If x is positive real number, find the rang..

Question 1 Report

If x is positive real number, find the range of values for which $\frac{1}{3 x}$ + $\frac{1}{2}$ x = $\frac{2 + 3 x}{6 x}$ > $\frac{1}{4 x}$

x > -

\frac{1}{6}

x > 0

0 < x < 6

0 < x <

\frac{1}{6}

Answer Details

$\frac{1}{3 x}$ + $\frac{1}{2}$ x = $\frac{2 + 3 x}{6 x}$ > $\frac{1}{4 x}$

= 4(2 + 3x) > 6x = 12x $^{2}$ - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x - 1 > 0

= x < 0, x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x - 1 > 0 = x > 0

x > $\frac{1}{6}$

Combining solutions in cases (1) and (2)

= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$

If x is positive real number, find the range of values for which 13x 1 3 x + 12 1 2 x = 2+3x6x 2 + 3 x 6 x > 14x 1 4 x