## Algebra and Trigonometry 10th Edition

$x \approx 12.49$ and $y=6.24$
We have: $36=(y-2(1))(x-2(2))$ $\implies 36=(x-4)(y-2)$ So, $y=\dfrac{36}{x-4}+2$ Now, $y=\dfrac{2x+28}{x-4}$ or, $xy=2x+4y+8$ or, $xy=2x+4(\dfrac{2x+28}{x-4}) +8$ Thus, $A=xy=\dfrac{2(x^3+4x+40)}{x-4}$ So, $x \approx 12.49$ and $y=\dfrac{2(12.49)+28}{12.49-4}=6.24$