Question

utilizando o método da substituição, obtenha a solução do sistema: 4x + y=6 5x - y=3


7x - 3y=5
12x - 5y=9


x + 4y=0
4x - 2y=9



4x - 3y=11
4x - 5y=5

Answer 1

$\begin{cases}7x-3y=5\\12x-5y=9\end{cases}$

Isolando uma variável em uma equação:

$7x-3y=5\\\\3y=7x-5\\\\\boxed{\boxed{y=\dfrac{1}{3}(7x-5)}}$

Substituindo na outra equação:

$12x-5y=9\\\\\\12x-\dfrac{5}{3}(7x-5)=9~~~~~~(\cdot3)\\\\\\36x-5(7x-5)=27\\\\36x-35x+25=27\\\\x=27-25\\\\\boxed{\boxed{x=2}}$

Agora, vamos achar y:

$y=\dfrac{1}{3}(7x-5)\\\\\\y=\dfrac{1}{3}(7\cdot2-5)\\\\\\y=\dfrac{1}{3}(14-5)\\\\\\y=\dfrac{1}{3}(9)\\\\\\\boxed{\boxed{y=3}}$


$\boxed{\boxed{S=(2,3)}}$
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$\begin{cases}x+4y=0\\4x-2y=9\end{cases}$

Isolando x na primeira equação:

$x+4y=0~~~\therefore~~~\boxed{\boxed{x=-4y}}$

Substituindo x na segunda equação:

$4x-2y=9\\\\4(-4y)-2y=9\\\\-16y-2y=9\\\\-18y=9\\\\\\y=-\dfrac{9}{18}=-\dfrac{9\div9}{18\div9}\\\\\\\boxed{\boxed{y=-\dfrac{1}{2}}}$

Achando x:

$x=-4y=-4\left(-\dfrac{1}{2}\right)=\dfrac{4}{2}~~~\therefore~~~\boxed{\boxed{x=2}}$


$\boxed{\boxed{S=\left(2,-\dfrac{1}{2}\right)}}$
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$\begin{cases}4x-3y=11\\4x-5y=5\end{cases}$

Isolando x na primeira equação:

$4x-3y=11~~~\therefore~~~4x=11+3y~~~\therefore~~~\boxed{\boxed{x=\dfrac{1}{4}(11+3y)}}$

Substituindo na segunda:

$4x-5y=5\\\\\\4\cdot\dfrac{1}{4}(11+3y)-5y=5\\\\\\11+3y-5y=5\\\\-2y=5-11\\\\-2y=-6\\\\2y=6\\\\\\y=\dfrac{6}{2}\\\\\\\boxed{\boxed{y=3}}$

Achando x:

$x=\dfrac{1}{4}(11+3y)\\\\\\x=\dfrac{1}{4}(11+3\cdot3)\\\\\\x=\dfrac{1}{4}(11+9)\\\\\\x=\dfrac{1}{4}(20)\\\\\\\boxed{\boxed{x=5}}$


$\boxed{\boxed{S=(5,3)}}$