[Guest]

Hello All,

I thought I could handle any sets of two equations with two variables, but the ones I am facing right now are quite challenging to me.

Given:

(1) t^2 + c = 92

(2) c^2 + t = 130

Find the values of c and t in positive integers.

My approach: First isolate the t, so I got t = 130 - c^2. Then I substituted the for t in the equation 1. So I got (130 - c^2)^2 + c = 92. I simplified the expression, and I got 130^2 - c^4 + c = 92. This led to 16900 - c^4 + c = 92 or 16808 = c^4 - c. At this point this equation was becoming strange looking to me. So I tried to "facotr" out the c. I ended up with 16808 = c (c^3 - 1). ==> c = 16808 and c^3 - 1 = 16808. or c^3 = 16809. I think I am going in the wrong direction. Any suggestions that will lead me to the POSITIVE values of c or t are greatly appreciated in ahead of time. Thanks.

[peace*out]

this is what i did. it might not be right, but its how i did it. tell me if you need more of an explanation of the steps i took.

oh, and yes, that IS orange highlighter.

Also: i didnt finish it, but i feel like i did do enough to, if it's right, allow you to finish it. once you get past the factoring. if you need me to do more, than i will, but for now, this is it.

Edited April 30, 2011 by peace*out

[benjer3]

You started out right, though you forgot that

(a-b)² = a² - 2ab + b²

Also remember that to solve through factoring as you did, one side must equal zero.

[curr3nt]

Remember the last one?

If t and c are positive integers then...

t^2 + c = 92 => t^2 < 92 => t < 10

c^2 + t = 130 => c^2 < 130 => c < 12

Now...what value of t (< 10) when subtracted from 130 gives a square of an integer?

edit - Depending on the math teacher they might like this solution or they might not. Depends if they are trying to drill a particular math concept in your head or not.

Edited May 1, 2011 by curr3nt

[Guest]

Thanks Peace-Out and Benjer3 for your suggestions and help. I don't think I can't get to the answers through your ways though. I think Curr3nt's way will lead me to the correct answers. However, I don't actually understand his reasoning though because he uses inequality and stuff which I've have trained to think like that before when solving equations. So I am just going to follow him blindly I guess. He asked, what value of t that is < 10? Where does he get the 10? Anyway, if t is to be less than 10, then t has to be at most 9. If t = 9, then I guess (9)(9) + c = 92. So c = 92 - 81 or 11. Suprisingly, the answers t = 9, and c =11 work and they check out.

(9)(9) + 11 = 92 ...

11(11) + 9 = 130 ...

Thank you, Curr3nt! But I still remain in the dark. I prefer to work out the solutions the traditional way just as Peace-Out and Benjer3 suggested.

[curr3nt]

Sorry, I guess I kind of skipped a step

t^2 + c = 92

t^2 = 92 - c

t^2 < 92 ( Since c is a positive integer t^2 has to be some number less than 92 )

t < sqrt(92)

t < 9.59 ( Which I just rounded up to 10 since we are dealing with integers )

Same reasoning for c^2 + t = 130

[curr3nt]

The "other" way...

t^2 + c = 92   =>  c = 92 - t^2

c^2 + t = 130  =>  t = 130 - c^2


Plug in t

(130 - c^2)^2 + c = 92

(16900 - 260c^2 + c^4) + c = 92


c^4 - 260c^2 + c + 16808 = 0


Plug in c

(92 - t^2)^2 + t = 130

(8464 - 184t^2 + t^4) + t = 130


t^4 - 184t^2 + t + 8334 = 0


Have fun with those...

Edited May 1, 2011 by curr3nt

[peace*out]

Thanks Peace-Out and Benjer3 for your suggestions and help. I don't think I can't get to the answers through your ways though. I think Curr3nt's way will lead me to the correct answers. However, I don't actually understand his reasoning though because he uses inequality and stuff which I've have trained to think like that before when solving equations. So I am just going to follow him blindly I guess. He asked, what value of t that is < 10? Where does he get the 10? Anyway, if t is to be less than 10, then t has to be at most 9. If t = 9, then I guess (9)(9) + c = 92. So c = 92 - 81 or 11. Suprisingly, the answers t = 9, and c =11 work and they check out.

(9)(9) + 11 = 92 ...

11(11) + 9 = 130 ...

Thank you, Curr3nt! But I still remain in the dark. I prefer to work out the solutions the traditional way just as Peace-Out and Benjer3 suggested.

yeah, sorry for the big numbers...

I like curr3nt's way though, i haven't learned it that way yet.

[Guest]

yeah, sorry for the big numbers...I like curr3nt's way though, i haven't learned it that way yet.

Thanks Peace*Out for continuing the discussion. But first I need to say this to Curr3nt, "Thank you, Curr3nt, for your great patience!"

I think, Peace*Out, you and I, we are used to solving equations in the "traditional" taught in a typical classroom. Going this route, I ended up with 4th degree polynomial equation and it is very horrifying. I think I finally UNDERSTAND Curr3nt's reasons.

Given: t*2 + c = 92; c^2 + t = 130 Solve for the positive values of t and c.

First I did not understand why Curr3nt said that t^2 is less than 92 or c^2 is less than 130. Then I recalled that the parts are always less than the whole. Now I got it.

From this point of view, then t*2 is less than 92. t^2 < 92. Take the square root of t^2, I got t; and take the square root of 92, I got 9.591663. So t has to be 9, if the equation t^2 < 92 is to be true. The same thing with the equation c^2 , 130. The square root of c^2 is c, and the square root of 130 is 11.401754. Again, c has to be 11 if the truth of the equation c^2 < 130 is to be upheld. Remember all of this is just "trial and eror" approach. If we continue to test t as 9 and c as 11, then the two original equations work at the end.

Thank you Peace*Out for allowing me to talk to you so that I can think out loud to myself. The "trial and error" method rules the world this time.