solve the following equation by ccramers rule 3x-2y=5/2;1/3x+3y=-4/3
Answers
Answer:
Answer: x=-\dfrac{1}{2},\ y=\dfrac{-1}{2}x=−
2
1
, y=
2
−1
Step-by-step explanation:
Given system: \begin{gathered}3x-2y=5/2\\\\1/3x+3y=-4/3\end{gathered}
3x−2y=5/2
1/3x+3y=−4/3
Using Cramer rule:
\begin{gathered}D=\begin{vmatrix}3 & -2\\ 1/3 & 3\end{vmatrix}=\dfrac{29}{3}\end{gathered}
D=
∣
∣
∣
∣
∣
3
1/3
−2
3
∣
∣
∣
∣
∣
=
3
29
\begin{gathered}D_1=\begin{vmatrix}5/2& -2\\ -4/3 & 3\end{vmatrix}=\dfrac{29}{6}\end{gathered}
D
1
=
∣
∣
∣
∣
∣
5/2
−4/3
−2
3
∣
∣
∣
∣
∣
=
6
29
\begin{gathered}D_2=\begin{vmatrix}3 & 5/2\\ 1/3 & -4/3\end{vmatrix}=\dfrac{-29}{6}\end{gathered}
D
2
=
∣
∣
∣
∣
∣
3
1/3
5/2
−4/3
∣
∣
∣
∣
∣
=
6
−29
Now, x=\dfrac{D_1}{D}=\dfrac{\dfrac{29}{6}}{\dfrac{29}{3}}=\dfrac{1}{2}x=
D
D
1
=
3
29
6
29
=
2
1
x=\dfrac{D_2}{D}=\dfrac{\dfrac{-29}{6}}{\dfrac{29}{3}}=\dfrac{-1}{2}x=
D
D
2
=
3
29
6
−29
=
2
−1
Hence, x=-\dfrac{1}{2},\ y=\dfrac{-1}{2}x=−
2
1
, y=
2
−1