Intersection of a line and a curve. 
Given that the line $y = 2x + 3$ intersects the curve $y = x^2 + 3x + 1$ at two separate points, I have to find these two points.

Here is what I did:
$$2x + 3 = x^2 + 3x + 1$$
$$0 = x^2 + 1x - 2$$
Using factorisation:
$$x = -2 \text{ or } 1$$
Substituting each values of $x$ obtained into the equation of the  straight line gives two points of intersections at $(-2, -1)$ and $(1, 5)$

Here is my issue:
Why does this work? 
Equating the curve and the straight line means they share a single similar value of $y$ while they clearly share two.
 A: Equating the two equations doesn't mean the curve and the line share a single value of $y$; it means that you're assuming they share a value of $y$, and then getting an equation for the corresponding shared value of $x$. This says nothing about how many shared pairs $(x, y)$ there might be.
A: What is an intersection?
An intersection is where both $y$ for each function are equal for the same $x$.
Consider $f(x)=g(x)$.
Your intersection(s) are ALL the points where you can plug (the same) $x$ into both $f(x)$ and $g(x)$ and get an equality.
Because $f(x)=g(x)$, $y=y$, if that makes sense.
And this can happen say $0,1,2...\infty$ times.
A: Define two systems of equations to be equivalent if they have precisely the same sets of solutions. Abstracting, "solving a system of equations" is the process of successively replacing a given system by equivalent systems until one reaches a "tautological" system whose solutions can be read off by inspection.
For example, the system
$$
\left.
\begin{aligned}
  y &= 2x + 3 \\
  y &= x^{2} + 3x + 1
\end{aligned}
\right\}
\tag{1}
$$
is equivalent, by subtracting the first equation from the second, to the system
$$
\left.
\begin{aligned}
  y &= 2x + 3 \\
  0 &= x^{2} + x - 2
\end{aligned}
\right\}
\tag{2}
$$
in which $y$ has been eliminated from the second equation. You solved the second equation using the quadratic formula, obtaining
$$
\left.
\begin{aligned}
  y &= 2x + 3 \\
  x &= -2\quad\text{or}\quad 1
\end{aligned}
\right\}
\tag{3}
$$
then implicitly used the first equation to deduce the corresponding value(s) of $y$.
"Why this works" should be apparent. From this perspective, it should be clear that the reasoning

Equating the curve and the straight line means they share a single similar value of $y$ while they clearly share two.

would point to a logical gap only if the equation $y = y$ had a unique solution. But the opposite is true: $y = y$ is a tautology; it has no non-solutions.

In general, any "reversible operation" on a system of equations yields an equivalent system. The following operations (non-exhaustive list!) are reversible is this sense:


*

*Adding a (constant) multiple of one equation to another equation.

*Multiplying an equation by a non-zero constant, or by a non-vanishing expression.

*Exchanging two equations.

*Replacing an equation $a = b$ with $f(a) = f(b)$ for some injective function $f$. (For equations involving real variables, this includes cubing or exponentiating both sides, squaring both sides when both sides are known to be non-negative, and so forth.)

*If $f$ and $g$ are functions, replacing $f(y) = g(y)$ with $f(\phi(x)) = g(\phi(x))$ for some injective function $\phi$.
Compare the first three with the Gaussian elimination algorithm for systems of linear equations in several variables.
A: In this case, you are actually not assuming that they share the same value for y. Here, you are equating at the point where both points intersect each other. It is easier to visualise this when you plot a graph online. And in this case, you are only able to equate the two together BECAUSE they intersect which allows you to find the points where they meet, and in this case, at two different intersections.
A: Let  $y=f(x)$ and $y=g(x)$ define two curves in the x-y plane.
When you set the two equations, $y=f(x)$ and $y=g(x)$ equal to one another (i.e. $f(x)=g(x)$ ), you are asking to find all values of $x$ for which the points $(x,f(x))$ and $(x,g(x))$ are the same points.  In other words, you are asking "Which point of the curve $y=f(x)$ is also a point on the curve $y=g(x)$?"
Depending on the functions $f$ and $g$, there may be zero, 1 or many points in common to both curves.  When you solve $f(x)=g(x)$, you get a collection of $x$ values that you can then plug into either of the equations $y=f(x)$ or $y=g(x)$ to determine the $y$-value for a specific $x$-value.  
For two different values of $x$, the resulting values of $y$ may be different, but for any given value of $x$ in the solution set, $f(x)$ and $g(x)$ will both yield the same $y$-value.
A: Here's how you can think about it.
You've got two curves with equations
$y = 2x + 3$ and $y = x^2 + 3x + 1$.
You're looking for a point of intersection, that is, a point $(x_0, y_0)$, for which both of these equations hold. Then for this point you have:
$y_0 = 2x_0 + 3$ 
$y_0 = x_0^2 + 3x_0 + 1$
Therefore:
$2x_0 + 3 = x_0^2 + 3x_0 + 1$
A: If I subtract a number from itself, I get zero. Conversely, if I subtract a number from any number, and I get a non-zero number, then the first number isn't equal to the number from which I subtracted it.
Therefore, for an x I select, if $x^2 + x - 2$ isn't zero, then your first equation does not hold at that x value. 
A: The term intersection of a line and a curve indicates set operations.

In fact when intersecting a line $\mathcal{l}: y=2x+3$ and a curve $\mathcal{C}: y=x^2+3x+1$ we consider the graphs of these functions
\begin{align*}
G(\mathcal{{l}})&=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big|\,y=2x+3\}\\
G(\mathcal{C})&=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big|\,y=x^2+3x+1\}\\
\end{align*}
and the intersection of these two sets
  \begin{align*}
G(\mathcal{l})\cap G(\mathcal{C})=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big|\,2x+3=x^2+3x+1\}
\end{align*}

