Question

Solve the inequality

$\frac{x - 4}{x + 2} \leqslant 2$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm \: \dfrac{x - 4}{x + 2} \leqslant 2$

Let's first define the domain of the above.

$\rm \: x \: \ne \: - \: 2$

So,

$\rm \: \dfrac{x - 4}{x + 2} \leqslant 2 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{x - 4}{x + 2} - 2 \leqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{x - 4 - 2(x + 2)}{x + 2} \leqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{x - 4 - 2x - 4}{x + 2} \leqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{ - x - 8}{x + 2} \leqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{ -( x + 8)}{x + 2} \leqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{ x + 8}{x + 2} \geqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \: \dfrac{(x + 8)(x + 2)}{(x + 2)^{2} } \geqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm \:(x + 8)(x + 2) \geqslant 0 \: \: \: and \: x \: \ne \: - \: 2$

$\rm\implies \:( - \infty , \: - 8] \: \cup \: ( - 2, \infty )$

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ADDITIONAL INFORMATION :-

If a and b are two real positive numbers such that a < b, then

$\rm \: (x - a)(x - b) < 0 \: \: \rm\implies \:a < x < b$

$\rm \: (x - a)(x - b) \leqslant 0 \: \: \rm\implies \:a \leqslant x \leqslant b$

$\rm \: (x - a)(x - b) > 0 \: \: \rm\implies \:x < a \: \: or \: \: x > b$

$\rm \: (x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x \leqslant a \: \: or \: \: x \geqslant b$

$\rm \: |x| < a \: \: \rm\implies \: - a < x < a$

$\rm \: |x| \leqslant a \: \: \rm\implies \: - a \leqslant x \leqslant a$

$\rm \: |x| > a \: \: \rm\implies \:x < - a \: \: or \: \: x > a$

$\rm \: |x| \geqslant a \: \: \rm\implies \:x \leqslant - a \: \: or \: \: x \geqslant a$

Answer 2

Step-by-step explanation:

→Replace the inequality sign with an equal sign, so that we can solve it like a normal equation.

$\sf \leadsto\dfrac{x-4}{x+2}=2$

→Multiply both sides by x+2.

⇝x-4=2(x+2)

→Subtract x from both sides.

⇝-4=2x+4-x

⇝x=−8

→Also note that x is undefined at -2.

x$\ne$−2

→From the values of xx above, we have these 3 intervals to test.

x≤−8

−8≤x≤−2

x≥−2

→Pick a test point for each interval.

For the interval x≤−8:

Let's pick x=−9. Then, $\tt\frac{-9-4}{-9+2}\le 2$

After simplifying, we get 1.857143≤2, which is true.

Keep this interval..

For the interval −8≤x≤−2:

Let's pick x=-7x=−7. Then, $\tt\frac{-7-4}{-7+2}\le 2$

After simplifying, we get 2.2≤2, which is false.

Drop this interval..

For the interval x≥−2:

Let's pick x=0x=0. Then, $\tt\frac{0-4}{0+2}\le 2$

After simplifying, we get −2≤2, which is true.

Keep this interval..

→Therefore,

•x≤−8,x≥−2

→Notice the equation contains x+2 in the denominator. Since any denominator must not equal zero, the domain is restricted to $\sf x+2\ne 0.$ Solving for x, we have:

$\tt x\ne -2$

→ Add the domain restrictions.

x≤−8,x>−2