Question

Resolva os sistemas de equações lineares a seguir:
a)
{
x + y = 6
3x + 4y = 22
}


b)
{
x − 3y − z = −6
x + 4y + 7z = 17
- x + 6y + 6z = 19

}

Answer 1

Resposta:

a) x = 2 e y = 4

b) x = (-1), y = 1 e z = 2

Explicação passo a passo:

$a)\ \left \{ {{x+y=6} \atop {3x+4y=22}} \right. \\\\\\\left \{ {{y=6-x} \atop {3x+4y=22}} \right. \\\\\\\left \{ {{y=6-x} \atop {3x+4(6-x)=22}} \right. \\\\\\\left \{ {{y=6-x} \atop {3x+24-4x=22}} \right. \\\\\\\left \{ {{y=6-x} \atop {-x+24=22}} \right. \\\\\\\left \{ {{y=6-x} \atop {x-24=(-22)}} \right. \\\\\\\left \{ {{y=6-x} \atop {x=24-22}} \right. \\\\\\\left \{ {{y=6-x} \atop {x=2}} \right. \\\\\\\left \{ {{y=6-2} \atop {x=2}} \right.$

$\left \{ {{y=4} \atop {x=2}} \right.$

$b)\ \left\{\begin{array}{ccccc}x&-3y&-z&=&-6\\x&+4y&+7z&=&17\\-x&+6y&+6z&=&19\end{array}\right\} \\\\\\-x+6y+6z+x-3y-z=19+(-6)\\\\3y+5z=19-6\\\\\bold{3y+5z=13}\\\\\\\\-x+6y+6z+x+4y+7z=19+17\\\\\bold{10y+13z=36}$

$\left \{ {{3y+5z=13} \atop {10y+13z=36}} \right. \\\\\\\left \{ {{10(3y+5z)=10(13)} \atop {3(10y+13z)=3(36)}} \right. \\\\\\\left \{ {{30y+50z=130} \atop {30y+39z=108}} \right. \\\\\\\left \{ {{30y+50z=130} \atop {30y=108-39z}} \right. \\\\\\\left \{ {{108-39z+50z=130} \atop {30y=108-39z}} \right. \\\\\\\left \{ {{108+11z=130} \atop {30y=108-39z}} \right. \\\\\\\left \{ {{11z=130-108} \atop {30y=108-39z}} \right. \\\\\\\left \{ {{11z=22} \atop {30y=108-39z}} \right.$

$\left \{ {{z=\frac{22}{11} } \atop {30y=108-39z}} \right. \\\\\\\left \{ {{z=2 } \atop {30y=108-39z}} \right. \\\\\\\left \{ {{z=2 } \atop {30y=108-39(2)}} \right. \\\\\\\left \{ {{z=2 } \atop {30y=108-78}} \right. \\\\\\\left \{ {{z=2 } \atop {30y=30}} \right. \\\\\\\left \{ {{z=2 } \atop {y=\frac{30}{30} }} \right. \\\\\\\left \{ {{z=2 } \atop {y=1}} \right. \\\\\\x+4y+7z=17\\\\x+4(1)+7(2)=17\\\\x+4+14=17\\\\x+18=17\\\\x=17-18\\\\x=(-1)$