2 Math problems?

Question

1.) The monthly costs associated with driving a Ford Explorer are the monthly loan payment plus the cost of filling up ur tank with gasoline. If you fill up 3 times a month your total money cost is $520. If you fill up 5 times a month, your total monthly cost is $600. How much is your monthly loan and how much does it cost everytime you fill up with gasoline?

2. A clothing store sells a shirt costing $20 for $33 and a jacket costing $60 for $93. If the markup policy of the store for items costing over $10 is assumed to be linear, write an equation that expresses retail price R in terms of cost C (wholesale price) What does the store pay for a suit that retails for $240.

10 pts to the best answer. please show me how i got 173 and 1/3 = x + b for one equation for the first ? and 120 =x + b for another equation to #1. and my equation to the second one is y = 3/4 X + 6 and 1/3. Am I going in the right direction.

Answer 1

loan = L
one fill up with gas = G
L + 3G = 520
L + 5G = 600
2G = 80
G = 40
L = 400


You find the slope from (R1 - R0)/(C1 - C0)
(93-33)/(60-20) = 60/40 = 1.5
Then you plug the slope into the equation to find the intercept
R = mC + b
93 = 1.5(60) + b
93 = 90 + b
3 = b
sooooo,
R = 1.5C + 3
240 = 1.5C + 3
237 = 1.5C
158 = C

Answer 2

1) Lets let $L = price of loan per month and $P be the price for one rank of petrol.

We know
$L + 3$P = 520
$L + 5$P = 600

If we subrtact the first equation from the second we get:
$L - $L +3$P - 5$P = -80
-2$P = -80
P = 40

So a tank of petrol costs $40 (I wish....)

If we sub this back into one of the equations:
$L + 3(40) = 520
$L = 400.

So the montly loan is $400 and one tank of petrol costs $40.

I dont know how you got the equations you said you got...

2) A nice trick that may make this a bit easier: Lets pretend that the Wholesale price is X and the retail price is Y in an equation. This means we have 2 points (20,33) and (60,93). Lets rind the equation using the 2 point formula.

(y - 33)/(x-20)=(93-33)/(60-20)
60y - 1320= 40x -800
6y-4x -52 =0

sooo

6R - 4C -52 = 0
when R = 240
1440 - 52 = 4C


6R -4C + 9 =0

Well... that didnt seem to be working. Ah well, I'm off to get something to eat...

Answer 3

Your answers at the bottom are confusing me... so I will just type my explaination.

#1. First you need to find out how much the gas costs. 600-520 =80 so 2 tanks of gas were $80. Which means one tank was $40. So if we multiply the $40 x 5= $200 for gas a month. Which means that since he paid $600 for the loan and gas if we subtract the gas from that $600 we end up with a Truck pmt of $400.

#2. awww I am too tired to do it now....

Answer 4

1) Well, set g as the amount of times you fill up,x as the cost of gas and c for the cost of the loan and t as total cost.
t = gx + c

So, given your problem, you have 2 equations.
520 = 3x + c
600 = 5x + c

Using linear combination to solve the system, youre left with.
80 = 2x
x = 40

Plug x back in either equation to solve for c
600 = 5(40) + c
600 = 200 + c
400 = c

So, your answer is that your monthly loan is 400 and it costs you 40 everytime you fill the tank.