Solving Simultaneous Equations: Step-by-Step Guide | x + 2y = 5, x² + y² = 36

  • Thread starter Thread starter Gringo123
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around solving simultaneous equations, specifically the equations x + 2y = 5 and x² + y² = 36. Participants are exploring methods to manipulate these equations to find the values of x and y.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to rearrange the first equation and seeks guidance on the next steps. Subsequent participants suggest substituting variables and expanding expressions, while others question the accuracy of the expansion and seek clarification on specific mathematical concepts.

Discussion Status

The discussion is active, with participants providing guidance on substitution and expansion. There is acknowledgment of a mistake in the expansion process, and further exploration of solving a quadratic equation is suggested. Multiple interpretations of the problem are being discussed, but no consensus has been reached.

Contextual Notes

Participants are addressing potential misunderstandings in algebraic manipulation and the notation for squared terms. There is an emphasis on ensuring clarity in the steps taken to solve the equations.

Gringo123
Messages
141
Reaction score
0
I have just tried to solve the following simultaneous equation.

x + 2y = 5
x squared + y squared = 36

I guess that the 1st step is to rearrange the 1st equation to give:

x = 5 - 2y

From that point onwards I have no idea what to do! Can anybody guide me through this?
Thanks a lot!
 
Physics news on Phys.org
Substitute x with (5 - 2y) in the second equation to give:
[tex] (5 - 2y)^2 + y^2 = 36[/tex]
That's how to solve almost all simultaneous equations; the idea is to eliminate one variable in one of the equations so as to obtain an equation solely in terms of one variable, which can then be solved.
Carry on.
 
so...
(5 - 2y) x (5 - 2y) + y squared = 36
multipying out the brackets gives us:
(25 - 10y - 10y + 4y) + y squared = 36
so...
(25 - 16y) + y squared = 36
I've had a long hrad think at this stage and I'm stuck again! Am I going in the right direction?
Thanks for your help!
 
Yes, you are going in the right direction...but you made a mistake in expanding the brackets.
[tex](5 - 2y)^2 + y^2 = 36[/tex]
[tex](25 - 20y + 4y^2) + y^2 = 36[/tex]
[tex]5y^2 - 20y + 25 = 36[/tex]
[tex]5y^2 - 20y - 11 = 0[/tex]
Now its just a matter of solving for y from this quadratic equation.
 
just going back to the expanding of the brackets, i worked it outr as follows:
(5 - 2y) x (5 - 2y)
so..
5 x 5 = 25
5 x -2y = -10y
-2y x 5 = -10y
-2y x -2y = 4y (why does this equal 4y squared?)
By the way, how do you type the little 2 tyat indicated squared?
Thanks again
 
because a minus x a minus is a plus. -2 x -2 = 4
y x y = y^2

so -2y x -2y = 4y^2
 
Gringo123 said:
By the way, how do you type the little 2 tyat indicated squared?

Hi Gringo123!

[NOPARSE]Type "y2",[/NOPARSE] and it comes out y2 :smile:

(or use the X2 tag just above the Reply box :wink:)
 

Similar threads

Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
3K
Replies
2
Views
2K
Replies
2
Views
2K