Question
Solve the following equations by Cramer’s method.
$3 x-2 y=\frac{5}{2} ; \frac{1}{3} x+3 y=-\frac{4}{3}$
$3 x-2 y=\frac{5}{2} ; \frac{1}{3} x+3 y=-\frac{4}{3}$
✓
Answer
$\begin{aligned} & 3 x -2 y =\frac{5}{2} \Rightarrow 6 x -4 y =5 \\ & \frac{1}{3} x +3 y =-\frac{4}{3} \Rightarrow \frac{ x +9 y }{3}=-\frac{4}{3} \Rightarrow x +9 y =-4 \\ & D =\left[\begin{array}{cc}6 & -4 \\ 1 & 9\end{array}\right]=(6 \times 9)-(-4 \times 1)=54+4=58 \\ & D _{ x }=\left[\begin{array}{cc}5 & -4 \\ -4 & 9\end{array}\right]=(5 \times 9)-(-4 \times-4)=45-16=29 \\ & D _{ y }=\left[\begin{array}{cc}6 & 5 \\ 1 & -4\end{array}\right]=(6 \times-4)-(5 \times 1)=-24-5=-29 \\ & x=\frac{D_x}{D}=\frac{1}{2}, y=\frac{D_y}{D}=\frac{(-29)}{58}=\frac{(-1)}{2} \\ & \therefore( x , y )=(1 / 2,-1 / 2)\end{aligned}$
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