What is the solution set for 2x-3-3x-7-0-

The correct option is C ( ∞, 32)∪(73, ∞) Given 2x 3/3x 7 gt;0 Above inequality holds if both numerator and denominator are less than 0 or gre ...

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Standard XII

Mathematics

Algebra of Derivatives

What is the s...

Question

What is the solution set for $\frac{2 x - 3}{3 x - 7} > 0$ ?

(- \infty, - \frac{1}{2}) \cup (\frac{2}{3}, \infty)

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(- \infty, - \frac{3}{2}) \cup (\frac{7}{4}, \infty)

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(- \infty, \frac{3}{2}) \cup (\frac{7}{3}, \infty)

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(- \infty, - \frac{1}{2}) \cup (\frac{7}{4}, \infty)

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Solution

The correct option is C $(- \infty, \frac{3}{2}) \cup (\frac{7}{3}, \infty)$
Given $\frac{2 x - 3}{3 x - 7} > 0$
Above inequality holds if both numerator and denominator are less than $0$ or greater than $0$
i.e $(2 x - 3) > 0$ and $(3 x - 7) > 0$ OR $(2 x - 3) < 0$ and $(3 x - 7) < 0$
If $x > \frac{7}{3}$ , then from given equation $x > \frac{3}{2}$ , which give us $x > \frac{7}{3}$

If $x < \frac{7}{3}$ , then from given equation $x < \frac{3}{2}$ , it give us $x < \frac{3}{2}$

Therefore, $x \in (- \infty, \frac{3}{2}) \cup (\frac{7}{3}, \infty)$

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Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Question 22

Which of the following is an example of distributive property of multiplication over addition for rational numbers.
(a) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = [- \frac{1}{4} \times \frac{2}{3}] + [- \frac{1}{4} \times (\frac{- 4}{7})]$
(b) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = [\frac{1}{4} \times \frac{2}{3}] - (\frac{- 4}{7})$
(c) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = \frac{2}{3} + (- \frac{1}{4}) \times \frac{- 4}{7}$
(d) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = {\frac{2}{3} + (\frac{- 4}{7})} - \frac{1}{4}$