Find $x$
$\frac{2 x - 3}{3 x - 7} > 0$

Open in App

Solution

$\frac{2 x - 3}{3 x - 7} > 0$
$=> 2 x - 3 > 0$
$=> 2 x > 3$
$=> x > \frac{3}{2}$

Suggest Corrections

Similar questions

\frac{4 - 3 x}{5} + \frac{7 - x}{3} + 4 \frac{1}{3} = 0

p

3 p - 2 x + 1 = 0.

\frac{3 x + 1}{6 x + 7} - \frac{x + 1}{2 x + 3} = 0

Solve : $\frac{4 - 3 x}{5} + \frac{7 - x}{3} + 4 \frac{1}{3} = 0$ Hence find the value of 'p' if $3 p - 2 x + 1 = 0$

x

x - [2 x - \frac{3 x - 4}{7}] = \frac{4 x - 27}{3} - 3

x

\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3}

Join BYJU'S Learning Program