Yazomi
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I’m a bit confused on this question- I realised the question didn’t mean -8a cannot be a value when it stated a is a positive constant. This part makes sense

But for the substitution part- how would you know which value sub into which equation because I’d get 4 different answers rather than 2 if I tried to do all possible variations
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Nagromicous
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(Original post by Yazomi)
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I’m a bit confused on this question- I realised the question didn’t mean -8a cannot be a value when it stated a is a positive constant. This part makes sense

But for the substitution part- how would you know which value sub into which equation because I’d get 4 different answers rather than 2 if I tried to do all possible variations
You wouldn't, because although x can take two different values, it can't take two different values simultaneously.
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Yazomi
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(Original post by Nagromicous)
You wouldn't, because although x can take two different values, it can't take two different values simultaneously.
How would you know which pair is the correct pair (if that makes sense)

If I sub it in one way I would get the correct answers

But if I switch them around I would get 14a and -16a which aren’t the answer. Is there a way to tell?
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Nagromicous
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(Original post by Yazomi)
How would you know which pair is the correct pair (if that makes sense)

If I sub it in one way I would get the correct answers

But if I switch them around I would get 14a and -16a which aren’t the answer. Is there a way to tell?
It's like I said, you are making x equal to two different values simultaneously. The way you tell is that for each case, you need to be consistent with your value of x. Either it is 2a for both brackets, or it is -8a for both brackets.
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Yazomi
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(Original post by Nagromicous)
It's like I said, you are making x equal to two different values simultaneously. The way you tell is that for each case, you need to be consistent with your value of x. Either it is 2a for both brackets, or it is -8a for both brackets.
I think I’m a bit confused with this part because on the mark scheme, it says
To use both values (highlighted as green). Like one of each rather than both x=2a etc🤔
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Nagromicous
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(Original post by Yazomi)
I think I’m a bit confused with this part because on the mark scheme, it says
To use both values (highlighted as green). Like one of each rather than both x=2a etc🤔
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I think I was confused as well by your working. The teacher probably shouldn't have marked your last answer as correct. The way you are substituting the values back in is completely wrong. It is really quite simple. Image
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Yazomi
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(Original post by Nagromicous)
I think I was confused as well by your working. The teacher probably shouldn't have marked your last answer as correct. The way you are substituting the values back in is completely wrong. It is really quite simple. Image
Ohhhhhh it finally made sense thanks for your help honestly!!!

I thought you need to consider the negative version of the modulus as well for |x+7a|-|x-7a|
Like how |x+3a|=5a had considered both positive as negative versions.
(Is there a reason why the bottom mod equations doesn’t need the negative versions though?)
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Nagromicous
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(Original post by Yazomi)
Ohhhhhh it finally made sense thanks for your help honestly!!!

I thought you need to consider the negative version of the modulus as well for |x+7a|-|x-7a|
Like how |x+3a|=5a had considered both positive as negative versions.
(Is there a reason why the bottom mod equations doesn’t need the negative versions though?)
When you multiply the contents of the modulus sign by -1, you are not changing anything, but if you multiplied the outside of the modulus sign by -1, you are making a material difference. You can only get away with it in the first part because you know 5a is non-negative (the question tells you explicitly). I maybe didn't do a good job explaining, but it's something simple enough to remember.
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Yazomi
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(Original post by Nagromicous)
When you multiply the contents of the modulus sign by -1, you are not changing anything, but if you multiplied the outside of the modulus sign by -1, you are making a material difference. You can only get away with it in the first part because you know 5a is non-negative (the question tells you explicitly). I maybe didn't do a good job explaining, but it's something simple enough to remember.
Yeah that kinda makes sense. Mods a difficult topic to get my head around T^T
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