Solve for x
x\in \left(-\infty,-\frac{1}{2}\right)\cup \left(3,\infty\right)
Graph
Share
Copied to clipboard
2x+1<0 x-3<0
For the quotient to be positive, 2x+1 and x-3 have to be both negative or both positive. Consider the case when 2x+1 and x-3 are both negative.
x<-\frac{1}{2}
The solution satisfying both inequalities is x<-\frac{1}{2}.
x-3>0 2x+1>0
Consider the case when 2x+1 and x-3 are both positive.
x>3
The solution satisfying both inequalities is x>3.
x<-\frac{1}{2}\text{; }x>3
The final solution is the union of the obtained solutions.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}