There seems to be something odd with this problem. By definition, all three sides of an equilateral triangle have the same length. So, the three sides must satisfy the following equations:
[tex] \begin{cases} 2x-7 = x+y-9 \\ x+y+9 = y+5 \\ 2x-7 = y+5 \end{cases} [/tex]
If we consider the second equation, we can cancel y, since it appears on both sides. We're left with
[tex] x+y+9 = y+5 \implies x+9 = 5 \implies x = -4 [/tex]
But if we plug this value for x in the top-left side, its lenght should be
[tex] 2(-4)-7 = -8-7 = -15 [/tex]
but negative lengths are not acceptable.
Moreover, if we plug this value for in the third equation, we have
[tex] 2x-7 = y+5 \implies -15 = y+5 \implies y = -20 [/tex]
So, if [tex] x = -4 [/tex] and [tex] x = -20 [/tex], the three sides should be
[tex] \begin{cases} 2x-7 = -15\\ y+5 = -15 \\ x+y-9 = -33 \end{cases} [/tex]
So, two of the sides are negative, which we can't accept, and the system is also inconsistent, meaning that the solution arising from two of its equations don't satisfy the third one.
Unless I'm missing something, I'd say that there's a typo in the exercise
