find the solutions that satisfy both x^2 + 4x-5>0 and x^2+6x-7<=0
$\displaystyle x^2 + 4x-5>0$
$\displaystyle (x+5)(x-1)>0$
Critical values $\displaystyle x=1$, $\displaystyle x=-5$
That isn't a solution. These are the possible "changes" - the boundaries of the inequalities. To test, choose one value on each side of every c.v.
One value greater than $\displaystyle 1$:
$\displaystyle x=2$
$\displaystyle 2^2+4(2)-5$
$\displaystyle =7$
So x can be greater than $\displaystyle 1$ as this fits the inequality.
One value within the range $\displaystyle -5<x<1$, such as $\displaystyle 0$
$\displaystyle 0^2+4(0)-5$
$\displaystyle =-5$
So values within the $\displaystyle -5<x<1$ range do not suit the inequality.
One value less than $\displaystyle -5$, such as $\displaystyle -10$
$\displaystyle (-10)^2+4(-10)-5$
$\displaystyle =55$
So values less than -5 work, and values greater than 1 work, as these are the only values which suit the inequality.
Can you do the same for the second inequality? Then you just need to find the values which are within the accepted range for both. Post your working if you get stuck.
An alternative (easier) method is to complete the square...
$\displaystyle \displaystyle x^2 + 4x - 5 > 0$
$\displaystyle \displaystyle x^2 + 4x + 2^2 - 2^2 - 5 > 0$
$\displaystyle \displaystyle (x + 2)^2 - 9 > 0$
$\displaystyle \displaystyle (x + 2)^2 > 9$
$\displaystyle \displaystyle |x + 2| > 3$
$\displaystyle \displaystyle x + 2 < -3$ or $\displaystyle \displaystyle x + 2 > 3$
$\displaystyle \displaystyle x < -5$ or $\displaystyle \displaystyle x > 1$.
Don't worry about, thanks for the help though I do understand it, I was just confruzzled by the fact that they asked you to satisfy both inequalities. I though they meant I to like substitute each inequality into each other, but that isn't possible (to my understanding, since each inequality has a different condition). But thanks again for the help (: