Will post random car photos for math homework help

Will post random car photos for math homework help.

While trick or treating in Clayton, NY (population 1,978), some friends play a counting game-they counted all the houses they visited while trick or treating (they all counted the same houses). Diego counted the houses by 5's, and had four houses left at the end. Madeline counted the houses by 7's, and had six houses left over at the end. Based on Diego's and Madeline's counts, how many houses did they visit? Is this the only possible answer? Explain why you are sure you found all possible counts.

Bump

What makes you think anyone here knows math?

fuck off to Veeky Forums

>Redcarbigtires

jesus fucking christ. No underage shits allowed here you faggot

You must be 18 to post here. If you're 18 or older and can't figure this question out, go kill yourself.

...

Someone has to be a computer science or engineering major.

We aren't all idiots.

>We aren't all idiots.

Wishful thinking
>captcha POLLING ARCHITECTS

bruh kill yourself.

Nice evo

Ok, I feel like spoonfeeding you for a bit
>Diego counted the houses by 5's, and had four houses left at the end
So the number can only end in either 4 or 9. Now start thinking on what you need with multiples of 7.

So we got something like
5*X + 4 = Y
7*Z + 6 = Y
I don't really remember how to go from here and if it's possible at all to solve it, haven't done math in aeons. I'll look into it tho, seems fun

Spoon feed me a bit more if you will.
How the hell did you get that they have to end in either a 4 or 9 from just that sentence?

No I won't. How young are you that you can't figure multiples of 5?

Interesting.

That may help me finish the problem. I appreciate it.

Because 5 multiplied by any whole number ends in either 5 or 0 , so adding 4 would make it 9 or 4.

Whoops.

Ok thanks.

Oh what the hell, we're already this far. Answer can be any multiple of 7 ending in 8 or 3, plus 6. This up to 1978, assuming one person per house.

Thank you. I shall work it out.

It's been almost 15 years since I last made this, but you have to turn
>5*X + 4 = Y
>7*Z + 6 = Y
Into
5*X + 4 = 7*Z + 6

Now work it from there

It's 69
>Is this the only possible answer?
No. Also 104 and a whole bunch of other numbers.

5*X=7*Z+6
5*X=7*Z-4
----------------
5*X=7*Z+2

I know if I move one over to the other side I have to subtract the reciprocal, but do I take all of 5*X, or 7*Z??

>5*X=7*Z+2

Oh yeah, this was it. Iirc now you can do
>X = (7Z+2)/5

Thanks for the guidance, I appreciate it.

7x+6=1978
5x+4=1978
Kramers rule
I want a midnight purple S30 on Watanabes please.

>7x+6=1978

Just drop the trip and save yourself the embarassment. This puts X at ~281.71

That simple huh??
Man, I've been thinking that it would be complex all this time.

Sorry I don't have a midnight purple..

>That simple huh??

No, that is flat out wrong. You don't get an integer in either of the variables this way.

Yeah, that's what I thought.
What did you have in mind?

D'oh fuck.
Disregard any math advice you get from me, but too late to take back the S30

This is 5th grade shit.

Solutions are in the form
7x + 6 = 5y + 4

Population is extraneous info.

Simplifying, find x and y that satisfy
5y - 7x = 2

So one solution is x = 4, y = 6. Checking that, 5y + 4 = 34 (4 left over), 7x + 6 = 34 (6 left over). There might be 34 houses. This is just one solution