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Worked Examples · Example 5

Q.Using Cramer's Rule, solve the system of equations 3x+2y=43x+2y=4 and x−3y=5x-3y=5.

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Forming DD

D=∣321−3∣=3(−3)−2(1)=−9−2=−11D=\begin{vmatrix}3&2\\1&-3\end{vmatrix}=3(-3)-2(1)=-9-2=-11

Since D≠0D\neq0, a unique solution exists.

Forming DxD_x

Dx=∣425−3∣=4(−3)−2(5)=−12−10=−22D_x=\begin{vmatrix}4&2\\5&-3\end{vmatrix}=4(-3)-2(5)=-12-10=-22

Forming DyD_y

Dy=∣3415∣=3(5)−4(1)=15−4=11D_y=\begin{vmatrix}3&4\\1&5\end{vmatrix}=3(5)-4(1)=15-4=11

Solving

x=−22−11=2y=11−11=−1x=\frac{-22}{-11}=2 \qquad y=\frac{11}{-11}=-1 …

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