Question

x²+y²=20
x+y=6

Como resolver?

Answer 1

Resposta:

$x_{1},y_{1} = (4,2)\\x_{2},y_{2}= (2,4)$

Explicação passo-a-passo:

$x^{2} + y^{2} = 20\\x=6\\x^{2} +2y=20\\x=6-y\\(6-y)^{2}+y^{2}=20\\ y=2\\y=4\\x=6-2\\y=4\\x=6-2\\x=6-4\\x=4\\x=2\\\\x_{1},y_{1}=(4,2)\\x_{2},y_{2}=(2,4)$

Answer 2

$\begin{cases}{x}^{2}+{y}^{2}=20\\ x+y=6\end{cases}$

$\begin{cases}{x}^{2}+{y}^{2}=20\\ y=6-x\end{cases}$

${x}^{2}+{y}^{2}=20 \\{x}^{2}+{(6-x)}^{2}=20$

${x}^{2}+36-12x+{x}^{2}=20\\{x}^{2}+{x}^{2}-12x+36-20=0$

$2{x}^{2}-12x+16=0\div2\\ {x}^{2}-6x+8=0$

$\boxed{\Delta={b}^{2}-4ac}$

$\Delta={(-6)}^{2}-4.1.8$

$\Delta=36-32$

$\Delta=4$

$\boxed{x=\frac{-b±\sqrt{\Delta}}{2a}}$

$x=\frac{-(-6)±\sqrt{4}}{2.1}$

$x=\frac{6±2}{2}$

$x'=\frac{6+2}{2}=\frac{8}{2}$

$\boxed{\boxed{x'=4}}$

$x''=\frac{6-2}{2}=\frac{4}{2}$

$\boxed{\boxed{x''=2}}$

$y=6-x \\ y=6-4$

$\boxed{\boxed{y'=2}}$

$y=6-x \\ y=6-2$

$\boxed{\boxed{y''=4}}$

S={(2,4),(4,2)}