Can someone help me solve this algebraically? (x+2)/(x-8)>(x)/(x+5)
=(x+2)(x+5)>x(x-8) x\[\neq\]8,-5 =x^2 + 7x+10>x^2-8x =15x>-10 =x>-2/3 where x\[\neq\]8
Thank you!
i think in interval notation it is \[(-5,-\frac{2}{3})\cup (8,\infty)\]
you cannot multiply both sides by the variable so solve these, because you don't know if it is positive or negative
you have to start with \[\frac{x+2}{x-8}-\frac{x}{x+5}>0\] and go from there
so i am afraid your first solution is incorrect. can you continue?
Ok. Well, your answer does look more like the examples in the book. I'm afraid I can't continue, I really don't know what i'm doing. What would I do next?
you have to subtract on the left. \[\frac{(x+2)(x+5)-x(x-8)}{(x-8)(x+5)}>0\] then multiply out in the numerator and combine like terms
\[\frac{x^2+7x+10-x^2+8x}{(x-8)(x+5)}>0\] \[\frac{15x+10}{(x-8)(x+5)}>0\]
now the zeros of the factors are at \[-5,-\frac{2}{3},8\]so you have to make 4 intervals
\[(-\infty, -5),(-5,-\frac{2}{3}),(-\frac{2}{3},8),(8,\infty)\]
then check on one interval by picking a number for x and substituting in the expression to see if it is positive or negative. it will change signs over each interval. if you do that you will see that the answer i wrote above is correct.
Wow. That's wonderful! Thank you! I will check to make sure it is.
easiest number to check is at x = 0 on the interval \[(-\frac{2}{3},8)\] if you replace x by 0 you get \[\frac{10}{-8\times 5}\] which is surely negative. you want positive because you have greater than zero, so pick the appropriate intervals
Thank you. Did you happen to see my other question? If the total weekly cost for a company to produce its product as a function of the number of units produced is given by the equation f(x)=2x+500, write an equation to represent the average cost per unit.
I'm suppose to write an equation and the answer someone else gave me was f(x)=2+(500/x). Does that look right?
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