For what #x# is #3/(x+3)>(4x)/(x-2)#?

1 Answer
Nov 7, 2015

Answer:

#x\in(-3, 2)#

Explanation:

To simplify the inequality, we are going to perform cross-multiplication of the denominator to avoid dealing with fractions. However, the sign of the inequality might be flipped if we multiply both sides by a negative expression. Hence, we consider #x# on 3 separate intervals, namely, #(-oo,-3)#, #(-3, 2)#, #(2, oo)#. Note that #x# cannot -3 or 2. Now, we wish to find the values of #x# which satisfies

#frac{3}{x+3}>frac{4x}{x-2}#

For the case of #x\in(-oo, -3)#, both #x+3# and #x-2# are negative.

#3(x-2)>4x(x+3)#

The sign remains the same as the product of 2 negative expressions is positive.

#3x-6>4x^2+12x#

#0>4x^2+9x+6#

#0>x^2+9/4x+3/2#

Completing the square,
#0>(x+9/8)^2+{15}/{64}#

Since the RHS is always #>=15/64#, no real value of #x# satisfies
#3(x-2)>4x(x+3)#.

Therefore, no value of #x\in(-oo, -3)# satisfies #frac{3}{x+3}>frac{4x}{x-2}#.

For the case of #x\in(-3, 2)#, #x+3# is positive while #x-2# is negative.

#3(x-2)<4x(x+3)#

The sign flips as the product of a positive and a negative expressions is negative.

Completing the square,
#0<(x+9/8)^2+{15}/{64}#

Since the RHS is always #>=15/64>0#, all real value of #x# satisfies
#3(x-2)>4x(x+3)#.

Therefore, all values of #x\in(-3, 2)# satisfies #frac{3}{x+3}>frac{4x}{x-2}#.

For the case of #x\in(2, oo)#, both #x+3# and #x-2# are negative.

#3(x-2)>4x(x+3)#

The sign remains the same as the product of 2 negative expressions is positive.

Completing the square,
#0>(x+9/8)^2+{15}/{64}#

Since the RHS is always #>=15/64#, no real value of #x# satisfies
#3(x-2)>4x(x+3)#.

Therefore, no value of #x\in(2, oo)# satisfies #frac{3}{x+3}>frac{4x}{x-2}#.

Combining the solutions on all 3 intervals, we get

#frac{3}{x+3}>frac{4x}{x-2}#

if and only if #x\in(-3, 2)#.