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Answer:$x=\frac{16}{3} \text { and } y=\frac{-2}{9}$

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

Given:

\begin{aligned} &2 x+3 y=10 \\ &x+6 y=4 \end{aligned}

Solution:

First D: determinant of the coefficient matrix

$\mathrm{D}=\left|\begin{array}{ll} 2 & 3 \\ 1 & 6 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$

\begin{aligned} &=(2)(6)-(3)(1) \\ &=12-3 \\ &=9 \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}{cc} 10 & 3 \\ 4 & 6 \end{array}\right| \\ &=60-12 \\ &=48 \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} 2 & 10 \\ 1 & 4 \end{array}\right| \\ &=8-10 \\ &=-2 \end{aligned}

Now,

$\begin{gathered} \mathrm{x}=\frac{D_{1}}{D}=\frac{48}{9}=\frac{16}{3} \\ \mathrm{y}=\frac{D_{2}}{D}=\frac{-2}{9} \end{gathered}$

Hence,$x=\frac{16}{3} \text { and } y=\frac{-2}{9}$

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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