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This is one of our many quizzes on Solving inequalities. Click on this to access more such competitive quizzes on Algebra.

Solving inequalities Quiz 2

mathematics

 Solve the given inequality and find the interval of $x: \left( {\dfrac{{x - 2}}{5}} \right) > \left( {\dfrac{{2 - x}}{3}} \right)$

 Cross multiply and further solve for like-equality.

Cross Multiplying the denominators $ \\ \left( {3x - 6} \right) > \left( {10 - 5x} \right) \\ 8x > 16 \\ x > 2 \\ $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cfed3ad1-6237-4708-a720-3b53191894372487572490835944234.png"/>

Solve the given equation and find all the satisfying values of $x : \dfrac{{(x - 10)}}{{(x + 8)}} \geqslant 0$, if $x$ is not equal to $ - 8$ .

 Check for the cases in the different range of $x$ where the given inequality is satisfied.

$\dfrac{{(x - 10)}}{{(x + 8)}} \geqslant 0 …..(i) \\ $ The given expression is equal to zero for $x = 10$. $ \\ $ The given expression will be positive when, $ \\ $ Case 1: $ \\ \left( {x - 10} \right) > 0\;\;\;and\;\;\;\left( {x + 8} \right) > 0 \\ \Rightarrow x > 10\;\;\;and\;\;\;x > - 8 \\ \Rightarrow x > 10 \\ $ Case 2: $ \\ \left( {x - 10} \right) < 0\;\;\;and\;\;\;\left( {x + 8} \right) < 0 \\ \Rightarrow x < 10\;\;\;and\;\;\;x < - 8 \\ \Rightarrow x < - 8 \\ $ Thus the final answer is $x \in \left( { - \infty , - 8} \right) \cup \left[ {10,\infty } \right)$ $ \\ $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/150d3c47-e074-4e85-a0a7-2a1b1d0356088081472565780894821.png"/>

 Solve the set of linear in equations and find the value of $x$ or their corresponding intervals: $(x - 2) \geqslant 0$, $(2x - 8) \geqslant 0$

 Solve the above two as linear equations and find the common interval.

$ x - 2 \geqslant 0 \\ x \geqslant 2 \\ $ And $ 2x - 8 \geqslant 0 \\ 2x \geqslant 8 \\ x \geqslant 4 \\ $ Taking the common interval between both equation $(1)$ and $(2)$ we get, $x \geqslant 4$ $ \\ $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d45c9d6d-4f1e-4091-8a2e-0924fd5b3c074601792840776771807.png"/>

Solve $\dfrac{{|x| - 5}}{{|x| - 3}} > 0$ for $x$, if $x$ is not equal to $3$

 Use different cases to find the range of $x$ which satisfies the inequality.

Given $\dfrac{{|x| - 5}}{{|x| - 3}} > 0$ $ \\ $ Case 1: $ \\ \left| x \right| - 5 > 0\;\;\;and\;\;\;\left| x \right| - 3 > 0 \\ \Rightarrow \left| x \right| > 5\;\;\;and\;\;\;\left| x \right| > 3 \\ \Rightarrow \left( {x > 5\;or\;x < - 5} \right)\;\;and\;\;\left( {x > 3\;or\;x < - 3} \right) \\ \Rightarrow x \in \left( { - \infty , - 5} \right) \cup \left( {5,\infty } \right) \\ $ Case 2: $ \\ \left| x \right| - 5 < 0\;\;\;and\;\;\;\left| x \right| - 3 < 0 \\ \Rightarrow \left| x \right| < 5\;\;\;and\;\;\;\left| x \right| < 3 \\ \Rightarrow \left( { - 5 < x < 5} \right)\;\;and\;\;\left( {\; - 3 < x < 3} \right) \\ \Rightarrow x \in \left( { - 3,3} \right) \\ $ Thus the final result is $x \in \left( { - \infty , - 5} \right) \cup \left( { - 3,3} \right) \cup \left( {5,\infty } \right)$.

Solve for value of $x$ for $\sin x > 0$

 Use range of $\sin x$ given as $ - 1 \leqslant \sin x \leqslant 1$

We know that $ - 1 \leqslant \sin x \leqslant 1$ $ \\ \sin x > 0 \\ 0 < \sin x \leqslant 1 \\ 0 < x \leqslant \dfrac{\pi }{2} \\ x \in \left( {0,\dfrac{\pi }{2}} \right] \\ $<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d2a483c8-d167-4dfa-a418-289313e560c53175438471424119830.png"/>