math??problems involving linear systems?

Question

hi if you explain these two examples thoroughly, i will click correct answer for you.

1.A sports club charges an intiation fee and a monthly fee. At the end of 5 months, a member had paid a total of $170. At the end of 10 months, she paid a total of $296. What is the initiation fee?

2. A crate of 36 grapefruit has a total mass of 4 kg. When 12 grapefruit are removed, the total mass is 3 kg. Determine the mass of the crate and the mass of one grapefruit.

Answer 1

These are examples of lines. To find the equation of a line, you either need two points, or a point and a slope.

Both of these are examples of lines with two points.

(5,170) and (10,296) are each on the line y=(126/5)x+44 The initiation fee is $44.

(36,4) and (24,3) are each on the line y=12x-12 The crate weighs 12 (no grapefruit) and one grapefruit weighs 1/12 kg.

Answer 2

1) 296 - 170 = 126
10 - 5 = 5
126/5 = 25.2 per month
25.2(10) = 252
296 - 252 = 44

2) 12 grapefruit weigh 1 kg, so 1 grapefruit weighs 1/12 kg
12 times 3 = 36, so 1 times 3 = 3. That means 36 grapefruit weigh 3 kg. Which means the crate is 1 kg

Answer 3

Qn1

Let I be cost of initiation fee
Let M be cost of monthly fee

I + 5M = 170
I + 10M = 296
=> 5M = 296 - 170 = 126
=> M = $25.20
=> I = $44

So inititation fee is $44

Qn2

Let c be mass of crate. Let g be mass of grapefruit.

c + 36g = 4
c + 24g = 3
=> 12g = 1
g = 1/12 kilograms
=> c = 1 kilogram

Therefore the mass of the crate is 1 kg and the mass of one grapefruit is 1/12 kg.

Answer 4

1. the init fee is $44

2. mass of crate = 1 kg
mass of 1 grapefruit = 1/12 kg

Answer 5

Let initiation fee = x

Monthly fee = y

x+5y = 170
x+10y = 296

So, 5y = 126
or y = $ 25.2

So, x = $ 44

Let mass of crate = x

mass of grapefruit = y

x + 36y = 4

x + 12 y = 3

So, 24y = 1

or y = 41.67 g

So, x = 3 - 12*41.67 g = 2.5 kg

Answer 6

Do your own homework.