Question

Solve this question from intervals

Answer 1

Question :

If $\sf \dfrac{2x - 3}{4} + 1 > 2 + \dfrac{1 + 4x}{3}$ Then find the interval in which x lies .

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 :

$\sf \dfrac{2x - 3}{4} + 1 > 2 + \dfrac{1 + 4x}{3}$

$\implies \sf\frac{2x - 3 + 4}{4} > \frac{6 + 1 + 4x}{3}$

$\implies \sf \frac{2x + 1}{4} > \frac{7 + 4x}{3}$

$\implies6x + 3 > 28 + 16x$

$\implies6x - 16x > 28 - 3$

$\implies - 10x > 25$

$\implies10x <- 25$

$\implies x < \frac{-25}{10}$

$\implies x < -2.5$

⇒x ∈(-∞, -2.5)

Therefore, the solution set of given inequation is the interval (-∞, -2.5)

Answer 2

Solution :-

$\frac{2x - 3}{4} + 1 > 2 + \frac{1 + 4x}{3} \\ \\ \implies \frac{2x - 3 + 4}{4} > \dfrac{6 + 1 + 4x}{3} \\ \\ \implies \dfrac{2x + 1}{4} > \dfrac{7 + 4x}{3} \\ \\ \: by \: cross \: multiplication\implies \\ \\ \implies3(2x + 1) > 4(7 + 4x) \\ \\ \implies6x + 3 > 28 + 16x \\ \\ \implies6x - 16x > 28 - 3 \\ \\ \implies - 10x > 25 \\ \\ \implies \: - x > \dfrac{25}{10} \\ \\ \implies \: x < - \frac{5}{2} \\ \implies \: x\in( - \infty , \frac{ - 5}{2} )$

Therefore,

${\purple{\boxed{\mid{\bold{The \: interval \: in \: which \: x \: lies ( - \infty , \frac{ - 5}{2} )}}}}}$