For this exercise, the addition method is most suitable, as $+4f+4f$ occurs in both equations.

Solve for $ee$

Plug $ee$ into one of both equations, for instance $\mathrm{I}\backslash mathrm\{I\}$

Write down the solution set

$\mathbb{L}=\{(e\mathrm{\mid }f)=(8\mathrm{\mid }3)\}\backslash mathbb\{L\}=\backslash \{(e|f)=(8|3)\backslash \}$

You get a very nice solution with the substitution method.

The equation $\mathrm{I}\backslash mathrm\{I\}$ is solved to $7y7y$ and in equation $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ there is $14y14y$. You can even use the substitution method without doing any transformation.

Transform equation $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$

Plug $7y7y$ from $\mathrm{I}\backslash mathrm\{I\}$ into $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$

Conclude and solve for $xx$

$\begin{array}{}\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{lrcl\}\backslash end\{array\}$

As you can see now, the entries in the last line are not the same, which results in a contradiction. Hence, the system has no solution, since no number can be assigned for $xx$, such that $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ is satisfied.

You best use the addition method, since you can see with a sharp eye (this is what math lessons shall train you for) that $2k2k$ is $44$ times $0.5k0.5k$.

Multiply the first equation by 4

As you can see now, both equations are identical. It follows that the system has an infinite number of solutions. Graphically, this corresponds to two straight lines lying on top of each other.

Solve the equation for one more variable to obtain the solution set.

$10m=14+2k\mathrm{\mid }:10m=1.4+0.2k10\; m\; =\; 14\; +\; 2\; k\; \backslash quad\; |\; :\; 10\; \backslash \backslash m\; =\; 1.4\; +\; 0.2\; k$

$\mathbb{L}=\{(m\mathrm{\mid }k)\mid m=1.4+0.2k\}\backslash mathbb\{L\}\; =\; \backslash left\backslash \{\; (m|k)\; \backslash ;\backslash big\; |\backslash ;\; m\; =\; 1.4\; +\; 0.2\; k\; \backslash right\backslash \}$