Math problem -- help/hints please (NOW SOLVED: includes solution)
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Nope.. it's a fraction (I don't want to give the answer straight away here, as it's the flaw in my method I'm trying to spot) It's ONLY an A Level problem; you'd think at the age of 38 I'd find this stuff elementary. But no, I was smarter at maths when I was 18!?? :doh:
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
So, if I'm not wrong: (x-4)/6 - 2x+1 = (3x-4)/2 ! -1 x/6 - 4/6 - 2x = 3x/2 - 2 - 1 ! +4/6 x/6 - 2x = 3x/2 - 3 + 4/6 x/6 - 12x/6 = 9x/6 - 18/6 + 4/6 ! -9x/6 -11x6 - 9x/6 = -14/6 ! *(-1) 20x/6 = 14/6 !*6 20x = 14 ! /20 x = 14/20 (The ! stands for the pipe followed by the operation I did in that step) Hope that is correct. :rolleyes:
------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
x-4 + 6(-2x + 1) ... etc. The error came in when you introduced the parenthesis around the second term on the left hand side. It changed the meaning of negative sign to include the one, which is not correct if you look closely at the original. Cheers!
"With sufficient thrust, pigs fly just fine."
Ross Callon, The Twelve Networking Truths, RFC1925
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Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.
I was wrong in my first attempt. I have changed the reply now. 7/11 is a shop supposedly, mostly run by Indians in US (I have never been to US so don't know). This is often used in stand up comic shows for jokes. That's what I was referring to.
"Fear no factor", Prime Numbers.
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.
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x-4 + 6(-2x + 1) ... etc. The error came in when you introduced the parenthesis around the second term on the left hand side. It changed the meaning of negative sign to include the one, which is not correct if you look closely at the original. Cheers!
"With sufficient thrust, pigs fly just fine."
Ross Callon, The Twelve Networking Truths, RFC1925
Of course!! Thanks buddy! :-)
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So, if I'm not wrong: (x-4)/6 - 2x+1 = (3x-4)/2 ! -1 x/6 - 4/6 - 2x = 3x/2 - 2 - 1 ! +4/6 x/6 - 2x = 3x/2 - 3 + 4/6 x/6 - 12x/6 = 9x/6 - 18/6 + 4/6 ! -9x/6 -11x6 - 9x/6 = -14/6 ! *(-1) 20x/6 = 14/6 !*6 20x = 14 ! /20 x = 14/20 (The ! stands for the pipe followed by the operation I did in that step) Hope that is correct. :rolleyes:
------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.
Bang on (although the 'book answer' is 7/10 - simplest fractional answer). I'm curious about your notation though, what's with the factorial -1?
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Bang on (although the 'book answer' is 7/10 - simplest fractional answer). I'm curious about your notation though, what's with the factorial -1?
Kevin Bewley wrote:
I'm curious about your notation though, what's with the factorial -1?
Oh, that should be a pipe, but I wasn't able to type it in the editor, so it's just the remark, what I did. Forgot to post that :rolleyes:
------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
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Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!
Other people have pointed this out already (so I'm not going to take the credit for solving this), but trying to stick to your calculations as closely as possible:
(x-4)/6 - 2x+1 = (3x-4)/2
My approach was to find the lcm, which is 6 and then do this:
1(x-4) - 6(2x-1) = 3(3x-4) <-- note 2x-1, not 2x+1 because previous line is -2x + 1, not - (2x + 1)
multiply out the brackets:
x-4 - 12x+6 = 9x-12 <-- from previous line 12x+6, not 12x-6
which yields:
20x = 14
or x = 7/10Jon CodeWrite