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cupoosta [38]
2 years ago
13

Math help @LIEMONDER

Mathematics
1 answer:
klio [65]2 years ago
5 0
First picture)
I: 5x+2y=-4
II: -3x+2y=12

add I+(-1*II):
5x+2y-(-3x+2y)=-4-12
8x=-16
x=-2

insert x=-2 into I:
5*(-2)+2y=-4
-10+2y=-4
2y=6
y=3

(-2,3)

question 6)
I: totalcost=115=3*childs+5*adults
II: 33=adults+childs
33-adults=childs

insert childs into I:
115=3*(33-adults)+5*adults
115=99-3*adults+5*adults
16=2*adults
8=adults

insert adults into II:
33-8=childs
25=childs

so it's the last option

question 7)
a) y<6 and y>2 can also be written as 2<y<6, so solution 3 exist for example
b) y>6 and y>2 can also be written as 2<6<y, so solution 7 exist for example
c) y<6 and y<2 inverse of b: y<2<6, so for example 1
d) y>6 and y<2: y<2<6<y, this is impossible as y can be only either bigger or smaller than 2 or 6

so it's the last option

question 8)
I: x+y=12
II: x-y=6

subtract: I-II:
x+y-(x-y)=12-6
2y=6
y=3

insert y into I:
x+3=12
x=9

(9,3)

question 9)
I: x+y=6
II: x=y+5

if you take the x=y+5 definition of II and substitute it into I:
(y+5)+y=6

which is the second option :)
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