Podemos resolver todos estes sistemas pelo método da adição, que consiste em somar as equações membro a membro anulando uma das incógnitas para descobrir o valor de outra
Pois vemos claramente que, podemos anular ''y'' deixando os coeficientes desta incógnita opostos e assim obtendo ''x'', essa é a ideia
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Letra A)
[tex]\begin{array}{l}\begin{cases}\sf x+y=11~~~~(\,I\,)\\\\\sf x-y=3~~~~~(\,II\,)\end{cases}\\\\\end{array}[/tex]
Somando ( l ) + ( ll ):
[tex]\begin{array}{l}\sf x+y+x-y=11+3\\\\\sf 2x=14\\\\\sf \dfrac{2x}{2}=\dfrac{14}{2}\\\\\!\boxed{\sf x=7}~~~~\sf(\,III\,)\\\\\end{array}[/tex]
Substituindo ( lll ) em ( l ):
[tex]\begin{array}{l}\sf x+y=11\\\\\sf 7+y=11\\\\\sf -\:7+7+y=11-7\\\\\!\boxed{\sf y=4}\\\\\end{array}[/tex]
O conjunto solução é:
[tex]\large\begin{array}{l}\sf S=\Big\{\Big(7~~,~~4\Big)\Big\}\end{array}[/tex]
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Letra B)
[tex]\begin{array}{l}\begin{cases}\sf x+y=6~~~~(\,I\,)\\\\\sf 2x+y=4~~~(\,II\,)\end{cases}\\\\\end{array}[/tex]
Para que os coeficientes de y sejam opostos, inverta os sinais de ( l ):
[tex]\begin{array}{l}\sf x+y=6~~\Longleftrightarrow~~\boxed{\sf -x-y=-6}\\\\\end{array}[/tex]
Somando este valor com ( ll ):
[tex]\begin{array}{l}\sf -x-y+2x+y=-6+4\\\\\!\boxed{\sf x=-2}\sf~~~~(\,III\,)\\\\\end{array}[/tex]
Substituindo ( lll ) em ( l ):
[tex]\begin{array}{l}\sf x+y=6\\\\\sf -2+y=6\\\\\sf 2-2+y=6+2\\\\\!\boxed{\sf y=8}\\\\\end{array}[/tex]
O conjunto solução é:
[tex]\large\begin{array}{l}\sf S=\Big\{\Big(-2~~,~~8\:\Big)\Big\}\end{array}[/tex]
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Letra C)
[tex]\begin{array}{l}\begin{cases}\sf 3x+y=5~~~~(\,I\,)\\\\\sf 2x+y=4~~~(\,II\,)\end{cases}\\\\\end{array}[/tex]
Para que os coeficientes de y sejam opostos, inverta os sinais de ( ll ):
[tex]\begin{array}{l}\sf 2x+y=4~~\Longleftrightarrow~~\boxed{\sf -2x-y=-4}\\\\\end{array}[/tex]
Somando este valor com ( l ):
[tex]\begin{array}{l}\sf -2x-y+3x+y=-4+5\\\\\!\boxed{\sf x=1}\sf~~~~(\,III\,)\\\\\end{array}[/tex]
Substituindo ( lll ) em ( ll ):
[tex]\begin{array}{l}\sf 2x+y=4\\\\\sf 2(1)+y=4\\\\\sf 2+y=4\\\\\sf -2+2+y=4-2\\\\\!\boxed{\sf y=2}\\\\\end{array}[/tex]
O conjunto solução é:
[tex]\large\begin{array}{l}\sf S=\Big\{\Big(1~~,~~2\Big)\Big\}\end{array}[/tex]
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Letra D)
[tex]\begin{array}{l}\begin{cases}\sf x-y=6~~~~(\,I\,)\\\\\sf x+y=-7~~(\,II\,)\end{cases}\\\\\end{array}[/tex]
Somando ( l ) + ( ll ):
[tex]\begin{array}{l}\sf x-y+x+y=6-7\\\\\sf 2x=-1\\\\\sf \dfrac{2x}{2}=\dfrac{-1~}{2}\\\\\!\boxed{\sf x=-\dfrac{1}{2}}~~~~\sf(\,III\,)\\\\\end{array}[/tex]
Substituindo ( lll ) em ( ll ):
[tex]\begin{array}{l}\sf x+y=-7\\\\\sf -\dfrac{1}{2}+y=-7\\\\\sf \dfrac{1}{2}-\dfrac{1}{2}+y=-7+\dfrac{1}{2}\\\\\!\boxed{\sf y=-\dfrac{13}{2}}\\\\\end{array}[/tex]
O conjunto solução é:
[tex]\large\begin{array}{l}\sf S=\Bigg\{\Bigg(-\dfrac{1}{2}~~,~-\dfrac{13}{2}\:\Bigg)\Bigg\}\end{array}[/tex]
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Att. Nasgovaskov
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