Question

por favor,quanto é:

a) {x+y/3=9
{x+y=15


b) {x+y/4=8
{x+y=20


c) { x+y=8
{ 2x+2y=16

Answer 1

Estas questões são sistema de equações.


$a)~~ \left \{ {x+{\frac{y}{3} =9} \atop {x+y=15}} \right. \\ \\ Isolaremos~o~x~da~segunda~equa\c{C} \~ao \\ \\\boxed{\begin{array}{c}\mathsf{x=15-y} \end{array}} \\ \\ Agora~substitu\'imos~na~primeira$
$(15-y)+ \frac{y}{3} =9~~~\rightarrow~~~15-y+ \frac{y}{3} =9 \\ \\ -y+ \frac{y}{3} =9-15~~~\rightarrow~~~-y+ \frac{y}{3}=-6 \\ \\ Igualamos~os~denominadores \\ \\ \frac{-3y}{3} + \frac{y}{3}= \frac{-18}{3} ~~~\rightarrow~~~-3y+y=-18~~~\cdot(-1) \\ \\ 3y-y=18 \\ 2y=18 \\ \\ \boxed{\begin{array}{c}\mathsf{y= \frac{18}{2}=9 } \end{array}} \\ \\ Utilizamos~a~equa\c{C} \~ao ~isolada~inicialmente~~ \boxed{\begin{array}{c}\mathsf{x=15-y} \end{array}} \\ \\ x=15-9 \\ x=6$




$b)~~ \left \{ {x+{\frac{y}{4} =8} \atop {x+y=20}} \right. \\ \\ Isolaremos~o~x~da~segunda~equa\c{C} \~ao \\ \\\boxed{\begin{array}{c}\mathsf{x=20-y} \end{array}} \\ \\ Agora~substitu\'imos~na~primeira$
$(20-y)+ \frac{y}{4} =8~~~\rightarrow~~~20-y+ \frac{y}{4} =8 \\ \\ -y+ \frac{y}{7} =8-20~~~\rightarrow~~~-y+ \frac{y}{4}=-12 \\ \\ Igualamos~os~denominadores \\ \\ \frac{-4y}{4} + \frac{y}{4}= \frac{-48}{4} ~~~\rightarrow~~~-4y+y=-48~~~\cdot(-1) \\ \\ 4y-y=48\\ 3y=48 \\ \\ \boxed{\begin{array}{c}\mathsf{y= \frac{48}{6}=16 } \end{array}} \\ \\ Utilizamos~a~equa\c{C} \~ao ~isolada~inicialmente~~ \boxed{\begin{array}{c}\mathsf{x=20-y} \end{array}} \\ \\ x=20-16 \\ x=4$



$c) \left \{ {{x+y=8} \atop {2x+2y=16}} \right. ~~n\~ao~da~para~resolver$

Espero ter ajudado com as letras A e B. Qualquer dúvida comente embaixo! :)