Elizabeth claims that she has proven that 7=5with the following:

7(a+b)-5(a+b)=7c-5c
7a+7b-5a-5b=7c-5c
7a+7b-7c=5a+5b-5c
7(a+b-c)/(a+b-c) = 5(a+b-c)/(a+b-c)
7=5
1.) Explain why this is an invalid proof.

This is an invalid proof because it contains a division by zero error.

In the last step of the proof, the expression is divided by (a+b-c). However, if (a+b-c) were to equal zero, then the division would not be defined mathematically. So, in order for this proof to be valid, it must be shown that (a+b-c) is not equal to zero. Without this information, the proof is incomplete and cannot be considered valid.

This is an invalid proof because it involves dividing both sides of the equation by (a+b-c), which assumes that (a+b-c) is not equal to zero. If (a+b-c) were equal to zero, then dividing by it would be dividing by zero, which is undefined in mathematics.

In this case, if (a+b-c)=0, then the step of dividing both sides by (a+b-c) would lead to a division by zero error. This error makes the proof invalid because it assumes a particular value for (a+b-c) without considering the possibility that it could be zero.

In mathematical proofs, it is essential to avoid division by zero since it leads to undefined results. In this case, the proof fails to consider the possibility of (a+b-c) being equal to zero, and therefore, it cannot be considered a valid demonstration that 7 equals 5.

First of all, where does the first line come from?

if it is true, then
7(a+b)-5(a+b)=7c-5c
2(a+b) = 2c
a+b = c
a+b-c = 0

so in your second last line you divided by zero, which is a mathematical No No.
any subsequent conclusion you draw would be bogus.