For this problem, we're going to take everything to one side and get it less than or equal to 0. We're going to get a common denominator and combine our numerators. So three X -, 5 X -2 X 5 -, a negative 2, positive seven. We're then going to find our four important pieces. Vertical asymptote comes from setting the denominator equal to 0 and solving the X intercept comes from setting the numerator equal to 0 in solving the horizontal asymptote. In this case the degree on top and degree on bottom is the same, so it's the leading coefficients and the Y intercept we get from setting X equal to 0. So if we put all these pieces onto our graph, we know our vertical asymptote X equal to 5th X intercept of seven halves. Remember, this is not to scale horizontal asymptote at -2 fifths and AY intercept at zero -7 halves. So when we're way out, when X is going out to Infinity, we know that our Y has to be approaching that asymptote when we get to that X intercept. That X intercept came from this numerator term, and if we look at its multiplicity, it's an understood one. So if the Y values are on one side are negative, they need to be positive on the other side. That vertical asymptote came from the denominator. It's got an understood multiplicity of one. So if the Y values on one side are positive, the Y values on the other side are negative. We have to go through the Y intercept and then when we go out to negative Infinity, we've got to get close to the asymptote again. So our solution set, we want to know where is the graph less than or equal to 0 or where are YS negative. So our YS are negative from negative Infinity to this asymptote here, which is at 2/5. And because it's an asymptote, we can't include it. So we're going to use a parenthesis union. It starts again at this X intercept. Because it's an intercept, it does equal 0 there. And we're going to use a bracket 7 halves to Infinity. So those are the locations where our Y values are less than 0. They're underneath the X axis.