Question

What is the solution set for $${\displaystyle \frac{2x-3}{3x-7}}>0$$?
1. $$(-\mathrm{\infty },{\displaystyle \frac{1}{2}})\cup ({\displaystyle \frac{2}{3}},\mathrm{\infty })$$
2. $$(-\mathrm{\infty },{\displaystyle \frac{3}{2}})\cup ({\displaystyle \frac{7}{4}},\mathrm{\infty })$$
3. $$(-\mathrm{\infty },{\displaystyle \frac{3}{2}})\cup ({\displaystyle \frac{7}{3}},\mathrm{\infty })$$
4. $$(-\mathrm{\infty },{\displaystyle \frac{-1}{2}})\cup ({\displaystyle \frac{7}{4}},\mathrm{\infty })$$

Answer 1

Hint: This problem is based on linear inequality. The inequality holds when both numerator as well as denominator is greater than or equal to 0. Therefore, we get two cases in this problem and by solving each case we get the value of x from that value we get the range of x.

Complete step-by-step solution:
In this problem, it is given that $${\displaystyle \frac{2x-3}{3x-7}}>0$$
It is a type of linear inequality, linear inequality means the linear equation is described in term of mathematics when two algebraic expressions are represented with symbols like$$>$$,$$<$$ or$$\le ,\ge$$ they can be called an inequality (general definition).
Above problem is given as $${\displaystyle \frac{2x-3}{3x-7}}>0$$
As you observe in this equation then you can observe that the equation can only be satisfied when the numerator as well as denominator must be positive that means it is greater than zero. Another case is that numerator as well as denominator both are negative that means it should be less than zero.
Case 1:
That is $$(2x-3)>0$$and $$(3x-7)>0$$
If$$x>{\displaystyle \frac{7}{3}}$$, then from given equation $$x>{\displaystyle \frac{3}{2}}$$which give us $$x>{\displaystyle \frac{7}{3}}$$
That means the coordinate becomes $$({\displaystyle \frac{7}{3}},\mathrm{\infty })$$
Case 2:
That is $$(2x-3)<0$$and $$(3x-7)<0$$
If$$x<{\displaystyle \frac{7}{3}}$$, then from given equation $$x<{\displaystyle \frac{3}{2}}$$which give us $$x<{\displaystyle \frac{3}{2}}$$
That means the coordinate becomes $$(-\mathrm{\infty },{\displaystyle \frac{3}{2}})$$
Therefore, $$x\in (-\mathrm{\infty },{\displaystyle \frac{3}{2}})\cup ({\displaystyle \frac{7}{3}},\mathrm{\infty })$$
So, the correct option is “option 3”.

Note: So far, everything you've learned about solving linear equations has also been applied to solving inequality. The only difference is that when multiplying or dividing by a negative number, you must also reverse the inequality sign. If both the numerator and denominator are less than or higher than 0, the inequality holds for the problem above. When dividing the coefficient of the variable on both sides of a linear inequality, keep in mind that if the coefficient is positive, the inequality's direction will not change. If the coefficient is negative, the inequality's direction will shift. Finally, you must plot all of the findings on a number line to obtain an interval solution.