Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.1 - Complex Numbers - 8.1 Problem Set - Page 426: 91


$y=2+i2\sqrt{3}$, $x=1-i\sqrt{3}$ $y=2-i2\sqrt{3}$, $x=1+i\sqrt{3}$

Work Step by Step

Working from the first equation, $2x+y=4$, we make $x$ the subject of the formula, $x=\frac{4-y}{2}$ We then sub this modified version of the first equation into the second equation , $xy=8$ to get $(\frac{4-y}{2})y=8$. Expanding this equation and moving all thew terms to the left hand side, we get: $-\frac{y^{2}}{2}+2y-8=0$ The method used in the introduction then uses the quadratic method to solve for the equation. Therefore, we get: $y=\frac{-2\pm\sqrt{2^{2}-4(-\frac{1}{2})(-8)}}{2(-\frac{1}{2})}\\=2\pm\frac{\sqrt{4-16}}{-1}\\=2\pm\sqrt{-12}\\=2\pm\sqrt{(12)(-1)}\\=2\pm\sqrt{12}\sqrt{-1}\\=2\pm i2\sqrt{3}$ From this, we can deduce the corresponding values of $x$. When $y=2+i2\sqrt{3}$, $x=\frac{4-(2+i2\sqrt{3})}{2}\\=1-i\sqrt{3}$ when $y=2-i2\sqrt{3}$, $x=\frac{4-(2-i2\sqrt{3})}{2}\\=1+i\sqrt{3}$
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