First slide

Introduction to linear inequalities

Question

$Solution of \frac{x - 7}{x + 3} > 2 is$

Moderate

A

$(- 3, \infty)$
B

$(- \infty, - 13)$
C

$(- 13, - 3)$
D

None of these

Solution

$\begin{array}{l} \frac{x - 7}{x + 3} > 2 \\ \Rightarrow \frac{x - 7}{x + 3} - 2 > 0 \\ \Rightarrow \frac{- x - 13}{x + 3} > 0 \\ \Rightarrow \frac{x + 13}{x + 3} < 0 \\ \Rightarrow (x + 13) (x + 3) < 0 (\sin c e \frac{a}{b} < 0 \Rightarrow a b < 0) \\ \Rightarrow - 13 < x < - 3 \end{array}$

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