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Old 12-02-2008, 07:21 PM
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Default Cramer's rule setup question

I am not sure how to solve this problem as the setup is different from other problems of this kind I have faced.All I need to know is how to set it up.If you notice the variable x is isolated on the left, usually the constant is in this position and I am not sure how to set this up.

4x= -5y+10
5x= -y-19
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Old 12-02-2008, 08:28 PM
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Hi

Quote:
Originally Posted by mr.confused View Post
I am not sure how to solve this problem as the setup is different from other problems of this kind I have faced.All I need to know is how to set it up.If you notice the variable x is isolated on the left, usually the constant is in this position and I am not sure how to set this up.

4x= -5y+10
5x= -y-19
Not sure about your question, but one possibility is:


4x= -5y+10
-> x = (-5y+10)/4

use this for your second equation 5x= -y-19

5*[(-5y+10)/4] = -y - 19

(-25y+50)/4 = -y - 19

-25y+50 = -4y - 76

-21y = -126

=> y = (-126)/(-21) = 6

Now you now y, can you find x?

You need to solve 4x= -5*(6)+10 OR 5x= -(6)-19

Remember, (6) is the solution for y. I just wrote 6 instead of y.

Regards
Rapha
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Old 12-02-2008, 08:54 PM
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Although you do have the correct answer for y, My problem is that I think I am supposed to use Cramer's rule to get determinants.It's a strange set up for Cramer's rule, I think.
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Old 12-02-2008, 09:17 PM
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Okay I figured it out finally:

Your supposed to to bring -5 and -y over the to the left side so that it looks like this:

4x+5y=10
5x+y=-19
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Old 12-03-2008, 12:25 PM
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Quote:
Originally Posted by mr.confused View Post
I am not sure how to solve this problem as the setup is different from other problems of this kind I have faced.All I need to know is how to set it up.If you notice the variable x is isolated on the left, usually the constant is in this position and I am not sure how to set this up.

4x= -5y+10
5x= -y-19

4x+5y=10
5x+y=-19

Find the determinant of the coefficient matrix:

D_c=\begin{vmatrix}4 & 5 \\ 5 & 1 \end{vmatrix}=-21

Now, find the determinants of the x and y matricies:

D_x=\begin{vmatrix}10 & 5 \\ -19 & 1 \end{vmatrix}=105

D_y=\begin{vmatrix}4 & 10 \\ 5 & -19 \end{vmatrix}=-126

x=\frac{D_x}{D_c}=\frac{105}{-21}=-5

y=\frac{D_y}{D_c}=\frac{-126}{-21}=6
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