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Q.Solve 7x−58x+3>4\dfrac{7x-5}{8x+3} > 4, x∈Rx \in \mathbb{R}.

Nagaland NbseNagaland Board of School Education (Class XI) 2024Subjective· 4mImportance★★★★★
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Move everything to one side to get a single rational expression >0>0 (or <0<0), then use a sign chart based on the critical points.

7x−58x+3>4\dfrac{7x-5}{8x+3} > 4

7x−58x+3−4>0\dfrac{7x-5}{8x+3} - 4 > 0

(7x−5)−4(8x+3)8x+3>0\dfrac{(7x-5)-4(8x+3)}{8x+3} > 0

7x−5−32x−128x+3>0\dfrac{7x-5-32x-12}{8x+3} > 0

−25x−178x+3>0\dfrac{-25x-17}{8x+3} > 0

Multiply both sides by −1-1 (reverse the inequality):

25x+178x+3<0\dfrac{25x+17}{8x+3} < 0

Critical points: 25x+17=0⇒x=−1725=−0.6825x+17=0\Rightarrow x=-\dfrac{17}{25}=-0.68;  8x+3=0⇒x=−38=−0.375\ 8x+3=0\Rightarrow x=-\dfrac{3}{8}=-0.375.

Since −1725<−38-\dfrac{17}{25} < -\dfrac{3}{8}, test the three intervals:

  • x<−1725x<-\dfrac{17}{25}: both factors negative ⇒\Rightarrow ratio positive (not <0<0) …

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