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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the given inequalities for real $x: 37 -(3x+5) \geq 9x -8 (x-3)$

$\begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,2]\\(C)\;(-4,\infty)\\(D)\; (-9,\infty]\end{array} $

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1 Answer

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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 inequality is $37-(3x+5) \geq 9x-8(x-3)$
=> $ 37-3x-5 \geq 9x-8x+24$
=> $ 32 -3x \geq x+24$
Adding 3x and subtracting 24 from both sides,
=> $32 -24 \geq x +3x$
=> $ 8 \geq 4x$
=> $ 2 \geq x$
Step 2:
All real number greater than or equal to 2 are solutions of the given inequality
The solution set is $(- \infty , 2]$
Hence B is the correct answer.
answered Jul 25, 2014 by meena.p
edited Jul 26, 2014 by meena.p
 
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