91Ó°ÊÓ

Number the given system of equations.

$\begin{array}{c}x−2y=10\left(1\right)\\ 2x−4y=12\left(2\right)\end{array}$

The first equation is: $x-2y=10$.

When $x=0$,

$\begin{array}{c}0−2y=10\\ −2y=10\\ y=−5\end{array}$

Therefore, when $x=0$, $y=-5$.

When $x=2$,

$\begin{array}{c}2−2y=10\\ −2y=8\\ y=−4\end{array}$

Therefore, when $x=2$, $y=-4$

Therefore, the equation $x-2y=10$ is the equation of the line passing through the points $\left(0,-5\right)$ and $\left(2,-4\right)$.

The graph of the equation $x-2y=10$ is:

The second equation is: $2x-4y=12$.

When $x=0$,

$\begin{array}{c}2x−4y=12\\ 2\left(0\right)−4y=12\\ 0−4y=12\\ −4y=12\\ y=−3\end{array}$

Therefore, when $x=0$, $y=-3$.

When $x=2$,

$\begin{array}{c}2\left(2\right)−4y=12\\ 4−4y=12\\ −4y=8\\ y=−2\end{array}$

Therefore, when $x=2$, $y=-2$

Therefore, the equation $2x-4y=12$ is the equation of the line passing through the points $\left(0,-3\right)$ and $\left(2,-2\right)$.

The graph of the equation $2x-4y=12$ is:

The graph of the equations $x-2y=10$ and $2x-4y=12$ is:

From the graph of the equations $x-2y=10$ and $2x-4y=12$, it can be noticed that the lines are parallel to each other and therefore there is no point of intersection of these lines.

The solution of the system of linear equations in two variables is given by the intersection point of the lines.

Therefore, there is no solution of the given system of equations.