Browse Questions

# Solve the inequality for real $x: 2(2x+3) -10 < 6(x-2)$

$\begin{array}{1 1} (A)\;(4,\infty)\\(B)\;(\infty,-6)\\(C)\;(-4,\infty)\\(D)\; (-9,\infty)\end{array}$

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
The given inequlaity is $2(2x+3) -10 < 6 (x-2)$
$=> 4x+6 -10 < 6 (x-2)$
$=> 4x-4 < 6x -12$
$=> 4x-4+12 < 6x-12+12$
Adding $-4x$ on both sides,
=> $4x+8-4x < 6x -4x$
=> $8 < 2x$
dividing by a positive number 2 on both sides, => $4 < x$
Step 2:
All numbers greater than $4$ satisfy the given inequality
The solution set is $(4 , \infty)$
Hence A is the correct answer.