Resolva os sistemas abaixo:
A.{2× +3y=8
{y=24
B.x+y=132
x=2y-12
C.{x-y=4
{x+y=20
D.x=y=7
x+y=173
E.{2x + 3y=34
{x - 3y=-19
Soluções para a tarefa
Respondido por
1
2 x+3.24=8
2 x+72=8
2 x=8-72
2 x=64
x=64/2
x=32
letra a
2 x+72=8
2 x=8-72
2 x=64
x=64/2
x=32
letra a
Respondido por
2
a. 2x+3y=8
y=24
1)2x+3y=8
2x+3(24)=8 ⇒Substitui o y
2x+72=8
2x=8-72
2x=-64
x=-64/2
x=-32
S={-32,24}
b. x+y=132
x=2y-12
1)x+y=132
(2y-12)+y=132⇒substitui o x
2y+y=132+12
3y=144
y=144/3
y=48
2) x=2y-12
x=2(48)-12⇒substitui o y
x=96-12
x=84
S={84,48}
c.x-y=4⇔x=4+y
x+y=20
1)x+y=20
(4+y)+y=20⇒substitui o x
2y=20-4
y=16/2
y=8
2)x=4+y
x=4+8⇒substitui o y
x=12
s={12,8}
E.2x + 3y=34
x - 3y=-19⇔x=-19+3y
1)2x+3y=34
2(-19+3y)+3y=34⇒substitui o x
-38+6y+3y=34
9y=34+38
y=72/9
y=8
2)x=-19+3y
x=-19+3(8)⇒substitui o y
x=-19+24
x=5
S={5,8}
y=24
1)2x+3y=8
2x+3(24)=8 ⇒Substitui o y
2x+72=8
2x=8-72
2x=-64
x=-64/2
x=-32
S={-32,24}
b. x+y=132
x=2y-12
1)x+y=132
(2y-12)+y=132⇒substitui o x
2y+y=132+12
3y=144
y=144/3
y=48
2) x=2y-12
x=2(48)-12⇒substitui o y
x=96-12
x=84
S={84,48}
c.x-y=4⇔x=4+y
x+y=20
1)x+y=20
(4+y)+y=20⇒substitui o x
2y=20-4
y=16/2
y=8
2)x=4+y
x=4+8⇒substitui o y
x=12
s={12,8}
E.2x + 3y=34
x - 3y=-19⇔x=-19+3y
1)2x+3y=34
2(-19+3y)+3y=34⇒substitui o x
-38+6y+3y=34
9y=34+38
y=72/9
y=8
2)x=-19+3y
x=-19+3(8)⇒substitui o y
x=-19+24
x=5
S={5,8}
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