Solve each of the following in equations and represent the solution set on

Given:

$\frac{5 x+8}{4-x}<2, x \in R$

Subtracting both the sides by 2

$\frac{5 x+8}{4-x}-2<2-2$

$\frac{5 x+8-2(4-x)}{4-x}<0$

$\frac{5 x+8-8+2 x}{4-x}<0$

$\frac{7 x}{4-x}<0$

Now dividing both the sides by 7

$\frac{7 x}{7(4-x)}<\frac{0}{7}$

$\frac{x}{4-x}<0$

Signs of x:

$x=0$

$x<0$

$x>0$

Signs of 4 – x:

$4-x=0 \rightarrow x=4$

(Subtracting 4 from both the sides, then dividing by -1 on both the sides)

$4-x<0 \rightarrow x>4$

(Subtracting 4 from both the sides, then multiplying by -1 on both the sides)

$4-x>0 \rightarrow x<4$

(Subtracting 4 from both the sides, then multiplying by -1 on both the sides)

At $x=4, \frac{x}{4-x}$ is not defined

Intervals satisfying the condition: < 0

x < 0 or x > 4

Therefore,

$x \in(-\infty, 0) \cup(4, \infty)$