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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the given inequality and show the graph of the solution on number line: $ \large\frac{x}{2} <\frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

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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 given inequality is : $ \large\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$
$=> \large\frac{x}{2} \geq \frac{5(5x-2)-3(7x-3)}{15}$
$=> \large\frac{x}{2} \geq \large\frac{25 x -10 -21 x +9}{15}$
$=> \large\frac{x}{2} \geq \large\frac{4x-5}{15}$
Multiplying both sides of the inequality by positive number 30.
$=> 15 x \geq 2(4x-1)$
$15 x \geq 8x-2$
Adding $-8x$ on both sides of the inequality
$=> 15 x - 8x \geq 8x -2 -8x$
=> $ 7x \geq -2$
Dividing both sides of the inequality by positive number 7.
$=> x \geq \large\frac{-2}{7}$
Step 2:
All number greater than or equal to $\large\frac{-2}{7}$from the solution for the given inequality .
The solution set is $[-2/7, \infty)$
The graphical representation is
answered Jul 28, 2014 by meena.p
 
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