cramer's rule e5q24 *
- Thread starter tamiatha
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Re: cramer's rule e5q24
Hello, tamiatha!
\(\displaystyle \text{[1] Find the determinant of the coefficients: }\;D \:=\:\begin{vmatrix}3 & \text{-}5 \\4 & 2 \end{vmatrix} \;=\;(3)(2) - (\text{-}5)(4) \;=\;6 + 20 \:=\:26\)
\(\displaystyle \text{[2] Find the }x\text{-determinant: }\:\text{Replace the }x\text{-coefficients with the constants.}\)
. . . \(\displaystyle \begin{vmatrix}21 & \text{-}5 \\ 2 & 2\end{vmatrix} \;=\;(21)(2) - (\text{-}5)(2) \;=\;42 + 10 \:=\:52\)
. . \(\displaystyle \text{Divide by }D\!:\quad x\:=\:\frac{52}{26} \quad\Rightarrow\quad\boxed{ x \:=\:2}\)
\(\displaystyle \text{[3] Find the }y\text{-determinant: }\:\text{Replace the }y\text{-coefficients with the constants.}\)
'. . . \(\displaystyle \begin{vmatrix}3 & 21 \\ 4 & 2\end{vmatrix} \;=\;(3)(2) - (21)(4) \;=\;6 - 84 \;=\;-78\)
. . \(\displaystyle \text{Divide by }D\!:\quad y \;=\;\frac{\text{-}78}{26} \quad\Rightarrow\quad\boxed{ y \;=\;-3}\)
Hello, tamiatha!
\(\displaystyle \text{[1] Find the determinant of the coefficients: }\;D \:=\:\begin{vmatrix}3 & \text{-}5 \\4 & 2 \end{vmatrix} \;=\;(3)(2) - (\text{-}5)(4) \;=\;6 + 20 \:=\:26\)
\(\displaystyle \text{[2] Find the }x\text{-determinant: }\:\text{Replace the }x\text{-coefficients with the constants.}\)
. . . \(\displaystyle \begin{vmatrix}21 & \text{-}5 \\ 2 & 2\end{vmatrix} \;=\;(21)(2) - (\text{-}5)(2) \;=\;42 + 10 \:=\:52\)
. . \(\displaystyle \text{Divide by }D\!:\quad x\:=\:\frac{52}{26} \quad\Rightarrow\quad\boxed{ x \:=\:2}\)
\(\displaystyle \text{[3] Find the }y\text{-determinant: }\:\text{Replace the }y\text{-coefficients with the constants.}\)
'. . . \(\displaystyle \begin{vmatrix}3 & 21 \\ 4 & 2\end{vmatrix} \;=\;(3)(2) - (21)(4) \;=\;6 - 84 \;=\;-78\)
. . \(\displaystyle \text{Divide by }D\!:\quad y \;=\;\frac{\text{-}78}{26} \quad\Rightarrow\quad\boxed{ y \;=\;-3}\)
the problem worked out entirely for me to understand