#### Тема: OECD.org - Employment

This worksheet generator makes worksheets for four different fraction topics: (1) equivalent fractions, (2) simplifying fractions, (3) converting fractions to mixed numbers, and (4) converting mixed numbers to fractions.

You can control the ranges for the whole number part, numerator and denominator, font size, number of problems, and more. You can force the answers not to be simplified, which is useful for grade 4 when students learn to write mixed numbers as fractions and vice versa but haven't yet learned about simplifying.

In other words, we want to move everything except x (or whatever name the variable has) over to the right hand side.

That was interesting. we thought we had found a solution, but when we looked back at the question and found it wasn't allowed!

In other words, we want to move everything except x (or whatever name the variable has) over to the right hand side.

That was interesting. we thought we had found a solution, but when we looked back at the question and found it wasn''''t allowed!

In other words, we want to move everything except x (or whatever name the variable has) over to the right hand side.

That was interesting. we thought we had found a solution, but when we looked back at the question and found it wasn''t allowed!

In other words, we want to move everything except x (or whatever name the variable has) over to the right hand side.

That was interesting. we thought we had found a solution, but when we looked back at the question and found it wasn''''''''''''''''t allowed!

The smartest way, ESPECIALLY since fractions cause you difficulties, is to clear the fractions away as your very first step. How? You might well ask that, eh? Find a number that can eliminate all the fractions if you multiply them by it. For instance, if you have 3/5 and 4/3, that number will be 15. To be snap-on, so to speak, and tricky looking, you want the lowest number that can do it (like 15). But to be honest, who cares, right? That was literally for the days when you weren t allowed to use a calculator. So if multiplying by 30 or 105 is clearer to you, or you notice 30 will do it before you realize 15 will, just go with it. multiply those two fractions by 15 and you get 15*(3/5) = 45/5 = 9 and 15*(4/3) = 60/3 = 20. So you end up with no fractions. Do it by 30 and you get 18 and 40, also NO fractions. So it hardly matters if you are tricky looking, you just want to get it done. Speaking of which: Your fractions are eigths and fourths so if you multiply by 8 (or 72 or 800) you make the fractions go away. For ease in multiplying, convert from those cr*ppy mixed fractions into "improper fractions." For instance: 2 3/8 = 2 + 3/8 2 = 2 * 8/8 = 16/8 so 2 + 3/8 = 16/8 + 3/8 = 19/8 Similarly, 5 3/4 = 23/4. So now you have: 19/8 + y = 23/4..multiply BOTH sides by 8 to clean out the fractions 8 * 19/8 + 8*y = 8 * 23/4..evaluate them 19 * 8/8 + 8y = 23 * 8/4..keep going 19 * 1 + 8y = 23 * 2..one more time 19 + 8y = 46 Now just solve that (subtract 19 from each side: 8y = 27 and divide each side by 8: y = 27/8) and you re done. So much easier once the nasty fractions are gone. Oh yeah, the slugs teaching you are using textbooks that HAVE to differ from yesterday s textbooks or why would schools buy them? So today s difference is that you can t ever leave something as an improper fraction, you must convert it to a mixed fraction or your priest will force a ton of pennance upon you. So the answer has one more step, converting 27/8 to 3 3/8. y = 3 3/8. Isn t it great you are getting taught poorly so that some textbook company can sell a ton of new textbooks and your parents think they ve done you well because "new" = "better"? Or. would being taught so it s understandable be better? Hmmm. Added, and nothing to do with the question, but: My GOD you re beautiful!!! Keep your chin up babe, college is even more fun!

Hi Elizabeth D, For 1), we have ((2/x) - (2/3)) / (x - 3). Notice that if you substitute x = 3 directly in, you will have zero over zero. This is where L Hospital s rule come in. If we have such a case of zero over zero, or a case where we have infinite over infinite, we can apply the L Hospital rule. According to the rule, what we do is to differentiate both the numerator and the denominator. The new derivatives will have the same limit as the old one. For example, since we have zero over zero now, we will apply L Hospital s rule and differentiate ((2/x) - (2/3)) as well as (x - 3). d/dx ((2/x) - (2/3)) = -2/x^2 d/dx (x - 3) = 1 Thus, the limit of ((2/x)-(2/3))/(x-3) as x approaches 3 is the same as the limit of (-2/x^2) / 1 as x approaches 3. The limit of (-2/x^2) / 1 as approaches 3 is clearly -2/9. Hence, we are done. For 2), you did not state what is f(x), so I have no idea how to solve it, since there are various different possible scenarios. For 3), we will have to split the limit into two parts. Notice that if x < -3, |x + 3| / (2x + 6) is the same as -(x + 3) / (2x + 6). But if x > -3, |x + 3| / (2x + 6) is the same as (x + 3) / (2x + 6) You may not have learnt about it, but we will need to apply one-sided limits in this case. -(x + 3) / (2x + 6) = -(x + 3) / 2(x + 3) = -1/2 (x + 3) / (2x + 6) = (x + 3) / 2(x + 3) = 1/2 Hence, the limit of |x + 3| / (2x + 6) as x approaches -3 from the left is -1/2 and the limit of |x + 3| / (2x + 6) as x approaches -3 from the right is 1/2. Cheers.

In other words, we want to move everything except x (or whatever name the variable has) over to the right hand side.

That was interesting. we thought we had found a solution, but when we looked back at the question and found it wasn''''''''t allowed!