how to solve x^2 less then or equal to 5x?

x^2 <= 5x

x^2 - 5x <= 0

x(x - 5) <= 0

this will be true when x is positive, and x - 5 is negative (between the zeros of the graph of y = x(x - 5) where the graph is below the x-axis

so 0 <= x <= 5

or, [ 0 , 5 ] in interval notation

x^2 <= 5x

x^2 - 5x < = 0

x(x - 5) < 0

x < 0 or x < 5

x < 5 satisfies x < 0 so x < 5 is the solution.

x must be 5 or less. just think about it-- what times itself is less than or equal to 5 times something? It has to be 5.