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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequalities and represent the solution graphically on number line. $ 5(2x-7) - 3(2x+3) \leq 0 ; 2x+19 \geq 6x+47 $

$\begin{array}{1 1}(A)\;[-7,11] \\(B)\;(-5,5)\\(C)\;(-23,2]\\(D)\;[2,3]\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 $'>'$ .
  • To represent solution of linear inequality involving one variable on a number line, if the inequality involves $\geq $ or $\leq$ are draw filled circle (0) on the number is included in the solution set.
  • If the inequality involves $'>'$ or $'<'$ we draw open circle (0) on the number line to indicate the number is excluded from the solution set.
Step 1:
The first inequality is $5(2x-7)-3(2x+3) \leq 0$
$=> 10x -35-6x-9 \leq 0$
$=> 4x -44 \leq 0$
Adding 44 on both sides of the inequality
$ => 4x -44 +44 \leq 44$
$=> 4x \leq 44$
Divide both sides by positive number 4 .
$=> x \leq 11$------(1)
Step 2:
The second inequality is $2x +19 \leq 6x+47$
Adding -47 to both sides of inequality .
Adding -2x to both sides of inequality
$=> 2x+19-47 -2x \leq 6x +47 -47 -2x$
$=> 19-47 \leq 4x$
$ -28 \leq 4x$
Dividing by positive number 4 on both sides of inequality.
$-7 \leq x$ -----(2)
Step 3:
From (1) and (2) it can be conclude that all numbers greater than or eqaual to -7 and less than or equal to 11 are solutions to given system of inequalities
The solution set is $[-7,11]$
The solution is represented graphically as the number line as follows,
answered Aug 5, 2014 by meena.p
 
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