## Precalculus (6th Edition) Blitzer

Let us assume $L$ be the length and $W$ be the width of the television viewing area such that: \begin{align} & {{L}^{2}}+{{W}^{2}}=100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & LW=48\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} The substitution method is used to solve the system. Now, the value of $L$ from the second equation is: $L=\frac{48}{W}$ (III) Substitute the value of L in equation (I): \begin{align} & \frac{{{48}^{2}}}{{{W}^{2}}}+{{W}^{2}}=100 \\ & 2804+{{W}^{4}}=100{{W}^{2}} \\ & {{W}^{4}}-100{{W}^{2}}+2304=0 \\ & \left( {{W}^{2}}-36 \right)\left( {{W}^{2}}-64 \right)=0 \end{align} Now, ${{W}^{2}}-36=0$ Or, ${{W}^{2}}-64=0$  If ${{W}^{2}}=36$ then $W=\pm 6$ and if ${{W}^{2}}=64$ then, $W=\pm 8$. The width cannot be negative so, $W=6,\ 8$. If $W=6$ then, \begin{align} & L=\frac{48}{6} \\ & =8 \end{align} If $W=8$ Then, \begin{align} & L=\frac{48}{8} \\ & =6 \end{align} Thus, the dimensions are 8 inches by 6 inches.