Keep in mind that both sides of the equations are the same (that's the point of an equation). From basic algebra we know that you can manipulate an equation by adding or subtracting the same value on both sides, so if you have
\begin{align}
x+y &= 5, \tag{1}\\
2x+y &= 8, \tag{2}
\end{align}
You can subtract the left hand side (LHS) of $(1)$ from the LHS of $(2)$, while subtracting the right hand side (RHS) of $(1)$ from the RHS of $(2)$ since the LHS and the RHS of each equation are the exact same thing. It's like you're subtracting $5$ from both sides, since $x+y$ really is equal to $5$. Hence we write $(2)$ as
\begin{align}
2x+y &= 8\\
2x + y - x - y &= 8 - 5\\
x &= 3,
\end{align}
which (since $x+y = 5$) implies $y$ must equal $2$.
You have here two of the fundamental ways to represent a line in $\mathbb R^2$. The first is described by a parametric representation that uses a point $\mathbf p_0$ on the line and a direction vector $\mathbf v$ parallel to the line. Every point on the line can then be expressed in the form $\mathbf p_0+\lambda\mathbf v$, $\lambda\in\mathbb R$.
The second line is described implicitly by an equation in point-normal form. There are various ways to interpret it geometrically, but I find this one easiest to visualize: The perpendicular line through the origin to a line $\mathscr l$ is completely characterized by a direction vector $\mathbf n$. Per the preceding paragraph, a parameterization of that perpendicular line is simply $\lambda\mathbf n$. If $\mathscr l$ is translated so that it passes through the origin, the translated line $\mathscr l'$ has the property that the position vector $\mathbf x'$ of every point on it is orthogonal to $\mathbf n$. That is, $\mathscr l'$ is the set of points that satisfy $\mathbf n\cdot\mathbf x'=0$. We can translate $\mathscr l$ in this way by fixing some point $\mathbf p$ on it and subtracting this from every point $\mathbf x$ on $\mathscr l$. Substituting into the orthogonality condition, an equation for $\mathscr l$ is therefore $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, or $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$. Observe that it doesn’t matter which point was chosen for $\mathbf p$—the last form of the point-normal equation says that the dot product of any point on $\mathscr l$ with the normal vector $\mathbf n$ is constant.
Multiplying both sides of that last equation by $\mathbf n/\|\mathbf n\|^2$, we have $${\mathbf n \cdot \mathbf x\over \mathbf n\cdot\mathbf n}\mathbf n = {\mathbf n \cdot \mathbf p\over \mathbf n\cdot\mathbf n}\mathbf n,$$ which says that every point of $\mathscr l$ has the same orthogonal projection onto the normal vector $\mathbf n$. The length of this projection is $|\mathbf n\cdot\mathbf p|/\|\mathbf n\|$, which is also the distance of the line from the origin. Thus, the constant term in the point-normal equation of a line is the distance to the origin multiplied by the length of the normal vector. The sign of the constant term tells you on which side of the origin relative to the direction $\mathbf n$ the line lies.
To convert from the point-direction form to point-normal form, you have to find a normal to the line, which is any vector perpendicular to the line’s direction vector, and vice-versa when converting from point-normal to point-direction. The constant term for the point-normal form is then found by computing the dot product of this normal vector with any point on the line, as described above.
It’s not really necessary to convert representations to solve this problem, though. The intersection of the two lines is a point on the first line that satisfies the equation of the second, so you could simply plug the generic expressions for $x$ and $y$ from the first into the second and solve for $\lambda$ without doing any conversion. I expect that your professor has you do the conversion because he wants you to practice solving systems of linear equations, which is a major motivation for linear algebra in the first place.
Best Answer
Suppose we have two linear equations, $$y_1=m_1x+b_1$$ and $$y_2=m_2x+b_2.$$
If we set $y_1=y_2$,then $m_1x+b_1=m_2x+b_2$. We then solve for $x$ and substitute that value into either equation to get the corresponding $y$ value. The result is the single intersection point of these two straight lines. It will exist provided $m_1 \neq m_2$. I have just made an online interactive plot for you to play with. Sometimes this visual representation is what helps a concept like this to be truly understood.
Try simultaneously solving various equations like these while at the same time dragging the sliders to the appropriate values in this demonstration I have constructed. I think with this, clarity will eventually come to you.
An answer to the actual question:
If we simply add the equations side by side, we get a third equation that has slope equal to the sum of slopes, and $y$ intercept equal to the sum of $y$ intercepts. You can see this by plotting $y_1+y_2$ in the demonstration. This would be a third line that does not necessarily pass through the intersect of the two equations. In order to find the solution to a system of equations, that being the point at which both equations are simultaneously true, we need to set equals to equals, ie. $x$ equal to $x$ or $y$ equal to $y$. In this manner we find the solution to the system.
I have two more plots to show you plot 1 is the first, and plot 2 is the other. I think I understand precisely why you are asking this question now after reflecting on what you must be studying. It has to do with operations on systems of equalities, and how after multiplying one equality by a constant, then adding it to the other, you arrive at an equation in one variable, and thus arrive at a solution. You see that the sum intersects when a scalar multiple of one equality is added to the other. If you can manage to add a scalar multiple of one equation to the other in such a way that one of the variables vanishes, then you will have that third equation, it being the sum of two equations (not the originals), and that third equation will indeed intersect at the intersection of the first two equations. Best of clarity and fortune in your studies. :)