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Eliminating a parameter

Casio

Member
Feb 11, 2012
86
I would appreciate a little help with a parametric equation please.

Example.

PHP:
x = t + 1
PHP:
y = 3t - 2
I want to eliminate t

PHP:
x = t + 1
so I change the sides

PHP:
t = x - 1
I now bring in the second equation

PHP:
y = 3t - 2 = 3(x - 1) - 2 = 3x - 5
PHP:
y = 3x - 5
The above I have understood, but my next equation I am getting confused with.

PHP:
2x = 2t - 5
PHP:
2y = 14t + 9
I want to eliminate t

I am more than likely incorrect but am thinking along the lines of;

PHP:
2x = 2t - 5
PHP:
x = t - 5
PHP:
t = x + 5
Therefore I can use x + 5 in place of t

In the next equation I think

PHP:
2y = 14t + 9
PHP:
12y = t + 9
PHP:
3y = t
PHP:
t = 3y
combine the equations

PHP:
t = 3y (x + 5)
PHP:
t = 3xy + 15
I not sure any help appreciated(Smile)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I would appreciate a little help with a parametric equation please.

Example.

PHP:
x = t + 1
PHP:
y = 3t - 2
I want to eliminate t

PHP:
x = t + 1
so I change the sides

PHP:
t = x - 1
I now bring in the second equation

PHP:
y = 3t - 2 = 3(x - 1) - 2 = 3x - 5
PHP:
y = 3x - 5
The above I have understood, but my next equation I am getting confused with.

PHP:
2x = 2t - 5
PHP:
2y = 14t + 9
I want to eliminate t

I am more than likely incorrect but am thinking along the lines of;

PHP:
2x = 2t - 5
PHP:
x = t - 5
What, exactly, did you do to go from one equation to the other?

PHP:
t = x + 5
Therefore I can use x + 5 in place of t

In the next equation I think

PHP:
2y = 14t + 9
PHP:
12y = t + 9
Now did "2y= 14t+ 9" become "12y= t+ 9"?

PHP:
3y = t
I think you tried to subtract 9 from both sides of 12y= t+ 9. But "subtract 9" is not the same as "subtract 9x".

PHP:
t = 3y
combine the equations

PHP:
t = 3y (x + 5)
PHP:
t = 3xy + 15
I not sure any help appreciated(Smile)
You see to have forgotten that the object was to eliminate t. You keep putting t back into the equations!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
The above I have understood, but my next equation I am getting confused with.

2x = 2t - 5

2y = 14t + 9

I want to eliminate t
Multiply the first equation by 7 to get:

14x = 14t - 35

then:

14t = 14x + 35

which may be substituted into the second equation to eliminate t.

CB
 

Casio

Member
Feb 11, 2012
86
It seems that some questions even with the best of intentions would be answered incorrectly by any person trying to help another forum members, if the question being asked had to be answered in accordance with the solutions of another previous question already answered, so my applogies for that as I did not fully understand the question I had been trying to answer.

The equation was;

2x = 2t - 5, 2y = 14t + 9

The task is to elliminate t

2x = 2t - 5

2t = 2x + 5

So at this point t is elliminated because I can now use 2x + 5 in it's place.

2y = 14t + 9

2y = 14(2x + 5) + 9

2y = 28x + 70 + 9

2y = 28x + 79

y = 28x + 79 / 2

Now in the previous part of the question (Not Shown) the midpoint rule used calculated that the coordinates recorded a mean figure at -5/2 and the y intercept was found to be 9/2

So by putting the solution above into the equation just worked out, we get;

y = 28(-5/2)+79 / 2

y = 9/2

Now whether I am 100% right or not remains to be seen, but if I had posted the complete question I think members would have not replied, as few have done!

But thanks anyway.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
It seems that some questions even with the best of intentions would be answered incorrectly by any person trying to help another forum members, if the question being asked had to be answered in accordance with the solutions of another previous question already answered, so my applogies for that as I did not fully understand the question I had been trying to answer.

The equation was;

2x = 2t - 5, 2y = 14t + 9

The task is to elliminate t

2x = 2t - 5

2t = 2x + 5

So at this point t is elliminated because I can now use 2x + 5 in it's place.
No you can't, you have at this point \(2t=2x+5\), so \(t=x+\frac{5}{2}\). Which is what you have to substitute for \(t\).

(and the last part of your last post is incomprehensible, it is always a good idea to post the real question and what you have done to get to where you are asking for help)

CB
 

Casio

Member
Feb 11, 2012
86
No you can't, you have at this point \(2t=2x+5\), so \(t=x+\frac{5}{2}\). Which is what you have to substitute for \(t\).

(and the last part of your last post is incomprehensible, it is always a good idea to post the real question and what you have done to get to where you are asking for help)

CB
Your right I missed the understanding.

2t = 2x + 5, so divide by sides by 2

t = x + 5

Now I can substitute x + 5 into the second equation.

2y = 14t + 9

2y = 14(x + 5) + 9

2y = 14x + 70 + 9

2y = 14x + 79.

Divide by sides by 2

y = 14x + 79 / 2

y = 14(-5/2) + 79 / 2

y = -35 + 79 / 2

y = 22
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Your right I missed the understanding.

