Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 Topic changed, help aprectiated loads! Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 (edited) . fail hey, topic has been changed as equations are what im struggling on, probably will be another topic change later, as my exam is tommorow and... lets say it doesnt look like ill be doing good so i have tried equations and been on My Maths and it didnt explain too well, so am i right in saying that when doing an equation like 6x + 7 = 4x + 19 you would take the 4x away from both sides, and then you're left with 2x+7 = 19. Yeah that's right, no idea what the hell I was doing just then Then you take away 7 from the 19 and the 7, then your left with 2x = 12 You then divide them both by 2. and get an answer of X=6 What i dont understand is, WHY do you take 7 from the 19 and the 7? is it because its the smallest out of the 2 numbers there? And WHY do you divide them both by 2? is it because 2 is the smallest number at the end or because 2 is the one with x beside it? Thanks for any answers, im really stressed about this exam because im in a credit maths class just now and have a feeling ill end up in foundation :/ Cheers You don't take 7 from the other side. It's hard to see it at first, but what you do to one side you have to do to other side exactly. Thuing is, you are taught that one goes to the other side, but it doesn't!! Example, 6x +7 = 4x +19 To get rid of the 4x you have to minus it from the -4x -4x right yeah? well then you have to minus it from the left too. Which gives you the 2x +7 = +19 Do you get it now? Edit: the layout didn't work out how I wanted to show you. weird. Edited February 23, 2011 by dann2707 Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 (edited) . fail Meant taking the 4x away because its the smallest x there . fail You don't take 7 from the other side. It's hard to see it at first, but what you do to one side you have to do to other side exactly. Thuing is, you are taught that one goes to the other side, but it doesn't!! Example, 6x +7 = 4x +19 To get rid of the 4x you have to minus it from the -4x -4x right yeah? well then you have to minus it from the left too. Which gives you the 2x +7 = +19 Do you get it now? Edit: the layout didn't work out how I wanted to show you. weird. Yeah i didnt explain that bit too well. So, you take away the 4x from itself and from the 6x because its the smallest number, then take awya the 7 from both the 19 and 7 because its the smallest number? And i can see the way you layed it out in the quote Edited February 23, 2011 by Kieran Morrison Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 It's got nothing to do with which is smaller. It's about getting to a level where you can find X on its own. And by doing that you need to deduct the sums from each other to get a final sum. I'll do something on paint see if it makes it easier. Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 It's got nothing to do with which is smaller. It's about getting to a level where you can find X on its own. And by doing that you need to deduct the sums from each other to get a final sum. I'll do something on paint see if it makes it easier. These are just simple equations that will be in the test and we have only been taught to take awy the smallest number so far, ive got equations done except for the bracketed ones so quick question if theres a sum like this- 6(x+7) = 8(x-9) do you put 6x down then times the 6 and 7 so 6x + 42 = 8x - 72? then you just do the normal equation is it? Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 Right. the red circled bits are the steps you do everytime. You remove the smaller number so you stil have a positive number, which is easier to work with Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 Right. the red circled bits are the steps you do everytime. You remove the smaller number so you stil have a positive number, which is easier to work with Okay, i understand cheers, thats only one of the things i dont understand in the book, if theres nothing on mymaths or the internet then ill come here to a last resort so im not bothering everyone with math questions Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 These are just simple equations that will be in the test and we have only been taught to take awy the smallest number so far, ive got equations done except for the bracketed ones so quick question if theres a sum like this- 6(x+7) = 8(x-9) do you put 6x down then times the 6 and 7 so 6x + 42 = 8x - 72? then you just do the normal equation is it? Yep thats it mate sorted. Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted February 23, 2011 Report Share Posted February 23, 2011 Note that it would still work out correct if you did it the other way, taking away the larger numbers. You'd get: -12 = -2x and so x = 6. It's just generally easier to stick with positive numbers. Quote Link to comment Share on other sites More sharing options...
dirt jumper jake Posted February 23, 2011 Report Share Posted February 23, 2011 hey, topic has been changed as equations are what im struggling on, probably will be another topic change later, as my exam is tommorow and... lets say it doesnt look like ill be doing good so i have tried equations and been on My Maths and it didnt explain too well, so am i right in saying that when doing an equation like 6x + 7 = 4x + 19 you would take the 4x away from both sides, and then you're left with 2x+7 = 19. Then you take away 7 from the 19 and the 7, then your left with 2x = 12 You then divide them both by 2. and get an answer of X=6 What i dont understand is, WHY do you take 7 from the 19 and the 7? is it because its the smallest out of the 2 numbers there? And WHY do you divide them both by 2? is it because 2 is the smallest number at the end or because 2 is the one with x beside it? Thanks for any answers, im really stressed about this exam because im in a credit maths class just now and have a feeling ill end up in foundation :/ Cheers To put it simply, you have to find out what x is, and what ever you do to one side of the eqaution HAS to be done exactly the same to the otherside Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted February 23, 2011 Report Share Posted February 23, 2011 To put it simply, you have to find out what x is, and what ever you do to one side of the eqaution HAS to be done exactly the same to the otherside Yeah, it's all about balancing the sides. Removing any kind of confusing unknown in the form of an 'x' or whatever: We all know 2+2 = 4 (ignore my avatar) That's an equation in it's basic form so we can manipulate it and it will still be true so long as we do the same thing to both sides so: 2+2+396.7 = 4+396.7. Works the same with more awkward functions: (2+2)2 = 42 and (2+2)/16 = 4/16 provided you do the same to both sides, it will always remain true. Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 (edited) Okay, just made one up to practise myself and it went horribly wrong lol 9x + 14 = 2x + 18 -2x from each side = 7x+14= 18 -14 from each side = 7x = 4 do i divide them both by 7, or divide them both by 4? what went wrong because when i type 4 divided by 7 into a calculator the answer is 0.57(loads more numbers) would the answer just be the first 2 digits after the point? (first 2 decimal places?) Also can anyone aswer this... if theres a sum like this- 6(x+7) = 8(x-9) do you put 6x down then times the 6 and 7 so 6x + 42 = 8x - 72? then you just do the normal equation is it? Cheers Edited February 23, 2011 by Kieran Morrison Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted February 23, 2011 Report Share Posted February 23, 2011 Okay, just made one up to practise myself and it went horribly wrong lol 9x + 14 = 2x + 18 -2x -2x = 7x+14= 18 -14 -14 = 7x = 4 do i divide them both by 7? Yes. You need to find x by itself. There are currently 7 x's so you need to divide both sides by 7 giving: x=4/7 edit: be careful of how many '=' signs you end up with. To be totally correct you should only have one 'equals' on each line so: 7x + 14 = 18 7x = 4 x = 4/7 edit 2: Also, as dann says you often won't get a nice number if you make something up like that. If an answer results in a fraction like that it's usually neater to leave it as it is- so the final answer is x = 4/7 rather than x = 0.57142857142857142857142857142857. Also can anyone aswer this... if theres a sum like this- 6(x+7) = 8(x-9) do you put 6x down then times the 6 and 7 so 6x + 42 = 8x - 72? then you just do the normal equation is it? Yes, exactly that- the 6 is multiplying each term within the first bracket and 8 is multiplying each of the terms in the second bracket. Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 Okay, just made one up to practise myself and it went horribly wrong lol 9x + 14 = 2x + 18 -2x from each side = 7x+14= 18 -14 from each side = 7x = 4 do i divide them both by 7, or divide them both by 4? what went wrong because when i type 4 divided by 7 into a calculator the answer is 0.57(loads more numbers) would the answer just be the first 2 digits after the point? (first 2 decimal places?) Also can anyone aswer this... if theres a sum like this- 6(x+7) = 8(x-9) do you put 6x down then times the 6 and 7 so 6x + 42 = 8x - 72? then you just do the normal equation is it? Cheers Yup. You nailed that until you got confused at the end... There are 7 x's right? you want 1 of them, so divide both sides by 7. 4/7 = blah blah It doesn't matter what number you got as you made it up, it's bound to not give you a direct answer. Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 (edited) Yup. You nailed that until you got confused at the end... There are 7 x's right? you want 1 of them, so divide both sides by 7. 4/7 = blah blah It doesn't matter what number you got as you made it up, it's bound to not give you a direct answer. Great, cheers for the help everyone. Now onto Algebraic notation: Expressions and formulae My whole night is going to be wasted by maths Edited February 23, 2011 by Kieran Morrison Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 topic changed hopefully for last time! Quote Link to comment Share on other sites More sharing options...
Kieran Morrison Posted February 23, 2011 Author Report Share Posted February 23, 2011 Anyone? Sorry to bump it so early, but this is tommorow and just need to get the areas done then im fine. just to check, is the area of a square just timesing the length by the height (so if it was 4 on each side it would be 16) so 16 is the answer? or is it 16cm squared? Cheers Quote Link to comment Share on other sites More sharing options...
dann2707 Posted February 23, 2011 Report Share Posted February 23, 2011 (edited) another topic change, probably an easier one for you guys How do i find area's for a square, triangle, rectangle and pentagon. Square = 4 cm all sides 4 x 4 = 16cm^2 ^2 = squared for the sake of ease rectangle = 4cm on 2 equal sides, 8 on other 2. 4 x 8 = 32cm^2 pentagon - dont really know how many equal sides not actually sure hahaa, pentagon has 5 sides, make 5 triangles? triangle - just need to know what to do. 1/2 x base x width. If its a right angled triangle it's easy to draw/ see this as you turn it into a 4 sided shape. Also, when do i need to use pi-r squared? when finding the area of a circle? and how do i use it?? That's the area of a circle. press the PI button x radius of the circle^2 Cheers You better rep me for spending my valuble time on you Edited February 23, 2011 by dann2707 1 Quote Link to comment Share on other sites More sharing options...
davidbarr Posted February 23, 2011 Report Share Posted February 23, 2011 to find the area of a square use- A=lxb It is the same for rectangles. to find the area of a triangle use- A=bxh then divide it by two. I'm not sure about pentagons. To find the area of a circle you need to find the radius (half of the diameter) then use the equation- A=pi x r* *=squared Quote Link to comment Share on other sites More sharing options...
JT! Posted February 24, 2011 Report Share Posted February 24, 2011 (edited) Squares / rectangles are simple, just multiply one side by the other. Answer will always be written like 8cm2 Triangles are pretty easy to. You do exactly the same thing as a square by you half the answer. The reason is a triangle will ways fit into a rectangle and take up exactly half of the room. With regard to the pentagon, it depends what information you are given about it. Also, if you're going to be asking more questions in here, don't keep editing the first post, just ask the question in a new reply, it makes it much easier for you and the people replying. Edited February 24, 2011 by JT! Quote Link to comment Share on other sites More sharing options...
monkeyseemonkeydo Posted February 24, 2011 Report Share Posted February 24, 2011 The area of a regular pentagon (5 equal length sides) can be found by separating it into 'slices' across the corners (most likely creating 5 isosceles triangles). If you have enough information you can then just find the area of one of these and multiply by 5. If more information is required you can split each of the individual triangles into two new right angled triangles and take it from there. Just depends what geometry you're given. Quote Link to comment Share on other sites More sharing options...
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