Find their respective speeds

JeffM said:
YES.

Now I shall show you I would do this problem if I found it difficult. (People like lev, Subhotosh, and me, who have been doing algebra problems for decades, do not find this a difficult problem and so take shortcuts, but when we find a problem to be hard, we do something like this.) I like Subhotosh's numbering.

(Step 1) Read the problem carefully and make sure you understand it well. This may allow you to take shortcuts. If it doesn't

(Step 2) Assign a distinct letter to each potentially relevant quantity and write them down. Try to use the notation to help your memory.

[MATH]d_1 = \text {distance travelled by car 1.}[/MATH]
[MATH]r_1 = \text {average rate of car 1.}[/MATH]
[MATH]t_1 = \text {time traveled by car 1.}[/MATH]
[MATH]d_2= \text {distance travelled by car 2.}[/MATH]
[MATH]r_2 = \text {average rate of car 2.}[/MATH]
[MATH]t_2 = \text {time traveled by car 2.}[/MATH]
[MATH]d_3 = \text {final distance between the cars.}[/MATH]
Notice that the distances are all subscripted d, all the rates are subscripted r, and all the times are subscripted t. It puts the least stress on my aged memory.

(Step 3): Look for clues, translate them into mathematical notation, and write the translations down as you find them. The whole point of this writing down is to relieve your memory while you are trying to think. This step is where most of the thinking comes in, and you do not want to be trying to remember what everything means while you read and think. Here is a tip: look for the easy stuff first.

[MATH]d_3 = 40.[/MATH] Easy, the problem explicitly tells you that in words.

[MATH]t_1 = 2 \text { and } t_2 = 2.[/MATH] Easy, the problem implicitly tells you in words

[MATH]r_1 = r_2 + 4.[/MATH] Easy, the problem explicitly tells you that in words.

We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.

[MATH]d_1^2 + d_2^2 = d_3^2.[/MATH]
We still need two more clues. But we know a general formula relating rates, times, and distances, which gives us those two extra two clues.

[MATH]r * t = d \implies r_1 * t_1 = d_1 \text { and } r_2 * t_2 = d_2.[/MATH]
This step is where the brain work occurs, but you have a goal. Seven relevant pronumerals requires seven independent equations, which means seven clues to find and translate. You are not working entirely in the dark.

When you are done, you should be looking at

[MATH]d_3 = 40,\\ t_1 = 2\\ t_2 = 2,\\ r_1 = r_2 + 4,\\ d_1^2 + d_2^2 = d_3^2,\\ r_1 * t_1 = d_1 \text { and,}\\ r_2 * t_2 = d_2.[/MATH](Step 4): Now it is mechanical. Just carefully solve the system equations by substitution

It is trivial to get to

[MATH]d_1^2 + d_2^2 = d_3^2 = 40^2 = 1600,\\ 2r_1 = d_1,\\ 2r_2 = d_2,\\,[/MATH]Now it is easy to get to

[MATH](2r_1)^2 + (2r_2^2) = 1600 \implies 4r_1^2 + 4r_2^2 = 1600 \implies r_1^2 + r_2^2 = 400.[/MATH]
Just two unknowns left. Do we know how they are related? We do so

[MATH](r_2 + 4)^2 + r_2^2 = 400 \implies r_2^2 + 8r_2 + 16 + r_2^2 = 400 \implies\\ 2r_2^2 + 8r_2 - 384 = 0 \implies r_2^2 + 4r_2 - 192 = 0.[/MATH]Can you solve that equation? If so, you can say what every relevant quantity is.

Then go on to step 5: check that the answers work.

It is not a formula. It is a procedure that requires persistence, thought, and care. It is a way of thinking step by step through complicated problems.
Click to expand...

Hi, can you explain what you meant here with
d_1^2 + d_2^2 = d_3^2.
I did not understand that.
 
eddy2017 said:
Sorry, when i git your message i had already gone to bed.
how (2x+8)^2 became 4x^2 +64
2x^2=4x^2+ 8^2
2x^2+ 64
Click to expand...
What are you studying now? Seems like you should be familiar with formulas such as (a + b)^2 = a^2 + b^2 + 2ab if you are expected to solve equations like 40^2==(2x)^2+ (2x+8)^2
 
I have seen it but I am not sure when to use it. I am studying math from 6 to 10th at this moment. Intermediate Algebra and geometry1, I think.
 
eddy2017 said:
I have seen it but I am not sure when to use it. I am studying math from 6 to 10th at this moment. Intermediate Algebra and geometry1, I think.
Click to expand...
Then you better learn these formulas:

