Why does the elimination method work?

Question

Consider solving the following system of equations:

\begin{cases} 2x−4y&=0, \\ 4x+4y&=−4. \end{cases}

We can use the substitution method or the elimination method, of which the latter has always been unclear to me: why does it work? I know that by eliminating one of the two variables you can solve for the variable you're left with, and consequently solve for the other one, but why do you get the point $(x,y)$ at which both lines intersect? Can anyone show me this graphically?

Answer 1

why do you get the point $(x,y)$ at which both lines intersect?

You are solving two equations, so two curves are involved (it can be more, of course, but for simplicity let's stick with two curves).

You are looking for points which are on both curves (lines in your example). The only way a point can be on both curves is if the curves intersect at those points.

Any other points on either line will not be on both lines (will not satisfy both equations at the same time).

So each equation represents one line/curve (or in more complex problems it can be curves or more complex structures in multidimensional spaces) and the set of values that satisfy all equations are, by definition, points on each line/curve.

Answer 2

Let's use different colors for individual equations in your system:

$$\color{red}{2x−4y=\phantom-0\tag {eq1}}\\ \color{blue}{4x+4y=-4\tag {eq2}}$$

If we express $\color{red}y$ from the first equation, we will obtain the equation of the same line $\color{red}{\mathrm{eq1}}$ (see the animation below), only in an other form:

$$\color{red}{y = {2 \over 4}x}$$

So for every real $\color{red}x$ you obtain $\color{red}{y = {2 \over 4}x}$ such that the point $\color{red}{(x,y)}$ is the point of the line $\color{red}{\mathrm{eq1}}$:

Now, when we substitute this $\color{red}{y = {2 \over 4}x}$ into the second equation

$$\color{blue}{4x+4y=-4\tag {eq2}}$$

then because $\color{red}{y = {2 \over 4}x}$, the point $(x, y)$ will belong to the line $\color{red}{\mathrm{eq1}}$, but because at the same time it fulfills the equation $\color{blue}{\mathrm{(eq2)}}$, it will be a point of the line $\color{blue}{\mathrm{eq2}}$, too.

Answer 3

This is a special case of a more general principle: if $S$ and $T$ are the solution sets to the equations $f(\mathbf x)=0$ and $g(\mathbf x)=0$, respectively, then the solution set of every linear combination $\lambda f(\mathbf x)+\mu g(\mathbf x)=0$ of these equations contains the common solutions $S\cap T$ of the original equations. You should verify this for yourself—the proof is fairly obvious.

In the case of a pair of linear equations, the solution to the system is the intersection of the two lines represented by the equations. Since any nontrivial linear combination of two linear equations is also linear, when you eliminate a variable by forming a linear combination of two equations, what you’re effectively doing is finding another line that passes through this intersection. Two distinct lines intersect in at most one point, so doing this doesn’t introduce any extraneous solutions. This interpretation generalizes to more linear equations with more variables, but instead of lines, each individual equation represents a hyperplane.