Math Problem

[adc90]

I was wondering if anyone could help me solve this math problem.

"If John starts at zero and adds 3 on his calculator and Sam starts at a hundred and subtracts seven at what number would they meet."

The only way that I could think to do this would be to use a number line but I was wondering if anyone knows
of a easier solution.

[Ad Bot]

[WRB852]

30.

[tails]

42.

[adc90]

30.

Thanks,  but I already figured it out, I was just wondering if anyone knew of a easier solution than a number line.

[WRB852]

It just makes sense. I didn't even think about it.

[Reactor Worker]

You could graph it using a graphing calculator.

Ok, here we go...

In order to use a graph we need to assign a value to X and Y. For our equation, X will equal the number of moves John must make. We then write 2 functions, one for each of them.

John starts at zero, and for every move he makes, his value increases by 3 which can be represented by the equation...

Y₁=0+3x

Sam starts at 100, but every move he makes is worth seven thirds (7/3) of every move John makes so his equation is...

Y₂=100-(7/3)x

You then (using your graphing calculator) enter in the 2 formula. Hit the Graph key and 2 lines should appear. Expand the graph using the setting under Window if necessary.

Now go to 2_nd, Table. Here is the tricky part. Remember that we used the # of Johns moves as the X value and that the equation for John was y=3x ? Well, we can use that information to figure out at what number they meet. Just scroll up or down the table until you find the point at which the Y₁ value is 3 times as large as the Y₂ value and the X value in that row is the number you are looking for.

So it looks like...

Code: [Select]

X     Y1      Y2

30    90      30

Answer = x = 30.

[blaman]

I'd go with reactors solution. 

[Wedge]

Am I the only person getting 10?

y=0+3x (John starts at 0 and gains 3)

y=100-7x (Sam starts at 100 and loses 7)

0+3x=100-7x
10x=100
x=10

I don't see where you get 7/3 from, but maybe I'm just missing something.

[Reactor Worker]

Am I the only person getting 10?

y=0+3x (John starts at 0 and gains 3)

y=100-7x (Sam starts at 100 and loses 7)

0+3x=100-7x
10x=100
x=10

I don't see where you get 7/3 from, but maybe I'm just missing something.

It looks like you are making the same mistake I originally made.

You see, by saying that sams equation is "y=100-7x" you are effectively saying that each move same makes is equal to 7 moves that John makes but it is not. Sam's moves are equal to 7/3'rds of each of John's moves.

[Otis Da HousKat]

Am I the only person getting 10?

y=0+3x (John starts at 0 and gains 3)

y=100-7x (Sam starts at 100 and loses 7)

0+3x=100-7x
10x=100
x=10

I don't see where you get 7/3 from, but maybe I'm just missing something.

Think of it this way.

John adds 3 to zero ten times. He gets to 30

Sam subtracts 7 from one hundred ten times. He gets 30.

The ten you are getting isn't the number they meet at, just how many steps were in each person's sequence before they meet.

[Reactor Worker]

Think of it this way.

John adds 3 to zero ten times. He gets to 30

Sam subtracts 7 from one hundred ten times. He gets 30.

The ten you are getting isn't the number they meet at, just how many steps were in each person's sequence before they meet.

So effectively, using 7x for same and 3x for John is a way of establishing the ratio of steps needed for each person. That would make it a valid method as long as you went on to say that they would meet at 30.

[Otis Da HousKat]

So effectively, using 7x for same and 3x for John is a way of establishing the ratio of steps needed for each person. That would make it a valid method as long as you went on to say that they would meet at 30.

Yes, Wedge just needed that extra step where he plugged ten back into the equations.

[Ghoul]

I would just take two calculators, set it up, and press the '=' button at the same time until the numbers match.

[Otis Da HousKat]

I would just take two calculators, set it up, and press the '=' button at the same time until the numbers match.

I guess that would work. :?

[UnDiViDeD]

Here's one of my homework problems: