Matemática, perguntado por alexsutt, 8 meses atrás

ME AJUDEM TENHO QUE ENTREGAR HOJE DE MANHÃ!!

2. Efetue os sistemas de duas equações pelo método da adição:

a) x + y = 35
x - y = 5

b) x + y = 42
x - y = 8

c) 2x +2y = 14
3x - 2y = 6

d) 5x + 3y = 21
2x - 3y = 14

Soluções para a tarefa

Respondido por lukascunhahp0vlu0
1

Resposta:

a) x = 20

b) x = 25

c) x = 4

d) x = 5

Explicação passo-a-passo:

a) x + y = 35

   x - y = 5

2x = 40

 x = 40/2

 x = 20

b) x + y = 42

   x-y = 8

2x = 50

 x = 50/2

 x = 25

c)2x + 2y = 14

  3x - 2y = 6

5x =20

 x = 20/5

 x = 4

d) 5x +3y = 21

   2x - 3y = 14

7x = 35

 x = 35/7

 x = 5

Respondido por gabrielhiroshi01
1

Explicação passo-a-passo:

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Resolvendo os sistemas:

a)

\left \{ {\big{x+y=35} \atop \big{x-y=5}} \right. \\\\\text{Somando as duas equa\c{c}\~oes:}\\\\x+x+y-y=35+5\\\\2x=40\\\\x=\dfrac{40}{2}\\\\\boxed{x=20} \\\\\text{Substituindo o x na primeira equa\c{c}\~ao:}\\\\x+y=35\\\\20+y=35\\\\y=35-20\\\\\boxed{y=15}\\\\\boxed{\boxed{S=\{(20,15)\}}}

b)

\left \{ {\big{x+y=42} \atop \big{x-y=8}} \right. \\\\\text{Somando as duas equa\c{c}\~oes:}\\\\x+x+y-y=42+8\\\\2x=50\\\\x=\dfrac{50}{2}\\\\\boxed{x=25} \\\\\text{Substituindo o x na primeira equa\c{c}\~ao:}\\\\x+y=35\\\\25+y=42\\\\y=42-25\\\\\boxed{y=17}\\\\\boxed{\boxed{S=\{(25,17)\}}}

c)

\left \{ {\big{2x+2y=14} \atop \big{3x-2y=6}} \right. \\\\\text{Somando as duas equa\c{c}\~oes:}\\\\2x+3x+2y-2y=14+6\\\\5x=20\\\\x=\dfrac{20}{5}\\\\\boxed{x=4} \\\\\text{Substituindo o x na primeira equa\c{c}\~ao:}\\\\2x+2y=14\\\\2.4+2y=14\\\\8+2y=14\\\\2y=14-8\\\\2y=6\\\\y=\dfrac{6}{2} \\\\\boxed{y=3}\\\\\boxed{\boxed{S=\{(4,3)\}}}

d)

\left \{ {\big{5x+3y=21} \atop \big{2x-3y=14}} \right. \\\\\text{Somando as duas equa\c{c}\~oes:}\\\\5x+2x+3y-3y=21+14\\\\7x=35\\\\x=\dfrac{35}{7}\\\\\boxed{x=5} \\\\\text{Substituindo o x na primeira equa\c{c}\~ao:}\\\\5x+3y=21\\\\5.5+3y=21\\\\25+3y=21\\\\3y=21-25\\\\3y=-4\\\\\boxed{y=\dfrac{-4}{3} }\\\\\boxed{\boxed{S=\bigg\{\bigg(5,\dfrac{-4}{3} \bigg)\bigg\}}}

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