solve-x-4-x-3-lt-2x-1-2-

Question Number 117027 by I want to learn more last updated on 09/Oct/20

solve : \frac{x - 4}{x - 3} < \frac{2 x - 1}{2}

Answered by bemath last updated on 09/Oct/20

\Rightarrow \frac{x - 4}{x - 3} - \frac{2 x - 1}{2} < 0

\Rightarrow \frac{2 x - 8 - (2 x - 1) (x - 3)}{2 (x - 3)} < 0

\Rightarrow \frac{2 x - 8 - ({2 x}^{2} - 7 x + 3)}{2 (x - 3)} < 0

\Rightarrow \frac{- {2 x}^{2} + 9 x - 11}{2 (x - 3)} < 0

\Rightarrow \frac{{2 x}^{2} - 9 x + 11}{2 (x - 3)} > 0

\Rightarrow \frac{x^{2} - \frac{9}{2} x + \frac{11}{2}}{x - 3} > 0

\Rightarrow \frac{{(x - \frac{9}{4})}^{2} - \frac{81}{16} + \frac{88}{16}}{x - 3} > 0

\Rightarrow \frac{{(x - \frac{9}{4})}^{2} + \frac{7}{16}}{x - 3} > 0

the solution x > 3

Commented by I want to learn more last updated on 09/Oct/20

Thanks sir

Answered by 1549442205PVT last updated on 09/Oct/20

\frac{x - 4}{x - 3} < \frac{2 x - 1}{2} \Leftrightarrow \frac{x - 4}{x - 3} - \frac{2 x - 1}{2} < 0

\frac{2 x - 8 - ({2 x}^{2} - 7 x + 3)}{2 (x - 3)} < 0

\Leftrightarrow \frac{- {2 x}^{2} + 9 x - 11}{x - 3} < 0 \Leftrightarrow \frac{{2 x}^{2} - 9 x + 11}{x - 3} > 0 (1)

{2 x}^{2} - 9 x + 11 = 2 {(x - \frac{9}{4})}^{2} + \frac{7}{8} > 0 \forall x \in R

Hence, (1) \Leftrightarrow x - 3 > 0 \Leftrightarrow x > 3

Thus, the given inequality has set of

roots is S = (3; + \infty)

Commented by bemath last updated on 09/Oct/20

gave kudos

Commented by I want to learn more last updated on 09/Oct/20

Thanks sir

Commented by 1549442205PVT last updated on 09/Oct/20

You are welcome .