cramer's rule e5q24 *

[tamiatha]

please explain cramer's rule to me
i textbook is confusing on the subject
i don't understand the first step of multiplying by a constant
my problem states:
solve the following using cramer's rule
3x-5y=21 and 4x+2y=2

 

[Loren]

Re: cramer's rule e5q24

http://www.purplemath.com/modules/cramers.htm

 

[soroban]

Re: cramer's rule e5q24

Hello, tamiatha!

$\displaystyle \text{Solve by Cramer's rule: }\;\begin{array}{ccc}3x-5y&=&21 \\ 4x+2y&=&2 \end{array}$

$\displaystyle \text{[1] Find the determinant of the coefficients: }\;D \:=\:\begin{vmatrix}3 & \text{-}5 \\4 & 2 \end{vmatrix} \;=\;(3)(2) - (\text{-}5)(4) \;=\;6 + 20 \:=\:26$


$\displaystyle \text{[2] Find the }x\text{-determinant: }\:\text{Replace the }x\text{-coefficients with the constants.}$

. . . $\displaystyle \begin{vmatrix}21 & \text{-}5 \\ 2 & 2\end{vmatrix} \;=\;(21)(2) - (\text{-}5)(2) \;=\;42 + 10 \:=\:52$
. . $\displaystyle \text{Divide by }D\!:\quad x\:=\:\frac{52}{26} \quad\Rightarrow\quad\boxed{ x \:=\:2}$


$\displaystyle \text{[3] Find the }y\text{-determinant: }\:\text{Replace the }y\text{-coefficients with the constants.}$

'. . . $\displaystyle \begin{vmatrix}3 & 21 \\ 4 & 2\end{vmatrix} \;=\;(3)(2) - (21)(4) \;=\;6 - 84 \;=\;-78$
. . $\displaystyle \text{Divide by }D\!:\quad y \;=\;\frac{\text{-}78}{26} \quad\Rightarrow\quad\boxed{ y \;=\;-3}$

 

[tamiatha]

Re: cramer's rule e5q24

big gigantic thank you
clear excellent explaination and the answer
sometimes i just need to see the problem worked out entirely for me to understand
now i am ready to do more like it