Question

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}$

| 2x - 3 | > - 2

Solve for x.

$\purple{\textsf{Answer:-}} (- [tex] \infty$ , $\infty$)

I want complete explanation of the problem with suitable steps.

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Answer 1

Answer:

2x-3>-2

2x-3+2

2x-1

x= -1/2

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Answer 2

Real numbers are positive, 0, or negative.

The absolute value function is the distance from the origin, which makes it positive or 0.

$|2x-3|\begin{cases} & 2x-3\ \text{for}\ x>\dfrac{3}{2} \\ & 0\ \text{for}\ x=\dfrac{3}{2}\\ & -2x+3\ \text{for}\ x<\dfrac{3}{2} \end{cases}$

Now let's find the answer stepwise.

Solution A

Since absolute value function is always positive or 0, it satisfies $|2x-3|\geq 0$ for real values of $x$. So the solution for $|2x-3|>-2$ are real values of $x$.

Solution B

(I) $x>\dfrac{3}{2}$

Given: $2x-3>-2$

$\implies 2x>-2+3$

$\implies 2x>1$

$\implies x>\dfrac{1}{2}$

From the given condition $x>\dfrac{3}{2}$, the solution is $\boxed{x>\dfrac{3}{2}}$.

(II) $x=\dfrac{3}{2}$

Given: $0>-2$

The inequality holds true. Hence $x=\dfrac{3}{2}$ is a solution.

(III) $x<\dfrac{3}{2}$

Given: $-2x+3>-2$

$\implies -2x>-2-3$

$\implies -2x>-5$

$\implies x<\dfrac{5}{2}$

From the given condition $x<\dfrac{3}{2}$, the solution is $\boxed{x<\dfrac{3}{2}}$.

Conclusion

$\implies x>\dfrac{3}{2} \ \text{or}\ x=\dfrac{3}{2} \ \text{or}\ x<\dfrac{3}{2}$

$\implies x\in \mathbb{R}$

Therefore, the solution of the absolute value inequality is $x\in \mathbb{R}$.