Solve for x
x\in [-\frac{1}{2},\frac{3}{5})
Graph
Share
Copied to clipboard
2x+1\leq 0 3-5x<0
For the quotient to be ≥0, 2x+1 and 3-5x have to be both ≤0 or both ≥0, and 3-5x cannot be zero. Consider the case when 2x+1\leq 0 and 3-5x is negative.
x\in \emptyset
This is false for any x.
2x+1\geq 0 3-5x>0
Consider the case when 2x+1\geq 0 and 3-5x is positive.
x\in [-\frac{1}{2},\frac{3}{5})
The solution satisfying both inequalities is x\in \left[-\frac{1}{2},\frac{3}{5}\right).
x\in [-\frac{1}{2},\frac{3}{5})
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}