2t = 2x + 5, so divide by sides by 2

t = x + 5
Wrong! I have already done this for you. If you divide 2x+5 by 2 you get x+(5/2)

CB
 

Casio

Member
Feb 11, 2012
86
Wrong! I have already done this for you. If you divide 2x+5 by 2 you get x+(5/2)

CB
I can see there is more than one way to get rid of "t", and I thank you for your assistance in this matter.

I also see that this method works;

2x = 2t - 5
2t = 2x + 5

as you pointed out multiply by 7

14x = 14t - 35

14t = 14x + 35

t = x + 35

2y = 14t + 9

2y = 14(x + 35)+9

2y = 14x + 44

y = 14x + 44 / 2 or y = 7x + 22

I am sure this method is correct as it seems to agree with the graph produced on the computer package.

Thanks again CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I can see there is more than one way to get rid of "t", and I thank you for your assistance in this matter.

I also see that this method works;

2x = 2t - 5
2t = 2x + 5

as you pointed out multiply by 7

14x = 14t - 35

14t = 14x + 35

t = x + 35

2y = 14t + 9

2y = 14(x + 35)+9

2y = 14x + 44

y = 14x + 44 / 2 or y = 7x + 22

I am sure this method is correct as it seems to agree with the graph produced on the computer package.

Thanks again CB
Learn to use brackets to make your expressions mean what you want them to mean, you want y = 14x + 44 / 2 to mean y=(14x+44)/2 but what you have does not mean that it means y=14x+(44/2)

CB
 

Casio

Member
Feb 11, 2012
86
Learn to use brackets to make your expressions mean what you want them to mean, you want y = 14x + 44 / 2 to mean y=(14x+44)/2 but what you have does not mean that it means y=14x+(44/2)

CB
I don't profess to understand what you wrote above?

I have looked at the graph and the intercept and the maths seems to show the same result, are you saying that it is still incorrect?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I don't profess to understand what you wrote above?

I have looked at the graph and the intercept and the maths seems to show the same result, are you saying that it is still incorrect?
I believe he is saying that your notation is ambiguous sometimes. For example y=14x+44/2 could mean \(\displaystyle y=\frac{14x+44}{2}\) or it could mean \(\displaystyle y=14x+\frac{44}{2}\). Using the order of operations in the usual sense y = 14x+44/2 would simplify to y=14x+22.
 

Mr Fantastic

Member
Jan 26, 2012
66
I don't profess to understand what you wrote above?

I have looked at the graph and the intercept and the maths seems to show the same result, are you saying that it is still incorrect?
Your final answer is correct but what CB is saying is that 14x + 44/2 does not mean 7x + 22, it means 14x + 22. The point is that you know what you mean but someone marking your work will not because it is wrong as written. CB is saying that you need to learn how to use brackets correctly, what you should have written was (14x + 44)/2 = 7x + 22. If you cannot see the difference then there is no point in further discussion on it.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I can see there is more than one way to get rid of "t", and I thank you for your assistance in this matter.

I also see that this method works;

2x = 2t - 5
2t = 2x + 5

as you pointed out multiply by 7
Yes, you could do that but there is no REASON to do it.
Do you understand why Captain Black multiplied by 7 in post #3?

14x = 14t - 35

14t = 14x + 35

t = x + 35

2y = 14t + 9

2y = 14(x + 35)+9

2y = 14x + 44

y = 14x + 44 / 2 or y = 7x + 22

I am sure this method is correct as it seems to agree with the graph produced on the computer package.

Thanks again CB
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I can see there is more than one way to get rid of "t", and I thank you for your assistance in this matter.

I also see that this method works;

2x = 2t - 5
2t = 2x + 5

as you pointed out multiply by 7

14x = 14t - 35

14t = 14x + 35
Yes, this is correct. You have multiplied each of the three terms by 7.

t = x + 35
??? And not the 7 as well as the original 2 have simply disappeared! There is no point in multiplying by 7 if you immediately get rid of the 7. I will ask again, what did you do to go from "14t= 14x+ 35"?

What you can do, algebraically, to get from 14t= 14x+ 35 to t= x+ ... is to divide both sides, and so each term, by 14. What does 35/14 reduce to? More simply, rather than multiplying by 7, you could go from 2t= 2x+ 5 by dividing both sides, and so each term, by 2: t= x+ 5/2. You were told that before.

2y = 14t + 9
Now, here is a very good reason for multiplying by 7 as you did before- you don't want t, you want 14t!

2y = 14(x + 35)+9
and now, you have thrown "multply by 7" away and repeating the mistake that was pointed out before- t is NOT equal to x+ 35. From 2t= 2x+ 35, dividing on both sides by 2 gives t= x+ 35/2. But, more important, you can replace 14t by what you had before, 14x+ 35:
2y= 14x+ 9= (14t+ 35)+ 9

2y = 14x + 44
Yes, this is correct but I have no idea how you gtot it! What you wrote was 2y= 14(x+ 35)+ 9= 14x+ 14(35)+ 9= 14x+ 490.

y = 14x + 44 / 2 or y = 7x + 22

I am sure this method is correct as it seems to agree with the graph produced on the computer package.

Thanks again CB
You seem to have great difficulty understanding what parentheses mean.