Square of the sum

Library: The square of the sum is one of the Factoring formulas. The square of the sum of two numbers is equal to the square of the first number, plus twice the product of the first and second numbers, plus the square of the second number.
onlinemschool.com onlinemschool.com

Square of the difference

Library: The square of the difference is one of the Factoring formulas. The square of the difference of two numbers is equal to the square of the first number, minus twice the product of the first and second numbers, plus the square of the second number.
onlinemschool.com onlinemschool.com

Difference of squares

Library: Difference of squares is one of the Factoring formulas. The difference of squares of the two numbers is equal to product of sum of these numbers and the difference of these numbers.
onlinemschool.com onlinemschool.com
 
Here's the work i had done so far until it was said to me there was a mistake in the square of the sum. Let's take it from here please.
Speed, distance timep w.p

Two people start from the same point and at the same time, both going through two perpendicular roads. Knowing that the speed of one of them is 4 km/h more than the other, and that after two hours both are 40 km away from each other, find their respective speeds.



We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.



Step 1 Read the problem carefully and make sure you understand it well. This may allow you to take shortcuts. If it doesn't

Step 2 Assign a distinct letter to each potentially relevant quantity and write them down. Try to use the notation to help your memory.

d1=distance travelled by car 1.

r1=average rate of car 1.

t1=time traveled by car 1.


d2=distance travelled by car 2

r2=average rate of car 2.

t2=time traveled by car 2.

d3=final distance between the cars.




  • Look for clues, translate them into mathematical notation, and write the translations down as you find them.
  • The whole point of this writing down is to relieve your memory while you are trying to think.
  • step is where most of the thinking comes in, and you do not want to be trying to remember what everything means while you read and think.
  • Here is a tip: look for the easy stuff first.

givens

d3=40. Easy, the problem explicitly tells you that in words.

t1=2 and t2=2. Easy, the problem implicitly tells you in words




r1= r2 +4. Easy, the problem explicitly tells you that in words.


x=40



I'll go with car 1 first.
distance1 that it travels in 2 hrs?
r1=d1
t1
d1= r1*2 (I am gonna label rate of speed as x)
d1=x*t1
d1=x * 2
d1=2x

I'll do the same with car 2
distance2 that it travels in 2 hrs?
r2= d2
t2
d2=r2*t2
d2=(x+4)* t2 (x+4 'cause one of the car's speed is 4kmp more than the other car's speed)
d2=(x+4)* 2 ( t=2) ( I'll distribute the 2 in to get rid of the parentheses, so
d2=2(x+4)
d2= 2x+ 8


[Then jeefM said: The length of the legs of this triangle are the distances that the two cars traveled and are named d1 and d2]

c^2= a^2+b^2
40^2==(2x)2+ (2x+8)^2
1600=4x^2+ 4x^2 +64
this is just to confirm if all the terms are accounted for in the equation above?. If there is nothing missing?
 
lev888 said:
Then you better learn these formulas:

Square of the sum

Library: The square of the sum is one of the Factoring formulas. The square of the sum of two numbers is equal to the square of the first number, plus twice the product of the first and second numbers, plus the square of the second number.
onlinemschool.com onlinemschool.com

Square of the difference

Library: The square of the difference is one of the Factoring formulas. The square of the difference of two numbers is equal to the square of the first number, minus twice the product of the first and second numbers, plus the square of the second number.
onlinemschool.com onlinemschool.com

Difference of squares

Library: Difference of squares is one of the Factoring formulas. The difference of squares of the two numbers is equal to product of sum of these numbers and the difference of these numbers.
onlinemschool.com onlinemschool.com
Click to expand...
Oh, thank you so much. I will learn them. As i posted my last post i saw your reply. Thanks a lot. Let's continue with the problem, please, i am eager to see how it is solved from where i stopped.
 
eddy2017 said:
40^2==(2x)2+ (2x+8)^2
1600=4x^2+ 4x^2 +64
this is just to confirm if all the terms are accounted for in the equation above?. If there is nothing missing?
Click to expand...
I am not sure why you keep reposting this summary. There is a mistake at the end - the square of a sum is NOT the sum of squares. You need to use a formula to do the expansion: (a + b)^2 = a^2 + b^2 + 2ab. I posted 3 links to explanation of 3 formulas you need to learn if you are studying algebra.
 
Wow, than you for such a good link!!!. I loved it!. A plethora of math info there.
Oh, okay, thanks. I'll do that right now.
One question: I use this formula when i have completed plugging the values in the pythagora theorem formula, right?.
 
okay, let me try
c^2= a^2+b^2
40^2==(2x)^+ (2x+8)^2
I will expand the above sum using the formula you gave me
(a + b)2 = a2 + b2 + 2ab
Just tell yes or no.
But this is the formula that appears on the link that you gave me.
(a + b)2 = a2 + 2ab + b2 (are they the same?)
 
