Question

×+y =12
3x-y= 20

×+y=4
×-y=8

3x+2y=-5
X-2y=-7

2×+y=6
2×+3y=2
Como resolver-las?

Answer 1

a)

$+\begin{cases}x+y =12\\3x-y= 20\end{cases}$

$4x=32\\x=\frac{32}{4}\\x=8$

$x+y=12\\8+y=12\\y=12-8\\y=4$

$s=\left\{(8,4)\right\}$

b)

$+\begin{cases}x+y =4\\x-y= 8\end{cases}$

$2x=12\\x=\frac{12}{2}\\x=6$

$x+y=4\\6+y=4\\y=4-6\\y=-2$

$s=\left\{(6,-2)\right\}$

c)

$+\begin{cases}3x+2y=-5\\x-2y=-7\end{cases}$

$4x=-12\\x=-\frac{12}{4}\\x=-3$

$-3-2y=-7\\2y=7-3\\2y=4$

$y=\frac{4}{2}\\y=2$

$s=\left\{(-3,2)\right\}$

d)

$\begin{cases} 2x+y=6\\2x+3y=2\end{cases}$

$\it{multiplicando \: a \: primeira \: equação } \\ \it{ por - 1 \: temos}$

$+\begin{cases}-2x-y=-6\\2x+3y=2\end{cases}$

$2y=-4\\y=-\frac{4}{2}\\y=-2$

$2x+y=6\\2x-2=6\\2x=6+2\\2x=8$

$x=\frac{8}{2}\\x=4$

$s=\left\{(4,-2)\right\}$