system of linear equations

**mathconfusesme** · May 19th 2008, 06:16 PM

Solve the following systems of equations

4x + y -2z = 0
2x - 3y + 3z = 9
-6x - 2y + z = 0

Simplify the following expressions

(6x^2 - 3x + 2) - (4x^2 + 2x - 5)

Simplify the following to the lowest term

x^3 - 1/x^4 - 1 * x^2 - 1/x^2 + x + 1

Solve the following equations

5/y+1 = 4/y+2

4/x^2 - x -12 + 1/x^2 - 9 = 2/x^2 - 7x + 12

Any help would be much appreciated! Thanks!!!

**Mathstud28** · May 19th 2008, 06:19 PM

Originally Posted by mathconfusesme

Solve the following systems of equations

4x + y -2z = 0
2x - 3y + 3z = 9
-6x - 2y + z = 0

Simplify the following expressions

(6x^2 - 3x + 2) - (4x^2 + 2x - 5)

Simplify the following to the lowest term

x^3 - 1/x^4 - 1 * x^2 - 1/x^2 + x + 1

Solve the following equations

5/y+1 = 4/y+2

4/x^2 - x -12 + 1/x^2 - 9 = 2/x^2 - 7x + 12

Any help would be much appreciated! Thanks!!!

The first one can be solved by so many ways...you couls solve for one variable in the first one, sub that in the second solve for another variable, and the sub that into the third and you will get the one value then sub back and get the others

You could use Cramer's Method

I will do this one for now

$\frac{5}{y+1}=\frac{4}{y+2}$

Cross-multiplying we get

$5(y+2)=4(y+1)\Rightarrow{5y+10=4y+4}$

Solving we get $y=-6$

**mathconfusesme** · May 19th 2008, 06:24 PM

that was faaaaaaaaaast, thank u soooooo much... i gave u thanks! ur good man

ur the MAN

**Krizalid** · May 19th 2008, 07:38 PM

Seven threads for this...

At least show what you've done.

**mathconfusesme** · May 19th 2008, 07:41 PM

sorry, i did my whole guide and am just stuck on these 3, i'm trying them as we speak. i'll post what i did in a moment. didn't mean to stike a nerve. just anxious about my final.

**Soroban** · May 20th 2008, 04:23 AM

Hello, mathconfusesme!

Did you expect anyone to read your last two problems?
Please learn to use parentheses!

Simplify: . $\frac{x^3 - 1}{x^4 - 1 }\cdot \frac{x^2 - 1}{x^2 + x + 1 }$

Factor: . $\frac{(x-1)(x^2+x+1)}{(x^2-1)(x^2+1)}\cdot\frac{x^2-1}{x^2+x+1}$

Reduce: . $\frac{x-1}{x^2+1}$

Solve: . $\frac{4}{x^2 - x -12} + \frac{1}{x^2 - 9} \;= \;\frac{2}{x^2 - 7x + 12}$

Factor: . $\frac{4}{(x+3)(x-4)} + \frac{1}{(x-3)(x+3)} \;=\;\frac{2}{(x-3)(x-4)}$

Multiply through by the LCD, $(x-3)(x+3)(x-4)$ :

. . $4(x-3) + (x-4) \;=\;2(x+3) \quad\Rightarrow\quad 4x - 12 + x -4 \;=\;2x + 6$

. . $3x \:=\:22 \quad\Rightarrow\quad\boxed{x \:=\:\frac{22}{3}}$

**masters** · May 20th 2008, 04:42 AM

Let's use Cramer's rule to solve the system. If you're unfamiliar with Cramer's rule, go here: Cramer's Rule

1. Find the determinant of the coefficient matrix.
[4 1 -2]
[2 -3 3] = 36
[-6 -2 1]

2. Find the determinant of the coefficient matrix with the x column replaced by the constant matrix; then divide by the coefficient determinant.

[0 1 -2]
[9 -3 3] = 27
[0 -2 1]

x = 27/36 = 3/4

3. Find the determinant of the coefficient matrix with the y column replaced by the constant matrix; then divide by the coefficient determinant.

[4 0 -2]
[2 9 3] = -72
[-6 0 1]

y = -72/36 = -2

4. Find the determinant of the coefficient matrix with the z column replaced by the constant matrix; then divide by the coefficient determinant.

[4 1 0]
[2 -3 9] = 18
[-6 -3 0]

z = 18/36 = 1/2

The solution to the system is (3/4, -2, 1/2)

**colby2152** · May 20th 2008, 05:07 AM

Originally Posted by mathconfusesme

Solve the following systems of equations

4x + y -2z = 0
2x - 3y + 3z = 9
-6x - 2y + z = 0

Multiply the third line by -1 and the nadd up the equations. This will get rid of the y's and z's. You will then be able to solve for x, and consequently find y and z.

Originally Posted by mathconfusesme

Simplify the following to the lowest term

x^3 - 1/x^4 - 1 * x^2 - 1/x^2 + x + 1

Did you mean this? $x^3 - \frac{1}{x^4 - 1} + \frac{x^2 - 1}{x^2} + x + 1$

Originally Posted by mathconfusesme

4/x^2 - x -12 + 1/x^2 - 9 = 2/x^2 - 7x + 12

Any help would be much appreciated! Thanks!!!

Likewise, did you mean to write $\frac{4}{x^2} - x - 12 + \frac{1}{x^2} - 9 = \frac{2}{x^2} - 7x + 12$ ??

**masters** · May 20th 2008, 06:17 AM

Is your second expression:
$x^3 - \frac{1}{x^4} - x^2 - \frac{1}{x^2} + x + 1$

If so, simplify by putting everything over the same denominator $x^4$

Then we get,

$\frac{x^7-x^6+x^5+x^4-x^2-1}{x^4}$

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wow, thanks sooooooo much

sorry

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