Well, I think they are the same. the order of the addends won't affect the result. I will continue and let you know if i get stuck.
 
eddy2017 said:
okay, let me try
c^2= a^2+b^2
40^2==(2x)^+ (2x+8)^2
I will expand the above sum using the formula you gave me
(a + b)2 = a2 + b2 + 2ab
Just tell yes or no.
But this is the formula that appears on the link that you gave me.
(a + b)2 = a2 + 2ab + b2 (are they the same?)
Click to expand...
If you are studying algebra you should know arithmetic. Is 1 + 2 = 2 + 1?

P.S. to avoid confusion when writing "a squared" please use ^2 or even better highlight "2" and click the x2 button.
 
eddy2017 said:
Hi, can you explain what you meant here with
d_1^2 + d_2^2 = d_3^2.
I did not understand that.
Click to expand...
That is the Pythagorean Theorem.

I used [MATH]d_1^2 \text { for } a^2,\ d_2^2 \text { for } b^2, \text { and } d_3^2 \text { for } c^2[/MATH]
I explained that in post 68.

I have some advice. Take half an hour a day, and reread your first-year algebra text carefully and slowly.. One reason that you are struggling so much in intermediate algebra is that your knowledge of beginning algebra is weak.
 
JeffM said:
That is the Pythagorean Theorem.

I used [MATH]d_1^2 \text { for } a^2,\ d_2^2 \text { for } b^2, \text { and } d_3^2 \text { for } c^2[/MATH]
I explained that in post 68.

I have some advice. Take half an hour a day, and reread your first-year algebra text carefully and slowly.. One reason that you are struggling so much in intermediate algebra is that your knowledge of beginning algebra is weak.
Click to expand...
Yes, you are right. As I think I have told you I started studying Math in earnest five months ago. I am studying Algebra 1 and geometry 1. Someone sent this problem for me, and as I love the challenge that an algebraic word problem poses, I decided to go ahead and presented it to you, but I know I'm in too deep waters here for me, but I want you to help me finish it, if you don't mind.
So far i have learned a few things new for me.
I learned how to work with the square of a sum which was totally new for me.
 
This is what I have done so far. I do not know what to do next. Give me a hint when you have the time. And thank you for your help!!




 

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eddy2017 said:
This is what I have done so far. I do not know what to do next. Give me a hint when you have the time. And thank you for your help!!
Click to expand...
Group and simplify similar terms so your equation looks like this:
ax2+bx+c=0
That's a quadratic equation. Know how to solve it?
 
On the square of the sum, let's understand it.

[MATH](a + b)^2 = (a + b) * (a + b).[/MATH]
[MATH]\text {Let } c = (a + b).[/MATH]
[MATH]\therefore (a + b)^2 = c(a + b) = ac + bc \text { because of the distributive axiom.}[/MATH]
[MATH]\therefore (a + b)^2 = a(a + b) + b(a + b) =\\ a * a + a * b + b * a + b * b = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.[/MATH]You should try proving the square of a difference yourself. And notice

[MATH](a + b)^2 = a^2 + 2ab + b^2 \implies a^2 + b^2 = (a + b)^2 - 2ab.[/MATH]
Now as for the problem itself, I am going to use my recommended notation, not yours. Go toward the end of post 64. Did you understand every step of how I got to

[MATH]r_1^2 + r_2^2 = 400.[/MATH]
And we were told that

[MATH]r_1 = r_2 + 4.[/MATH]
[MATH]\therefore (r_2 + 4)^2 + r_2^2 = 400 \implies \\ r^2 + 8r_2 + 16 +r_2^2 = 400 \implies \\ 2r_2^2 + 8r_2 - 384 = 0 \implies \\ r_2^2 + 4r_2 - 192 = 0.[/MATH]That is called a quadratic equation. It may have two solutions, one of which may or may not be irrelevant. Do you know how to solve such equations?
 
I can solve a quadratic equation. I'll take some time doing it, but I know you will be there for me if I get stuck. I'll get back to you when I have something worked out. Thank you all a lot for your help.
 
eddy2017 said:
I can solve a quadratic equation. I'll take some time doing it, but I know you will be there for me if I get stuck. I'll get back to you when I have something worked out. Thank you all a lot for your help.
Click to expand...
To clarify things: after you simplify the equation in post 93 you'll get the equation at the end of post 97.
 
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