Question

solve the following inequation (4+2x)/3>=x/2-3


class 11 modern math
chapter 6
question no 14​

Answer 1

ANSWER:

To Solve:

• (4 + 2x)/3 ≥ x/2 - 3

Solution:

We are given that,

$\implies \dfrac{4+2x}{3}\geq\dfrac{x}{2}-3$

Taking LCM in RHS,

$\implies \dfrac{4+2x}{3}\geq\dfrac{x-6}{2}$

On cross-multiplication,

$\implies 2(4+2x)\geq3(x-6)$

$\implies 8+4x\geq3x-18$

Transposing 3x to LHS,

$\implies 8+4x-3x\geq-18$

$\implies 8+x\geq-18$

Transposing 8 to RHS,

$\implies x\geq-18-8$

$\implies x\geq-26$

Hence, value of x ∈ [-26, ∞).

Answer 2

Step-by-step explanation:

$\color{darkviolet} \[ \begin{array}{l} \text{We have \( \tt\dfrac{4+2 x}{3} \geq \tt \dfrac{x}{2}-3 \)} \\ \\ \tt\Longrightarrow \dfrac{4+2 x}{3}-\dfrac{x}{2} \geq \dfrac{x}{2}-3-\dfrac{x}{2} \\ \\ \tt \Longrightarrow \dfrac{2(4+2 x)-3 x}{6} \geq-3 \\ \\ \tt\Longrightarrow \dfrac{2(4+2 x)-3 x}{6} \times 6 \geq-3 \times 6 \\ \\ \tt\Longrightarrow 8+4 x-3 x \geq 6 \times-3 \\ \\ \tt \Longrightarrow 8+x \geq-18 \\ \\ \tt \Longrightarrow 8+x-8 \geq-18-8 \\ \\ \tt \Longrightarrow x \geq-26 \\ \\ \text{Hence, we have solution as \( \tt \: x \in[-26 \), \( \infty) \)}\end{array} \]$