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How do you solve for x when the equations are as follows...m∠2 = ((x2 − 1)(x + 1))°m∠8 = (184 − x2(x + 1))°

How do you solve for x when the equations are as follows...m∠2 = ((x2 − 1)(x + 1))°m-example-1
User Nir Berko
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1 Answer

28 votes
28 votes

The given equations are:


\begin{gathered} m\angle2=((x^2-1)(x+1))^(\circ) \\ m\angle8=(184-x^2(x+1))^(\circ) \end{gathered}

Notice from the figure that angles 2 and 6 are corresponding angles, hence, they must have equal measures. It follows that:


m\angle6=((x^2-1)(x+1))^(\circ)

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:


\begin{gathered} m\angle6+m\angle8=180^(\circ) \\ \Rightarrow(x^2-1)(x+1)+184-x^2(x+1)=180 \end{gathered}
\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}

The value of x is 3.
User Nachti
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