# Thread: If t ties cost d dollars...

1. ## If t ties cost d dollars...

solving this algebraically , If t ties cost d dollars, how many dollars would t+1 ties cost ??

2. ## re: If t ties cost d dollars...

Originally Posted by omar1
solving this algebraically , If t ties cost d dollars, how many dollars would t+1 ties cost ??

one tie would cost d/t dollars ... so, how much would t+1 ties cost?

3. ## Re: If t ties cost d dollars...

okay....
this is how I would do it:
one tie = d/t
so t+1 must be (d+t)/t
got it? you add one to each side .

hope it helps.

4. ## Re: If t ties cost d dollars...

Originally Posted by IBstudent
okay....
this is how I would do it:
one tie = d/t
so t+1 must be (d+t)/t
got it? you add one to each side .

hope it helps.
and that would be incorrect

5. ## Re: If t ties cost d dollars...

:O
but look:
t= d/t
t+1 = d/t +1
t+1 = (d+t)/t
where is the wrong bit?

Edit: even substituding works:
say 50 ties each costing 5 dollars
10 = 50/5
10+1= (50+5)/5
11=55/5
11=11

Thanks

6. ## Re: If t ties cost d dollars...

$t$ ties cost $d$ dollars

1 tie costs $\frac{d}{t}$ dollars

adding the cost for t ties and 1 tie ...

$d + \frac{d}{t}$ which does not equal $\frac{d+t}{t}$

note ...

$d + \frac{d}{t} = \frac{dt}{t} + \frac{d}{t} = \frac{dt+d}{t} = \frac{d(t+1)}{t} = \frac{d}{t}(t+1)$

... which is the cost of one tie times (t+1) ties

7. ## Re: If t ties cost d dollars...

Originally Posted by IBstudent
:O
but look:
t= d/t
Well, your very first equation is wrong!

t+1 = d/t +1
t+1 = (d+t)/t
where is the wrong bit?

Edit: even substituding works:
say 50 ties each costing 5 dollars
10 = 50/5
10+1= (50+5)/5
11=55/5
11=11

Thanks

8. ## Re: If t ties cost d dollars...

Originally Posted by HallsofIvy
Well, your very first equation is wrong!
Ouch!
Thank you for pointing that out though...

9. ## Re: If t ties cost d dollars...

Originally Posted by skeeter
$t$ ties cost $d$ dollars

1 tie costs $\frac{d}{t}$ dollars

adding the cost for t ties and 1 tie ...

$d + \frac{d}{t}$ which does not equal $\frac{d+t}{t}$

note ...

$d + \frac{d}{t} = \frac{dt}{t} + \frac{d}{t} = \frac{dt+d}{t} = \frac{d(t+1)}{t} = \frac{d}{t}(t+1)$

... which is the cost of one tie times (t+1) ties