-5)/(x+3)>0#?

Isaiah Anthony · Accepted Answer

#0 < (4x-5)/(x+3) = ((4x+12) - 17)/(x+3) = (4(x+3) -17)/(x+3) = 4 - 17/(x+3).#

Add #17/(x+3)# to both sides to get:

#17/(x+3) < 4#

There are now two permissible cases:

Case 1: When #x < -3#, #x + 3 < 0#, so if we multiply both sides by #(x+3)# we have to reverse the inequality to get:

#17>4(x+3)=4x+12#

Subtract 12 from both sides to get:

#5>4x#.

Divide both sides by 4 to get:

#5/4>x#, i.e. #x < 5/4#.

Since in this case we already know #x<-3#, this condition is already fulfilled.

Case 2: When #x > -3#, #x + 3 > 0#, so we can multiply both sides by #(x+3)# without reversing the inequality to get:

#17<4(x+3)=4x+12#

Subtract 12 from both sides to get:

#5<4x#.

Divide both sides by 4 to get:

#5/4 5/4#.

If #x > 5/4# then it satisfies #x > -3#

So in Case 2, we just require #x > 5/4#.

Combining the 2 cases, we find that #x < -3# or #x > 5/4#.

Note that #x = -3# is not allowed due to the resulting division by 0, which is undefined.

Genesis Allen · Answer

undefined To solve the inequality (4x - 5) / (x + 3) > 0:

1. Find the critical points by setting the numerator and denominator equal to zero and solving for x.
2. Determine the sign of the expression in each interval created by the critical points.
3. Identify the intervals where the expression is greater than zero.

The critical points are where the numerator (4x - 5) and the denominator (x + 3) are equal to zero.

Numerator:
4x - 5 = 0
x = 5/4

Denominator:
x + 3 = 0
x = -3

Now, we have three intervals: (-∞, -3), (-3, 5/4), and (5/4, ∞).

Test a value in each interval to determine the sign of the expression:
- Choose x = -4 for (-∞, -3)
- Choose x = 0 for (-3, 5/4)
- Choose x = 2 for (5/4, ∞)

For x = -4:
(4(-4) - 5) / (-4 + 3) = (-21) / (-1) = 21 > 0, so the interval (-∞, -3) is positive.

For x = 0:
(4(0) - 5) / (0 + 3) = (-5) / 3 < 0, so the interval (-3, 5/4) is negative.

For x = 2:
(4(2) - 5) / (2 + 3) = (3) / 5 > 0, so the interval (5/4, ∞) is positive.

Therefore, the solution to the inequality is:
x ∈ (-∞, -3) U (5/4, ∞).

How do you solve #(4x-5)/(x+3)&gt;0#?

How do you solve #(4x-5)/(x+3)>0#?