First slide

Introduction to linear inequalities

Question

The range of values of x for which the inequality $\frac{x - 1}{4 x + 5} < \frac{x - 3}{4 x - 3}$ holds is

Easy

A

$(\frac{- 4}{3}, \frac{5}{8})$
B

$(\frac{- 4}{3}, \frac{1}{2})$
C

$(\frac{- 5}{4}, \frac{3}{4})$
D

$(3, 8)$

Solution

The given inequation is $\frac{x - 1}{4 x + 5} < \frac{x - 3}{4 x - 3}$
$\Rightarrow \frac{x - 1}{4 x + 5} - \frac{x - 3}{4 x - 3} < 0$

$\Rightarrow \frac{(x - 1) (4 x - 3) - (x - 3) (4 x + 5)}{(4 x + 5) (4 x - 3)} < 0$

$\Rightarrow \frac{18}{(4 x + 5) (4 x - 3)} < 0$

$\Rightarrow (4 x + 5) (4 x - 3) < 0$

$\Rightarrow 4 x + 5 < 0, 4 x - 3 < 0$

$\Rightarrow x \in (\frac{- 5}{4}, \frac{3}{4})$

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