Question

$\text{For the given inequality find the value of}x?$

Answer 1

For the given inequality we will solve one bye one

$a)x+5>3x+7$
Now, we know the property that, adding or subtracting any constant or variable does not change it's sign.
$\therefore$ Subtracting $7$ from both sides of the inequality, we get:
$x+5-7>3x+7-7\Rightarrow x-2>3x$
Now, again subtracting $x$ from both sides, we get:
$x-2-x>3x-x\Rightarrow -2>2x\Rightarrow -1>x$
$\Rightarrow x\in (-\infty ,-1)$

$b)3x-5\le 5x-4$
Now, we know the property that, adding or subtracting any constant or variable does not change it's sign.
$\therefore$ Adding $4$ on both sides of the inequality, we get:
$3x-5+4\le 5x-4+4\Rightarrow 3x-1\le 5x$
Now, again subtracting $3x$ on both sides, we get:
$3x-1-3x\le 5x-3x\Rightarrow -1\le 2x\Rightarrow -\frac{1}{2}\le x$
$\Rightarrow x\in [-\frac{1}{2},\infty )$

$c)2x+5<11x-1$
Now, we know the property that, adding or subtracting any constant or variable does not change it's sign.
$\therefore$ Adding $1$ on both sides of the inequality, we get:
$2x+5+1<11x-1+1\Rightarrow 2x+6<11x$
Now, again subtracting $2x$ from both sides, we get:
$2x+6-2x<11x-2x\Rightarrow 6<9x\Rightarrow \frac{2}{3}<x$
$\Rightarrow x\in (\frac{2}{3},\infty )$

$d)9x-21\ge 5x+23$
Now, we know the property that, adding or subtracting any constant or variable does not change it's sign.
$\therefore$ Adding $21$ on both sides of the inequality, we get:
$9x-21+21\ge 5x+23+21\Rightarrow 9x\ge 5x+44$
Now, again subtracting $5x$ on both sides, we get:
$9x-5x\ge 5x+44-5x\Rightarrow 4x\ge 44\Rightarrow x\ge 11$
$\Rightarrow x\in [11,\infty )$