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anonymous
 5 years ago
Bryce insists that (2+x)/8 is equivalent to (1+x)/4. What mistakes do you think is being made and how could you demonstrate to Bryce that the two expressions are not equivalent?
anonymous
 5 years ago
Bryce insists that (2+x)/8 is equivalent to (1+x)/4. What mistakes do you think is being made and how could you demonstrate to Bryce that the two expressions are not equivalent?

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0(1+x)/4 *2/2=2(1+x)/4

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oops that should be 8 on the bottm

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0(1+x)/4 *2/2=2(1+x)/8 2(1+x) is not the same as 2+x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2+x}{8}=\frac{1+x}{4}\rightarrow \frac{8+4x}{8}=1+x \rightarrow 8+4x=8+8x \] \[4x=8x \rightarrow x=2x\] Which doesn't make sense since this says the variable x is equal to twice its value

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0they are only equal when x=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you can also graph these two and show that they aren't the same line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do that as well

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you can also show him a counterexample to his logic

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0by saying if we let x be 1 here what happens we ger 3/8 but over we get 2/4 2/4 does not equal to 3/8 sorry bryce

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The math i did earlier covers all examples for any value of x, Which demonstrate that the two expressions are not equal when x is greater than or less than 0, it is only true when x=0 as myininaya stated earlier
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