Amanda 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," Amanda explains. Using complete sentences, demonstrate to Amanda how the equation is still solvable.

Question

texaschic101 · Answer

http://openstudy.com/study#/updates/52b32ce2e4b03e81ff95262c

anonymous · Answer

@jim_thompson5910

jim_thompson5910 · Answer

any ideas?

anonymous · Answer

I made an equation for it such as 3x^2+x/x

anonymous · Answer

@jim_thompson5910

jim_thompson5910 · Answer

are you able to solve that equation?

anonymous · Answer

I did and got the answer 3x+1 but how would this respond to this question

jim_thompson5910 · Answer

well if you had something like $$\Large \frac{3x^2+x}{x-1} = 0$$ then multiplying both sides by x-1 gives you 

$$\Large \frac{3x^2+x}{x-1} = 0$$

$$\Large 3x^2+x = 0(x-1)$$

$$\Large 3x^2+x = 0$$

jim_thompson5910 · Answer

what do you get when you solve $$\Large 3x^2+x = 0$$

anonymous · Answer

x=0 or x=1/3

jim_thompson5910 · Answer

close

jim_thompson5910 · Answer

x = 0 is correct

anonymous · Answer

-1/3?

jim_thompson5910 · Answer

good

jim_thompson5910 · Answer

that proves just because you have a variable in the denominator, it doesn't mean the equation isn't solveable

anonymous · Answer

oh ok! that would be the solution?

jim_thompson5910 · Answer

those two x values, yes

jim_thompson5910 · Answer

their existence shows us that the initial claim is false