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 :)
Trial and error
tried 8 x 6 =48
i used 10 x 5 to get closer to 100
10 x 5= 50
50+ 48= 98 all I need is 6 which is 2 x 3
8 x 6 = 48
5 x 10= 50
3 x 2 = 6
Total = 104
Hope this helps
Answer:
w < 82/5
Step-by-step explanation:
5w + 8 < 90
5w < 82
w < 82/5
Since he starts with (10x+7) dollars and is buying two pairs of jeans that each cost (4x-5) dollars you should subtract the cost of the jeans from his starting amount.
(10x+7)-2(4x-5)=
10x+7-8x+10=
2x+17
