From the first inequality:

\[2x-2\ge x-8 \Rightarrow x\ge -6\]

From the second inequality:

\[-2x+7<-5 \Rightarrow -2x<-12 \Rightarrow x>6\]

Intersection:

\[x>6\]

\((2x-6)(x+1)=2(x-3)(x+1)\).

Since \(2>0\), solve \((x-3)(x+1)<0\).

The roots are \(-1\) and \(3\). The product is negative between the roots.

\[-1

\[-1\le -2x+3<5\]

\[-4\le -2x<2\]

Divide by \(-2\) and reverse the inequality signs:

\[2\ge x>-1\]

So \(-1

Factorise:

\[x(x+10)\ge 0\]

The zeros are \(-10\) and \(0\). The product is non-negative outside the interval between the roots.

\[x\le -10 \cup x\ge 0\]

From the first inequality:

\[3x-5\ge x-15 \Rightarrow 2x\ge -10 \Rightarrow x\ge -5\]

From the second inequality:

\[-2x+1>-1 \Rightarrow -2x>-2 \Rightarrow x<1\]

Intersection:

\[-5\le x<1\]