A ship travels between two ports. The cost of fuel is C=$100(ax+b/x+10)

Where x is the average speed of the ship in knots (x > 0) and a and b are constants. If the ship travels at 4 knots the cost of fuel is $9000 but at 6 knots it is $7000.

a) Form two linear equations consist of a and b.
b) Find the values of a and b.
c) If there are two ports that are 1000 knots apart, what is the cost of the fuel.

Since you failed it twice already... if someone hands you an answer you are going to fail a third time because you don't understand it.

How do YOU go about solving it first?

We HELP you with your homework... we don't DO it for you

Link to the rules for homework. https://www.askmehelpdesk.com/math-sciences/announcement-font-color-ff0000-u-b-read-first-expectations-homework-help-board-b-u-font.html

C = 100\(\frac{ax+b}{x+10}\)

a) Put first C = 9000, x = 4, then a second time using C = 7000 and x = 6.

You'll get two equations.

b) Solve those two equations simultaneously.

c) That questions seems wrong... knots is a measure of speed and two ports cannot be separated by 'some speed'...

I guess that you have to put x = 1000 and use the values of a and b you got earlier for that.

If I can understand the question I will not post it in here already.

If i can understand the question i will not post it in here already.

I agree, but the issue is, you should have posted what you have attempted. This is the usual procedure. Since you were new here, I didn't really impose it, but if you need more help, posting first what you have attempted would be something I'd encourage you to do.

That said, smoothy was right and his 'thumbs down' was not deserved.

Okay? :)

henrypoo does not find this helpful : at least give some guide
Firt off. You broke the rules for homework... YOU did take the time to read that link I provided to you Before the rude reddie... didn't you?

Then you get rude and break another rule about WHEN you are allowed to give a not helpful. Site rules Specifically say you can ONLY do that for inaccurate information. What I gave was completely accurate.

You never attempted to even try the question... I asked you how you would try to solve it...

Unknown gave you hints of how to start to solve it... but you want us to hand you the answer without you doing ANY work. Which means you learn nothing and WILL fail it a third time.

If you can't try... we aren't going to do it for you. That's why you have homework... its to learn how to do something.
Incidentally... I'm reporting this to site admins.

But how to solve the equation. Without b how to find a??

What did you get for part (a) ?

for part a is 9000=100(ax+b/x+10)

for part b is 7000=100(a(4)+b/6+10)

this is where I got stuck. I gt the 2 equation but for the second qns they ask me to find the value of a and b!

sorry for part a is 9000=100(a(4)+b / 4+10)

part b is 7000=100(a(6)+b / 6+10)

Right, can you simplify it a bit? Work out the denominator, then cross multiply for both.

Note that it's still in part (a). Part (b) is not involved yet.

9000= (400a+100b / 1400)

I don't know how to cross multiply cause I never learn this before.

Part (a)

C=\frac{100(ax+b)}{x+10}

Put C = 9000, and x = 4;

9000=\frac{100(a(4)+b)}{(4)+10}

What is 4+10 in the denominator on the right?
Can you expand the numerator on the right?

That's the first part in (a)

Second part in (a), put C = 7000, x = 6

7000=\frac{100(a(6)+b)}{(6)+10}

So the same thing as part one.

so for part a is it like that?

9000 = 100(4a+b) / 14

Yes the first part in part (a) is that. Now multiply both sides by 14, what do you get?

Multiply both side as in 14 times 9000,100,4a and b?

No, like this:

From this:

9000 = \frac{100(4a + b)}{14}

to this:

14(9000) = 14\(\frac{100(4a + b)}{14}\)

Can you work this out?

12,600= 1400(4a+b) / 196 ?

Not totally true...

What is 2 times 1/2 ?

1?

Right, now,

14\(\frac{100(4a + b)}{14}\)

1400(4a+b) / 196?

Then, why did you tell me that:

2\(\frac12\) = 1

?

If you did what you just did to 14 up there, you'd have:

2\(\frac12\) = \frac24

:eek:

So, some confusion from your part? :confused:

So it should be 1400(4a+b) / 14

Yes, now you got it.

And what can you do now? Do you see why I chose to multiply both sides by 14? You can now reduce the fraction.

Just like you reduce 200/2 to 100/1 to finally 100.

So its (4a+b) / 100?

Try again...

\frac{1400(4a + b)}{14} = \(\frac{1400}{14}\)(4a+b)

(100)(4a+b)?

Right! :)

Now let's do another one, still to solve this problem.

12\ 600 = 100(4a+b)

Can you divide both sides by 100? :)

126=(1)(4a+b)?

Good! Now, save this for later. The other one now.

7000 = \frac{100(6a + b)}{16}

Can you try putting it in the form of the one you just did?

can you show me the final step for 126 = (1)(4a+b)?
so that the other one I know how to do.

the other part is 112 = (1)(6a+b)

the other part is 112 = (1)(6a+b)

Geesh, it's only now that I notice that you made a wrong calculation earlier, 9000*14 = 126,000, not 12,600.

Not quite. I get:

1120 = 6a + b

(after distributing 1 into the brackets)

Okay, now we have:

1260 = 4a + b
1120 = 6a + b

We can get a and b from there. Subtract both equations:


1260 = 4a + 1b
1120 = 6a + 1b -
= a + b

sorry sorry. To sum up the other question. I have done the 2 equation.

0= 4a+b -126
0= 6a+b - 112

from the above I can find a and b using stimultaneuos equation?

Right, a correction is to be made through, it's 1260 and 1120 and one last thing, you posted in the wrong thread :p

Ok, I'm moving those posts.