Question

Please Help!! George tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes the denominator zero, and you can't divide by zero," George explains. Using complete sentences, demonstrate to George how the equation is still solvable

Answer 1

As long as the variable in the denominator does not combine in some way with an integer to equal 0, then there can be a variable in the denominator. Sometimes there is a "hole" in the graph of the function where it is undefined, but there will be infinitely many other options for a positive or negative value in the denominator.

Answer 2

ok... here is a simple example

1x−1


looking at the equation

the denominator will be zero when x = 1

and as stated in the question dividing by zero is undefined.

What you do is restrict the domain, or the x values that you can use...


I this example you say you can input all real x values, except x = 1

that means you can solve it or graph it... but the restriction will cause a gap..

hope it makes sense.

Answer 3

thank you

Answer 4

@IMStuck wait is that the answer?