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Answer:$\mathrm{x}=2 \text { and } \mathrm{y}=\frac{-2}{a}$

Hint: Use Cramer’s rule to solve a system of two equations in two variables.

Given:

\begin{aligned} &3 x+a y=4 \\ &2 x+a y=2, a \neq 0 \end{aligned}

Solution:

First D: determinant of the coefficient matrix

\begin{aligned} \mathrm{D} &=\left|\begin{array}{ll} 3 & a \\ 2 & a \end{array}\right| & \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right) \\ &=3 \mathrm{a}-2 \mathrm{a} \\ &=\mathrm{a} \end{aligned}

Now, $D\neq 0$. If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} \mathrm{D}_{1} &=\left|\begin{array}{ll} 4 & a \\ 2 & a \end{array}\right| \\ &=4 \mathrm{a}-2 \mathrm{a} \\ &=2 \mathrm{a} \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} &\mathrm{D}_{2}=\left|\begin{array}{ll} 3 & 4 \\ 2 & 2 \end{array}\right|\\ &\text { (2) }\\ &\begin{aligned} &=(2)(3)-(4) \\ &=6-8 \\ &=-2 \end{aligned} \end{aligned}

Now,

\begin{aligned} \mathrm{x}=\frac{D_{1}}{D}=\frac{2 a}{a}=2 \\ \mathrm{y}=\frac{D_{2}}{D}=\frac{-2}{a} \end{aligned}

Hence,$\mathrm{x}=2 \text { and } \mathrm{y}=\frac{-2}{a}$

Concept: Cramer’s rule for system of two equations.

Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

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