# Solving Simultaneous Equations

• May 28th 2008, 04:37 AM
tomjkear
Solving Simultaneous Equations
Please can someone help me with the section of the following equation?

I don't understand how the answers are produced by "solving simultaneously".

http://img514.imageshack.us/img514/1...eenshotir2.png

I would be very thankful if someone can help me. I have an exam on Friday and I can't make sence of this.

Thanks,
TK
• May 28th 2008, 05:31 AM
mr fantastic
Quote:

Originally Posted by tomjkear
Please can someone help me with the section of the following equation?

I don't understand how the answers are produced by "solving simultaneously".

http://img514.imageshack.us/img514/1...eenshotir2.png

I would be very thankful if someone can help me. I have an exam on Friday and I can't make sence of this.

Thanks,
TK

Streamline things a bit by solving:

x + 3y = -0.8944 .... (9)

x - y = -2.4149 .... (10)

(9) - (10): 4y = -0.8944 + 2.4149 = ....... Therefore y = 0.3801.

Now substitute the value of y into (10), say, and solve for x ......
• May 28th 2008, 05:45 AM
Soroban
Hello, TK!

Quote:

I don't understand how the answers are produced by "solving simultaneously".

http://img514.imageshack.us/img514/1...eenshotir2.png

Where did this problem come from? .Those variables are ugly!

Let: .$\displaystyle X \:=\: u^n_{P(a)} \qquad Y \:=\: u^n_{Q(a)}$

The equations become: .$\displaystyle \begin{array}{cccc}X + 3Y &=& -0.8944 & {\color{blue}(9)} \\ X - Y &=& -2.4149 & {\color{blue}(10)} \end{array}$

Subtract (9) - (10): .$\displaystyle 4Y \:=\:1.5205 \quad\Rightarrow\quad\boxed{ Y \:=\:0.380125}$

Add (9) + 3·(10): .$\displaystyle 4X \:=\:-8.1391\quad\Rightarrow\quad\boxed{ X \:=\:-2.034775}$

Too slow again . . . Nice job, Mr. F!
.
• May 28th 2008, 05:55 AM
mr fantastic
Quote:

Originally Posted by Soroban
[snip]
Too slow again . . . Nice job, Mr. F!
.

Well, you will set out your solution all nice and pretty (Rofl)
• May 28th 2008, 05:59 AM
tomjkear
Quote:

Originally Posted by Soroban
Hello, TK!

Where did this problem come from? .Those variables are ugly!

Let: .$\displaystyle X \:=\: u^n_{P(a)} \qquad Y \:=\: u^n_{Q(a)}$

The equations become: .$\displaystyle \begin{array}{cccc}X + 3Y &=& -0.8944 & {\color{blue}(9)} \\ X - Y &=& -2.4149 & {\color{blue}(10)} \end{array}$

Subtract (9) - (10): .$\displaystyle 4Y \:=\:1.5205 \quad\Rightarrow\quad\boxed{ Y \:=\:0.380125}$

Add (9) + 3·(10): .$\displaystyle 4X \:=\:-8.1391\quad\Rightarrow\quad\boxed{ X \:=\:-2.034775}$

Too slow again . . . Nice job, Mr. F!
.

Its from a degree level Computer Science exam about the analysis of collisions between particles. Thanks for helping me, I will try to go over it now and check it makes sense.

TK.
• May 28th 2008, 06:14 AM
tomjkear
Quote:

Originally Posted by tomjkear
Its from a degree level Computer Science exam about the analysis of collisions between particles. Thanks for helping me, I will try to go over it now and check it makes sense.

TK.

So do you always subtract first to find a single variables value then add that into the following equation?

I see how it works in this example but I am keen to know how it works for other equations.

Thanks again.
• May 28th 2008, 06:23 AM
mr fantastic
Quote:

Originally Posted by tomjkear
So do you always subtract first to find a single variables value then add that into the following equation?

I see how it works in this example but I am keen to know how it works for other equations.

Thanks again.

You need to study how to solve linear equations simultaneously. The method used by Soroban and myself is called the elimination method. Read this: Systems of Linear Equations: Solving by Addition / Elimination

They could also have been solved using the substitution method. Read this: Systems of Linear Equations: Solving by Substitution