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  1. Write a system of equations to represent the areas of the two rectangles in the diagram.

    \(\begin{align}
     A &= w(w + 1) \\
     A &= \left( {10 - w} \right)\left( {w + 5} \right)
     \end{align}\)


  2. For what value(s) of \(w\) are the two areas equal?

    Substitute the second equation in for \(A\), and solve for \(w\).

    \(\begin{align}
     A &= w(w + 1) \\
     \left( {10 - w} \right)\left( {w + 5} \right) &= w\left( {w + 1} \right) \\
     5w - w^2 + 50 &= w^2 + w \\
     0 &= 2w^2 - 4w - 50
     \end{align}\)

    \[\begin{align}
     w &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
      &= \frac{{ - \left( { - 4} \right) \pm \sqrt {\left( { - 4} \right)^2 - 4\left( 2 \right)\left( { - 50} \right)} }}{{2\left( 2 \right)}} \\
      &= \frac{{4 \pm \sqrt {416} }}{4} \\
      &= \frac{{4 \pm 4\sqrt {26} }}{4} \\
      &= 1 \pm \sqrt {26}  \\
     \end{align}\]

    The solution \(1 -\sqrt{26}\) is negative, so it is not a valid solution. The value of \(w\) that will make the two areas equal is \(1 + \sqrt{26}\).


 For more examples on solving systems of equations algebraically, see pp. 440 – 451 of Pre-Calculus 11.