Question

Solve them all.....​

Answer 1

Answer: (All answers are embedded in the explanation)

P.s consider being more generous on rewards when asking such lengthy questions ;)

Step-by-step explanation:

1) $\frac{2x-7}{3x-7} >0$

A strictly greater than sign implies that the terms in numerator and denominator must have the same sign. So we need to solve for two cases: when both the terms are strictly positive, and when both are strictly negative.

Case 1: When both terms are positive-

⇒(2x-7)>0 and (3x-7)>0

⇒x>7/2 and x>7/3

⇒x>7/2 (naturally, any number greater than 7/2 will be greater than 7/3, but not vice versa)

Thus, we get x∈(7/2, ∞) -----(1)

Case 2: When both are negative-

⇒(2x-7)<0 and 3x-7<0

⇒x<7/2 and x<7/3

⇒x<7/3 (reverse the reasoning used above)

Thus we get x∈(-∞, 7/3) -----(2)

From (1) and (2), the possible solution set for the given inequality is

Solution 1: x ∈ (-∞, 7/3) ∪ (7/2, ∞)

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