Question 962176
You need a common denominator.
{{{x^2+4x-5=(x+5)(x-1)}}}
{{{x^2+6x+5=(x+5)(x+1)}}}
So a common denominator would be {{{(x-1)(x+1)(x+5)}}}
{{{(x+2)/(x^2+4x-5) - 3/ (x^2+6x+5)=((x+2)(x+1))/((x-1)(x+1)(x+5))-(3(x-1))/((x-1)(x+1)(x+5))}}}
{{{(x+2)/(x^2+4x-5) - 3/ (x^2+6x+5)=(x^2+3x+2-3x+3)/((x-1)(x+1)(x+5))}}}
{{{(x+2)/(x^2+4x-5) - 3/ (x^2+6x+5)=(x^2+5)/((x-1)(x+1)(x+5))}}}
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Restrictions occur when the denominator equals zero so,
{{{x<>-5}}}
{{{x<>-1}}}
{{{x<>1}}}