Question

Solve the following in-equation and represent the solution set on the number line:

4x – 19 < (3x/5) -2 ≤ (-2/5) + x, x ε R answer jldi kro 5 min main bhejna hai plss plss ​

Answer 1

Question :

Solve the following in-equation and represent the solution set on the number line:

4x – 19 < (3x/5) -2 ≤ (-2/5) + x, x ε R

Theory :

Solution of linear equations :

1. Same number may be added to (or subtracted from ) both sides of an inequation without changing the sign of inequality.
2. The sign of inequality is reversed when both sides of an inequation are multipled or divided by a negative number .

Solution :

We have to solve :

$\tt\:4x-19<\dfrac{3x}{5}-2\leqslant\dfrac{(-2)}{5}+x$

Now ,

$\sf\:4x-19<\dfrac{3x}{5}-2$

$\sf\implies\:4x-\dfrac{3x}{5}<-2+19$

$\sf\implies\dfrac{20x-3x}{5}<-2+19$

$\sf\implies\dfrac{17x}{5}<17$

Now Multiply both sides by $\sf\dfrac{5}{17}$ then ,

$\sf\implies\:x<5...(1)$

And ,

$\sf\dfrac{3x}{5}-2\leqslant\dfrac{(-2)}{5}+x$

$\sf\implies\:\dfrac{2}{5}-2\leqslant\dfrac{(-3x)}{5}+x$

$\sf\implies\:\dfrac{2-10}{5}\leqslant\dfrac{-3x+5x}{5}$

$\sf\implies\:\dfrac{-8}{5}\leqslant\dfrac{2x}{5}$

Then ,

$\sf\implies-4\leqslant\:x$

$\sf\implies\:x\geqslant-4...(2)$

From Equation (1) & (2)

x ∈ [-4,5)

Therefore, the solution set of given in-equation is the interval [-4,5)