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 Pre-Calculus Pre-Calculus Math Forum

 May 19th, 2018, 07:05 PM #1 Newbie   Joined: May 2018 From: notimportant Posts: 9 Thanks: 0 Could you please help me to solve this? Could you please help me to solve these inequations..? I have many more and won't be able to solve them all  .. I need help. Last edited by skipjack; May 20th, 2018 at 11:13 AM. May 19th, 2018, 07:29 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 If there are specific questions you have regarding specific problems we can possibly help. We don't do your homework for you. May 19th, 2018, 08:17 PM   #3
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 Originally Posted by romsek If there are specific questions you have regarding specific problems we can possibly help. We don't do your homework for you.
The homework is around 100 problems, I only posted 7. May 19th, 2018, 08:22 PM   #4
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 Originally Posted by moredumbimpossible The homework is around 100 problems, I only posted 7.
These are all pretty straightforward linear inequalities.

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{a-1}{3}; \\ \\ &3(a+2) \leq 4(a-1); \\ \\ &3a+6 \leq 4a - 4; \\ \\ &10 \leq a \end{align*}

None of the rest of them are any more complicated. May 19th, 2018, 08:53 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 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. Last edited by JeffM1; May 19th, 2018 at 09:06 PM. May 19th, 2018, 10:13 PM #6 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 I ordinarily would never comment on someone else's method but ... really? $\dfrac{4x-3}{3} < \dfrac{2x+4}{4}$ $4(4x-3) < 3(2x+4)$ $16x-12 < 6x + 12$ $10x < 24$ $x < \dfrac{24}{10} = \dfrac{12}{5}$ sometimes it pays to just keep it simple. Thanks from Denis May 20th, 2018, 09:05 AM #7 Newbie   Joined: May 2018 From: notimportant Posts: 9 Thanks: 0 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...  Last edited by skipjack; May 20th, 2018 at 11:12 AM. May 20th, 2018, 10:19 AM #8 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 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. Last edited by skipjack; May 20th, 2018 at 11:15 AM. May 20th, 2018, 10:27 AM #9 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 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. Thanks from romsek May 21st, 2018, 10:07 AM   #10
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 Originally Posted by romsek 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.
I checked on a book, and I solved those applying notable products.
(a+b)(a-b)...
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.. Tags solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 06:11 AM

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