Answer
$ф$
Work Step by Step
$x^{2}+y^{2} =9$ Equation $(1)$
$x+y=5 $ Equation $(2)$
From Equation $(2)$,
$y=5-x $
Substitute $y=5-x $ in Equation $(1)$,
$x^{2}+y^{2} =9$
$x^{2}+(5-x)^{2} =9$
Using $(a-b)^{2} = a^{2}-2ab+b^{2}$
$(5-x)^{2} =5^{2}-2(5)(x)+x^{2} = 25-10x+x^{2}$
$x^{2}+ 25-10x+x^{2} =9$
$2x^{2} -10x + 25 =9$
$2x^{2} -10x + 25 -9 = 0$
$2x^{2} -10x + 16 = 0$
Using quadratic formula, $a=2, b = -10, c=16$
$x = \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
$x = \frac{10±\sqrt ((-10)^{2}-4(2)(16))}{2(2)}$
$x = \frac{10±\sqrt (100-128)}{4}$
$x = \frac{10±\sqrt (-28)}{4}$
Since $\sqrt (-28)$ is not a real number, there is no real solution.
Solution set is empty.
