# Thread: system of linear equations

1. ## system of linear equations

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!!!

2. 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 $y=-6$

3. ## wow, thanks sooooooo much

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

4. Seven threads for this...

At least show what you've done.

5. ## sorry

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.

6. 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, $(x-3)(x+3)(x-4)$:

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

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

7. 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)

8. 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 $\frac{4}{x^2} - x - 12 + \frac{1}{x^2} - 9 = \frac{2}{x^2} - 7x + 12$ ??

9. 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,

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