Comments on: Mr Brewin’s Y11 – Past Paper for Easter http://www.shsmaths.com/2010/04/mr-brewins-y11-past-paper-for-easter/ Tue, 07 Feb 2012 21:11:26 +0000 hourly 1 http://wordpress.org/?v=3.0.4 By: Mr Brewin http://www.shsmaths.com/2010/04/mr-brewins-y11-past-paper-for-easter/comment-page-1/#comment-6716 Mr Brewin Mon, 26 Apr 2010 19:37:01 +0000 http://www.shsmaths.com/?p=1511#comment-6716 Yeah, what she said ;-) 15 is LCM of 3 and 5, so it becomes [15 * (3x+9) / 3] - [15 * (2x +4)5] = 60 And that will then go to 5(3x+9) - 3(2x+4) = 60 OK? Yeah, what she said ;-)

15 is LCM of 3 and 5, so it becomes [15 * (3x+9) / 3] – [15 * (2x +4)5] = 60

And that will then go to 5(3x+9) – 3(2x+4) = 60

OK?

]]> By: Mrs Tibble http://www.shsmaths.com/2010/04/mr-brewins-y11-past-paper-for-easter/comment-page-1/#comment-6715 Mrs Tibble Mon, 26 Apr 2010 19:33:31 +0000 http://www.shsmaths.com/?p=1511#comment-6715 Hi Lily, Are those dividing lines under two terms? ie is it (3x + 9)/3 - (2x + 4)/5 = 4 If so, multiply every term by 15 to get rid of the denominators, then solve in the normal way. JT Hi Lily,

Are those dividing lines under two terms? ie is it

(3x + 9)/3 – (2x + 4)/5 = 4

If so, multiply every term by 15 to get rid of the denominators, then solve in the normal way.

JT

]]> By: Lily Bridger http://www.shsmaths.com/2010/04/mr-brewins-y11-past-paper-for-easter/comment-page-1/#comment-6714 Lily Bridger Mon, 26 Apr 2010 18:50:52 +0000 http://www.shsmaths.com/?p=1511#comment-6714 Sir im doing some revision on solving equations and i can't answer the fraction equation: 3x+9/3 - 2x+4/5 = 4 Please can you explain how to answer this question? Sir im doing some revision on solving equations and i can’t answer the fraction equation:
3x+9/3 – 2x+4/5 = 4
Please can you explain how to answer this question?

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