$\displaystyle \frac{5x^2 - 20x - 36}{x-5} \leq 4x $
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Originally Posted by BadMaterial $\displaystyle \frac{5x^2 - 20x - 36}{x-5} \leq 4x $ $\displaystyle \frac{5x^2 - 20x - 36}{x-5}\leq 4x$ $\displaystyle \frac{5x^2 - 20x - 36}{x-5}- 4x\leq 0$ $\displaystyle \frac{x^2 - 36}{x-5}\leq 0$ Now solve it.
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