The given equations are:
[tex]\begin{gathered} m\angle2=((x^2-1)(x+1))^{\circ} \\ m\angle8=(184-x^2(x+1))^{\circ} \end{gathered}[/tex]
Notice from the figure that angles 2 and 6 are corresponding angles, hence, they must have equal measures. It follows that:
[tex]m\angle6=((x^2-1)(x+1))^{\circ}[/tex]
Notice from the figure that the angles 6 and 8 form a linear pair, hence, they must be supplementary, that is, the sum of their measures must be 180º:
[tex]\begin{gathered} m\angle6+m\angle8=180^{\circ} \\ \Rightarrow(x^2-1)(x+1)+184-x^2(x+1)=180 \end{gathered}[/tex][tex]\begin{gathered} \text{ Expand the parentheses:} \\ \Rightarrow x^2(x+1)-1(x+1)+184-x^2(x+1)=180 \\ \Rightarrow x^2(x)+x^2(1)-1(x)+(-1)(1)+184-x^2(x)+(-x^2)(1)=180 \\ \Rightarrow x^3+x^2-x-1+184-x^3-x^2=180 \\ \text{ Collect like terms:} \\ \Rightarrow x^3-x^3+x^2-x^2-x-1+184=180 \\ \Rightarrow-x+183=180 \\ \Rightarrow-x=180-183 \\ \Rightarrow-x=-3 \\ \text{ Multiply both sides of the equation by }-1: \\ \Rightarrow x=3 \end{gathered}[/tex]
The value of x is 3.
