Question

Resolve cada um dos seguintes sistemas de equações:

171.7.
1 - x+y/2 = 3
1/2 (x-3) = x+y

171.8.
5 (a-2b) = 5
a/2 - b/3 = 1/6

171.9.
1 - x-y/4 = -(x+7/4)
3x/2 - 1-y/3 = x-y

Answer 1

Olá

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1)
Antes de resolvermos o sistema, primeiro temos que arruma-lo

$\displaystyle \left \{ {{1- \frac{x+y}{2} =3} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { \frac{1}{2}(x-3) =x+y}} \right. \\ \\ \\ \left \{ {{1- (\frac{x}{2}+ \frac{y}{2} ) =3} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { \frac{1}{2}x -\frac{3}{2} =x+y}} \right. \\ \\ \\ \left \{ {{- \frac{x}{2}- \frac{y}{2} =3-1} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { \frac{1}{2}x -x-y = \frac{3}{2} }} \right.$

$\displaystyle \left \{ {{- \frac{1}{2}x- \frac{1}{2}y =2} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { -\frac{1}{2}x-y = \frac{3}{2} }} \right.$

Agora podemos resolver... 
Usando o método da adição.

$\displaystyle \left \{ {{- \frac{1}{2}x- \frac{1}{2}y =2} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { -\frac{1}{2}x-y = \frac{3}{2}~~~~~~ ~~~~ ~\cdot(-1) }} \right. \\ \\ \\ \displaystyle \left \{ {{- \frac{1}{2}x- \frac{1}{2}y =2} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \atop { \frac{1}{2}x+y =- \frac{3}{2} }} \right. \\ \\ \\ \text{Somando as duas equacoes} \\ \\ \\ \frac{1}{2} y= \frac{1}{2} \\ \\ \\ y=\frac{ \frac{1}{2} }{ \frac{1}{2} } \\ \\ \\ \text{multiplica pelo inverso da segunda}$

$\displaystyle y= \frac{1}{2} \cdot \frac{2}{1} \\ \\ \boxed{y=1} \\ \\ \\ \text{Volta em uma das equacoes e substitui o valor de y e encontra o valor} \\ \text{do x} \\ \\ \\ \frac{1}{2}x+y =- \frac{3}{2} \\ \\ \frac{1}{2}x+ 1=- \frac{3}{2} \\ \\ \frac{1}{2} x=- \frac{5}{2} \\ \\ \\ x= \frac{- \frac{5}{2} }{ \frac{1}{2} } \\ \\ \\ x= -\frac{5}{2} \cdot \frac{2}{1} \\ \\ \\ \\\boxed{x=-5}$



2)


$\displaystyle \left \{ {{5(a-2b)=5} \atop { \frac{a}{2} - \frac{b}{3} = \frac{1}{6} }} \right. \\ \\ \\ \left \{ {{5a-10b=5} \atop { \frac{1}{2}a - \frac{1}{3}b = \frac{1}{6} }} \right. \\ \\ \\ \text{Multiplica a segunda equacao por -10, com isso eliminaremos o 'a'} \\ \\ \\ \left \{ {{5a-10b=5} \atop { -5a + \frac{10}{3}b = -\frac{5}{3} }} \right \\ \\ \\ \text{soma as equacoes} \\ \\ \\ - \frac{20}{3} b= \frac{10}{3} ~~~~~ ~~ \cdot(-1) \\ \\ \frac{20}{3} b=- \frac{10}{3}$

$\displaystyle b= \frac{- \frac{10}{3} }{ \frac{20}{3} } \\ \\ \\ b= -\frac{10}{\diagup\!\!\!\!3} \cdot \frac{\diagup\!\!\!\!3}{20} \\ \\ \\ b=- \frac{10}{20} \\ \\ \\ \boxed{b=- \frac{1}{2} } \\ \\ \\ \text{Substitui o valor de 'b' em qualquer uma das duas equacoes}$

$\displaystyle 5a-10b=5 \\ \\ 5a-10\cdot (-\frac{1}{2} )=5 \\ \\ \\ 5a+5=5 \\ \\ \\ 5a=5-5 \\ \\ \\ \boxed{a=0}$



3)


$\displaystyle \left \{ {{1- \frac{x-y}{4} =-(x+ \frac{7}{4} )} \atop { \frac{3x}{2}- \frac{1-y}{3} =x-y}} \right. \\ \\ \\ \left \{ {{1- (\frac{x}{4}- \frac{y}{4} ) =-x- \frac{7}{4} } \atop { \frac{3x}{2}- (\frac{1}{3}- \frac{y}{3} ) =x-y}} \right. \\ \\ \\ \left \{ {{- \frac{1}{4}x+ \frac{1}{4}y+x =- \frac{7}{4}-1 } \atop { \frac{3}{2}x- \frac{1}{3}+ \frac{1}{3}y   =x-y}} \right.$

$\displaystyle \left \{ {{- \frac{1}{4}x+ \frac{1}{4}y+x =- \frac{7}{4}-1 } \atop { \frac{3}{2}x+ \frac{1}{3}y -x+y = \frac{1}{3} }} \right. \\ \\ \\ \left \{ {{ \frac{3}{4}x+ \frac{1}{4}y =- \frac{11}{4} } \atop { \frac{1}{2}x+ \frac{4}{3}y = \frac{1}{3} }} \right. \\ \\ \\ \text{Multiplica a primeira equacao por } -\frac{16}{3} ,\text{ com isso conseguiremos }\\\text{eliminar o 'y'}$

$\displaystyle \left \{ {{ -4x -\frac{4}{3}y = \frac{44}{3} } \atop { \frac{1}{2}x+ \frac{4}{3}y = \frac{1}{3} }} \right. \\ \\ \\ \text{soma as duas equacoes} \\ \\ \\ - \frac{7}{2}x=15 \\ \\ \\ \frac{7}{2} x= -15 \\ \\ \\ x= \frac{-15}{ \frac{7}{2} } \\ \\ \\ x= -15 \cdot \frac{2}{7} \\ \\ \\\boxed{ x=- \frac{30}{7} }$

$\displaystyle \frac{1}{2}x+ \frac{4}{3} y= \frac{1}{3} \\ \\ \\ \frac{1}{2} \cdot( \frac{-30}{7} )+ \frac{4}{3} y= \frac{1}{3} \\ \\ \\ - \frac{15}{7} + \frac{4}{3} y= \frac{1}{3} \\ \\ \\ \frac{4}{3} y= \frac{52}{21} \\ \\ \\ y= \frac{ \frac{52}{21} }{ \frac{4}{3} } \\ \\ \\ y= \frac{52}{21} \cdot \frac{3}{4} \\ \\ \\ \boxed{y= \frac{13}{7} }$