what is the solution of $\frac{1}{x}+\frac{1}{3}<\frac{1}{5}$?\n$-\frac{15}{2}<x<0$\n$-\frac{15}{2}leq…

Answer:

$x<-\frac{15}{2}\text{ or }x > 0$

Explanation:

Step1: Combine fractions

$\frac{1}{x}+\frac{1}{3}<\frac{1}{5}$ becomes $\frac{3 + x}{3x}<\frac{1}{5}$.

Step2: Cross - multiply

$5(3 + x)<3x$ (note we need to consider sign of $x$ later). Expanding gives $15+5x<3x$.

Step3: Solve for $x$

Subtract $5x$ from both sides: $15<3x - 5x$, so $15<-2x$. Then divide by $- 2$ and reverse the inequality sign, getting $x<-\frac{15}{2}$.

Step4: Consider domain and critical points

The original inequality $\frac{1}{x}+\frac{1}{3}<\frac{1}{5}$ has a critical point at $x = 0$ (since $\frac{1}{x}$ is undefined at $x = 0$). Also, when we cross - multiplied we assumed $3x>0$. When $x>0$, the inequality $\frac{3 + x}{3x}<\frac{1}{5}$ still holds for some values. Testing intervals, we find that $x>0$ is also part of the solution set. So the solution is $x<-\frac{15}{2}\text{ or }x > 0$.