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  4. The set of values of x which satisfy the inequations 5x + 2 < 3x + 8 and (x+2/x-1) < 4 is

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Q. The set of values of $x$ which satisfy the inequations $5x + 2 < 3x + 8$ and $\frac{x+2}{x-1} < 4$ is

Linear Inequalities

Solution:

We have, $5x + 2 < 3x + 8$ and $\frac{x+2}{x-1} < 4$
$\Rightarrow \quad x < 3$ and $\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)^{2}} < 4$, $x \ne 1 $
$\Rightarrow \quad x < 3$ and $\left(x + 2\right)\left(x - 1\right) < 4x^{2} - 8x + 4$, $x \ne 1$
$\Rightarrow \quad x < 3$ and $3x^{2} - 9x + 6 > 0$, $x \ne 1 $
$\Rightarrow \quad x < 3$ and $x^{2} - 3x + 2 > 0$, $x \ne 1$
$\Rightarrow \quad x < 3$ and $\left(x - 1\right)\left(x - 2\right) > 0$, $x \ne 1$
$\Rightarrow \quad x < 3$ and $(x < 1$ or $x > 2)$
$\Rightarrow \quad x\in\left(-\infty,\,1\right)\cup\left(2,\,3\right)$.