Linear Systems and Matrices / Translations of Conics

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thoffman

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Solve the system of linear equations and check any solution algebraically.

#20)

5x - 3y + 2z = 3
2x + 4y - z = 7
x - 11y + 4z = 3

Okay so I put this in my T1-84 in matrix a with the proper coefficent values.
I use rref([A]) and this is the solution set I get-

1 0 5/26 0
0 1 -9/26 0
0 0 0 1

Generally the solution is in the last line of solutions. Anyhow, I am very confused on how to solve this by hand to get a solution. Also, is this a special case? A solution and walkthrough would be appreciated.

Write the form or the partial fraction decomposition of the rational rexpression. Do not solve for constants.

#50)

x^2 - 3x + 2
----------------
4x^3 + 11x^2

As long as the highest degree is on the bottom it is not an irrational partial fraction right? I would like the solution and walkthrough on this as well.

Find the center, foci, and vertices of the ellipse, and sketch its graph.

#38)

9x^2 + 4y^2 - 36x + 8y + 31 = 0

I seperate these by terms correct?

(9x^2 - 36x ) + (4y^2 + 8y ) = -31

Do I factor this next?

9(x^2 - 4x) + 4(4^2 + 2y) = -31

If this is the correct track, I am lost from here.

I would like the solution and walkthrough on this as well.

Thanks in advance!
 
There is no soluition to #20. The system is inconsistant. That is what the rref tells us.

For the last complete the squares:
\(\displaystyle 9\left( {x^2 - 4x + 4} \right) + 4\left( {y^2 + 2y + 1} \right) = - 31 + 36 + 4\)
 
thoffman said:
Solve the system of linear equations and check any solution algebraically.

#20)

5x - 3y + 2z = 3
2x + 4y - z = 7
x - 11y + 4z = 3

Okay so I put this in my T1-84 in matrix a with the proper coefficent values.
I use rref([A]) and this is the solution set I get-

1 0 5/26 0
0 1 -9/26 0
0 0 0 1

Last row of the solution matrix being 0 indicates that the determinant of the solution matrix is "0" and ther is no solution.

Generally the solution is in the last line of solutions. Anyhow, I am very confused on how to solve this by hand to get a solution. Also, is this a special case? A solution and walkthrough would be appreciated.

Write the form or the partial fraction decomposition of the rational rexpression. Do not solve for constants.

#50)

x^2 - 3x + 2
---------------- =
4x^3 + 11x^2

Code: 
 x^2 - 3x + 2
 ---------------- =
 x^2(4x + 11)

Ax + B          C
-------- + ----------
  x^2       4x + 11
As long as the highest degree is on the bottom it is not an irrational partial fraction right? I would like the solution and walkthrough on this as well.

Find the center, foci, and vertices of the ellipse, and sketch its graph.

#38)

9x^2 + 4y^2 - 36x + 8y + 31 = 0

I seperate these by terms correct?

(9x^2 - 36x ) + (4y^2 + 8y ) = -31
Code: 
[(3x)^2 - 2*3x*6 + 6^2] +[(2y)^2 + 2*2y *2 + 2^2] = -31 + 36 + 4

(3x - 6)^2 + (2y + 2)^2 = 3^2

Now continue...

Do I factor this next?

9(x^2 - 4x) + 4(4^2 + 2y) = -31

If this is the correct track, I am lost from here.

I would like the solution and walkthrough on this as well.

Thanks in advance!
Click to expand...
 
Hello, thoffman!

Here's the third one . . .

38) Find the center, foci, and vertices of the ellipse, and sketch its graph.
. . \(\displaystyle 9x^2\,+\, 4y^2\,-\,36x\,+\,8y\,+\,31\:=\:0\)
Click to expand...

We have: \(\displaystyle \:9x^2\,+\,-36x\,+\,4y^2\,+\,8y \:=\:-31\)

. . . .\(\displaystyle 9(x^2\,-\,4x\;\;\;) + 4(y^2\,+\,2y\;\;\;) \;=\;-31\)

Complete the square:
. . \(\displaystyle 9x^2(x^2-\,4x\,\)+ 4\(\displaystyle )\,+\,4(y^2\,+\2y\,\)+ 1\(\displaystyle )\;=\;-31\,\)+ 36 + 4

And we have: \(\displaystyle \:9(x\,-\,2)^2\,+\,4(y\,+\,1)^2\:=\:9\)

Divide by 9: \(\displaystyle \L\:\frac{(x\,-\,2)^2}{1}\,+\,\frac{(y\,+\,1)^2}{\frac{9}{4}} \;=\;1\)


The center of the ellipse is: \(\displaystyle \:\fbox{C(2,\,-1)}\)

The major axis is vertical.
. . We have: \(\displaystyle \:a\,=\,\frac{3}{2},\;b\,=\,1\)

The vertices are \(\displaystyle \frac{3}{2}\) units above and below the center: \(\displaystyle \:\fbox{\left(2,\,\frac{1}{2}\right),\;\left(2,\,-\frac{5}{2}\right)}\)

The foci are \(\displaystyle c\) units above and below the center,
. . where: \(\displaystyle \:c^2\:=\:a^2\,-\,b^2\)
We have: \(\displaystyle \:c^2\:=\:\left(\frac{3}{2}\right)^2\,-\,1^2\:=\;\frac{5}{4}\;\;\Rightarrow\;\;c\:=\:\pm\frac{\sqrt{5}}{2}\)

Hence, the foci are: \(\displaystyle \:\fbox{\left(2,\:-1\,\pm\,\frac{\sqrt{5}}{2}\right)}\)

 
Hello again, thoffman!

Here's the second one . . .

50) Write the form of the partial fraction decomposition of the rational rexpression.
Do not solve for constants.

. . \(\displaystyle \L\frac{x^2\,-\,3x\,+\,2}{4x^3\,+\,11x^2}\)
Click to expand...

We have: \(\displaystyle \L\:\frac{(x\,-\,1)(x\,-\,2)}{x^2(4x\,+\,11)} \;=\;\frac{A}{x}\,+\,\frac{B}{x^2}\,+\,\frac{C}{4x\,+\,11}\)


Note: If the quadratic is prime (cannot be factored), we use \(\displaystyle Ax\,+\,B\).

. . . . This quadratic is \(\displaystyle x^2\:=\:x\cdot x\), so we have repeated linear factors.

 
Thanks for the help everyone! Another quick question in response to #38. Since it is an ellipse shouldn't it have 4 vertices (2 on the major and 2 on the minor)? Would the minor axis vertices be (3,-1) and (1,-1)?
 
The vertices (V and V1) mark where the major axis intersects the ellipse. ...
 
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