# Thread: Cramer's rule setup question

1. ## 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

2. Hi

Originally Posted by mr.confused
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

3. 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.

4. 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

5. Originally Posted by mr.confused
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:

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

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

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

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