distance

irspow said:
That’s fine what you did there, now isolate and solve for t. To be clear

30t-120=0
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Why equal to 0. Where does the 0 come from?. It was not there at the beginning (at#11) when I equaled both equations. Please, can you explain that before I solve for t?.
 
eddy2017 said:
Why equal to 0. Where does the 0 come from?. It was not there at the beginning (at#11) when I equaled both equations. Please, can you explain that before I solve for t?.
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We are dealing with equations, you don’t lose the equal sign while manipulating the equations algebraically.

We started with:

75(t-1)=45(t+1) distribute

75t-75=45t+45 now get the t terms on one side and the constants on the other side.

75t-45t=45+75 which is

30t=120 divide both sides by 30 (here you put everything on the left side of the equation as 30t-120=0)

t=4

Then just plug this t into either original distance equation

d=75(t-1)

d=75(4-1)

d=75(3)

d=225
 
irspow said:
We are dealing with equations, you don’t lose the equal sign while manipulating the equations algebraically.

We started with:

75(t-1)=45(t+1) distribute

75t-75=45t+45 now get the t terms on one side and the constants on the other side.

75t-45t=45+75 which is

30t=120 divide both sides by 30 (here you put everything on the left side of the equation as 30t-120=0)

t=4

Then just plug this t into either original distance equation

d=75(t-1)

d=75(4-1)

d=75(3)

d=225
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But did you get rid of the - sign in 120 (-120). How was that?. Thanks for the bit about not losing the sign when working the equations that have the = sign. That is great to know.
 
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You had

75t-75=45t+45 whatever you do to one side of the equation you must do to both sides of the equation ( or it is no longer an equation)

what you did: subtract 45t from both sides

30t-75=45. then you subtracted 45 from both sides

30t-120=0. you lost the =0 part

To solve for t you should have added 120 to both sides (this is where the -120 goes away)

30t=120 then divide both sides by 30

t=4
 
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We got in the weeds for a bit there, I think making this problem seem more complex than it really is. The straight forward process is not that difficult.

We know both drivers went the same distance at different speeds and times. We know distance equals velocity times time. And we are given these different velocities. So:

d=vt

75(t-1)=45(t+1) distribute

75t-75=45t+45 (add 75 to both sides)

75t=45t+120 (subtract 45t from both sides)

30t=120 (divide 30 from both sides)

t=4, We can now use this t in either original distance equation to solve for d.

75(t-1)=75(4-1)=75*3=225 or

45(t+1)=45(4+1)=45*5=225

If any of that doesn’t make sense, let me know. :)
 
eddy2017 said:
Like t is the time that it will take him to get to the event driving at different speeds, so there will be two different t's?.
'cause, i know
d=rt
I have the rates of speed, I need to find the time and the distance, right?.
Can this be set up like this?
d=rt
d=45(t+1) time plus the extra hour it takes him to get there driving at this speed)
d=75(t-1)
time minus the hour it takes him to get there driving at this speed)
d=75(t-1)
Is this good?.
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No, it is NEVER good if you do not define YOUR variables.

You can do it! What does your t mean? Think what t-1 and t + 1 means and that should help you know what t means.
 
irspow said:
We got in the weeds for a bit there, I think making this problem seem more complex than it really is. The straight forward process is not that difficult.

We know both drivers went the same distance at different speeds and times. We know distance equals velocity times time. And we are given these different velocities. So:

d=vt

75(t-1)=45(t+1) distribute

75t-75=45t+45 (add 75 to both sides)

75t=45t+120 (subtract 45t from both sides)

30t=120 (divide 30 from both sides)

t=4, We can now use this t in either original distance equation to solve for d.

75(t-1)=75(4-1)=75*3=225 or

45(t+1)=45(4+1)=45*5=225

If any of that doesn’t make sense, let me know. :)
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Excellent . It makes full sense. Thanks for the corrections and nice explanations. I was going to ask you where the 0 was dropped, when I saw your post already anticipating my question. Appreciated!.
 
Jomo said:
No, it is NEVER good if you do not define YOUR variables.

You can do it! What does your t mean? Think what t-1 and t + 1 means and that should help you know what t means.
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Yes!. Definitely. Thank you Jomo
 
irspow said:
d=vt so you are getting there. You said you know velocities and times and we know distances are the same so you can solve for time:

75(t-1)=45(t+1)

Can you solve the above for t?
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No, eddy is getting nowhere until he defines and understands his own variable. This is a must to be able to answer the question. Even if he solves for t and say gets t = 2 hrs the issue is what does t=2 hrs even means?
 
Jomo said:
No, eddy is getting nowhere until he defines and understands his own variable. This is a must to be able to answer the question. Even if he solves for t and say gets t = 2 hrs the issue is what does t=2 hrs even means?
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Yes, you're right. But I think I understood the whole process.
 
P
Jomo said:
How can you solve for t if there is no equation??
I had this:
30t-120=0 it was here when I was advised to drop the 0 and solve for t.
If it can't be done, then, what should I do right there?
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eddy2017 said:
But did you get rid of the - sign in 120 (-120). How was that?. Thanks for the bit about not losing the sign when working the equations that have the = sign. That is great to know.
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If you subtract two quantities(like 30t and 120) and the result is 0, then the two quants must be equal. That is 30t = 120.
 
eddy2017 said:
Yes, you're right. But I think I understood the whole process.
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How can you understand anything you wrote if you do not know what your variable represents. I can not stress enough how important it is to know what your variable means. How can your equations make any sense to you or anyone if you do not know what your variable means.
 
That is why i thought. But irslow told me I could not lose the 0. I am not blaming him I am just stating what he told me to do.
 
eddy2017 said:
I dont understand. Please, Jomo, I asked you a question at 33. I need to know the answer to understand this.
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eddy, start from the beginning. Do not write down any equations at all. First decide what your variable t represents. Do not move past there until you know what your variable means. If you insist I will tell you but you really should figure it out on your own. Think about where t+1 and t-1 came from. What are you adding and subtracting 1 from?
 
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