Isn't 7(x+1)/(x+1)=7? Can't we cancel out (x+1) from both numerator and denominator?
By considering an equation y=7(x+1)/(x+1), when we cancel out (x+1) from both numerator and denominator, value of y becomes 7. But if value of x is equal to -1, then value of y becomes 0/0 which is not defined. And it is impossible for 2 different values of variable y.
What is the correct order to solve it?
Answers
Whenever there's a possibility for the denominator to be zero, we are not allowed to cancel the suspicious terms.
Here, unless it is given that (x + 1) ≠ 0, we are not allowed to cancel (x+1) from the numerator and denominator as x could be anything. So, y can hold the value 7 if it was stated with a condition: Find y, where (x + 1) ≠ 0. That is the reason why quite a few problems are stated with such condition(s).
(x+1)/(x+1) = 1 is true for almost every x which makes y equals 7. But it is not true for x = -1 as division by 0 is undefined . So, 7 can not be considered as the value of y for all the x's that exist.
The equation can not be solved with the given info, but can be restated in quite a lot ways [...where (x+1) should not be cancelled ]
y = 7(x+1)/(x+1) or y(x+1) = 7(x+1) or y = x(7-y) + 7 etc.
while making use of 0 in any equation , there is possibility for a contradiction. like 0/0 = ∞ and any other no./0= ∞ too , but 0 ≠ any other no.