How do you solve 1x25x=x+7x1 and check for extraneous solutions?

1 Answer
Jan 17, 2017

{367}

Explanation:

Recall that x25x can be written as x(x5).

1x(x5)=x+7x1

Put on a common denominator:

1x(x5)=(x+7)(x5)x(x5)x(x5)x(x5)

1=x2+7x5x35(x25x)

1=x2+7x5x35x2+5x

36=7x

x=367

This is not extraneous, as x0,5.

Hopefully this helps!