Question

How to do both 14) and 16) b/c I have no idea. It is solving rational equations.

Answer 1

Just like at any other time, to add/subtract fractions you need a common denominator.

14)

1/(x^2+2x)+(x-1)/x=1  so we need a common denominator of x(x^2+2x)

[1(x)]/(x(x^2+2x))+[(x-1)(x^2+2x)]/(x(x^2+2x))=[1(x(x^2+2x))]/(x(x^2+2x))

now if you multiply both sides of the equation by x(x^2+2x) you are left with:

x+(x-1)(x^2+2x)=x(x^2+2x)

x+x^3+2x^2-x^2-2x=x^3+2x^2

x^3+x^2-x=x^3-2x^2

x^2-x=-2x^2

3x^2-x=0

x(3x-1)=0, x=0 is an extraneous solution as division by zero is undefined. So the only real solution is:

x=1/3

...

16)

(r+5)/(r^2-2r)-1=1/(r^2-2r)  the common denominator we need r^2-2r so

[r+5-1(r^2-2r)]/(r^2-2r)=1/(r^2-2r), multiplying both sides by r^2-2r yields:

r+5-r^2+2r=1

-r^2+3r+5=1

-r^2+3r+4=0

r^2-3r-4=0

(r-4)(r+1)=0, r^2-2r cannot equal zero, r(r-2)=0, r cannot equal 0 or 2...

r=-1 or 4