Could you please help me to solve this? http://i.imgur.com/o5CLFI7.jpg Could you please help me to solve these inequations..? I have many more and won't be able to solve them all :eek::eek:.. I need help. 
If there are specific questions you have regarding specific problems we can possibly help. We don't do your homework for you. 
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In all cases you can do the algebra to reduce it to a simple statement of inequality. For example the first one $\begin{align*} &\dfrac{a+2}{4} \leq \dfrac{a1}{3}; \\ \\ &3(a+2) \leq 4(a1); \\ \\ &3a+6 \leq 4a  4; \\ \\ &10 \leq a \end{align*}$ None of the rest of them are any more complicated. 
You can always deal with an inequation by solving the corresponding equation. Example. $\text {In what interval or intervals is } \dfrac{4x  3}{3} < \dfrac{2x + 4}{4}.$ Start by solving $\dfrac{4x  3}{3} = \dfrac{2x + 4}{4} \implies 16x  12 = 6x + 12 \implies$ $10x = 24 \implies x = 2.4.$ Is 2.4 in the interval you want? No because we are looking for less than, not less than or equal. If we were looking for less than or equal, 2.4 would be in an interval. You have now divided the number line into two parts, x < 2.4 and x > 2.4. Pick a number in each part and test. Well 2 is less than 2.4. $\dfrac{4 * 2  3}{3} = \dfrac{5}{3} = \dfrac{20}{12}.$ $\dfrac{2 * 2 + 4}{4} = \dfrac{8}{4} = \dfrac{24}{12}.$ $\dfrac{20}{12} < \dfrac{24}{12}.$ So numbers less than 2.4 are in the interval you want. And 3 is greater than 2.4. $\dfrac{4 * 3  3}{3} = \dfrac{9}{3} = \dfrac{36}{12}.$ $\dfrac{2 * 3 + 4}{4} = \dfrac{10}{4} = \dfrac{30}{36}.$ $\dfrac{36}{12} > \dfrac{30}{12}.$ So numbers greater than 2.4 are NOT in the interval you want. Thus the answer is $(\ \infty,\ 2.4).$ CAUTION We had one point of equality in this problem so we divided the number line into 2 parts, and we had to test each part. But if we had two points of equality, we would divide the number line into three parts and would need to test each part. If we have n points of equality, we divide the number line into n + 1 parts and must test a number in each part. EDIT: Alternatively, you can try to solve a linear inequality directly. In my example, $\dfrac{4x  3}{3} < \dfrac{2x + 4}{4} \implies \dfrac{12(4x  3)}{3} < \dfrac{12(2x + 4)}{4} \implies$ $16x  12 < 6x + 12 \implies 10x < 24 \implies x < 2.4.$ That is quicker, but it gets tricky for more complex inequalities. 
I ordinarily would never comment on someone else's method but ... really? $\dfrac{4x3}{3} < \dfrac{2x+4}{4}$ $4(4x3) < 3(2x+4)$ $16x12 < 6x + 12$ $10x < 24$ $x < \dfrac{24}{10} = \dfrac{12}{5}$ sometimes it pays to just keep it simple. 
Thanks that doesn't look that hard. Maybe you can explain how to solve notable products like this. This is the last one, I swear... :cool:http://i.imgur.com/UL46iTT.jpg 
There's nothing to solve here. These are just expressions that I guess they want you to multiply out. You can do this; you're just lazy. Get to it. 
Romsek will probably say I am being too complex. You can solve these particular problems directly and easily using basic algebra and the laws of exponents, but you can also use a method that is particularly helpful for more complex cases. Let's take the last problem as an example. Direct method. $\left ( \dfrac{x^2}{2}  \dfrac{x}{3} \right ) \left ( \dfrac{x^2}{2} + \dfrac{x}{3} \right ) = $ $ \left ( \dfrac{3x^2}{3 * 2}  \dfrac{2x}{2 * 3} \right ) \left ( \dfrac{3x^2}{3 * 2} + \dfrac{2x}{2 * 3} \right ) = $ $\dfrac{3x^2  2x}{6} * \dfrac{3x^2 + 2x}{6} =$ $\dfrac{9x^4 + 6x^3  6x^3  4x^2}{36} = \dfrac{9x^4  4x^2}{36} = \dfrac{9x^4}{36}  \dfrac{4x^2}{36} = \dfrac{x^4}{4}  \dfrac{x^2}{9}.$ Basic algebra. Or you can do this, formally or in your head. $u = \dfrac{x^2}{2} \text { and } v = \dfrac{x}{3}.$ $\therefore \left ( \dfrac{x^2}{2}  \dfrac{x}{3} \right ) \left ( \dfrac{x^2}{2} + \dfrac{x}{3} \right ) =$ $( u  v)(u + v) = u^2  v^2 =$ $\left ( \dfrac{x^2}{2} \right )^2  \left ( \dfrac{x}{3} \right )^2 = \dfrac{x^4}{4}  \dfrac{x^2}{9}.$ There are sometimes many correct ways to attack a problem. 
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(a+b)(ab)... It seems like you're the only one here so I'll assume that you are some sort of moderator too.. Please feel free to close this useless account. Bye.. 
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