Solving Systems of Linear Equations Using Elimination
Systems of Linear equations:
A system of linear equations is just a set of two or more linear equations.
In two variables $\left(x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\right)$ , the graph of a system of two equations is a pair of lines in the plane.
There are three possibilities:
- The lines intersect at zero points. (The lines are parallel.)
- The lines intersect at exactly one point. (Most cases.)
- The lines intersect at infinitely many points. (The two equations represent the same line.)
How to Solve a System of Linear Equations Using The Elimination Method (aka The Addition Method, aka The Linear Combination Method)
- Step $1$ : Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the $x$ -terms or the $y$ -terms cancel out.
- Step $2$ : Then solve for $x$ (or $y$ , whichever's left) and substitute back to get the other coordinate.
Now, how do we know that a linear equation obtained by the addition of the first equation with a scalar multiplication of the second is equivalent to the first?
Let us take an example. Consider the system
$\begin{array}{l}3x+2y=3\\ x-y=-4\end{array}$ .
Consider the equation obtained by multiplying the second equation by a constant $m$ and then adding the resultant equation with the first one.
That is, $\left(3x+2y\right)+m\left(x-y\right)=\left(3\right)+m\left(-4\right)$ .
What we need to prove is that this equation is equivalent to the equation $3x+2y=3$ .
We have $x-y=-4\Rightarrow m\left(x-y\right)=-4m$ .
Since $m\left(x-y\right)=-4m$ , subtract $m\left(x-y\right)$ from the left side and $-4m$ from the right side of the equation $\left(3x+2y\right)+m\left(x-y\right)=\left(3\right)-4m$ which will retain the balance.
$\left(3x+2y\right)+m\left(x-y\right)-m\left(x-y\right)=\left(3\right)-4m-\left(-4m\right)$
Cancelling common terms we get, $3x+2y=3$ which is equivalent to the first equation.
Therefore, the systems of equations $\begin{array}{l}3x+2y=3\\ x-y=-4\end{array}$ and $\begin{array}{l}\left(3x+2y\right)+m\left(x-y\right)=3+m\left(-4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-y=-4\end{array}$ are equivalent.
In general, for any system of equations $K=L$ and $P=Q$ , it ca be shown that $K+mP=L+mQ$ is equivalent to $K=L$ .
Example:
Solve the system $\{\begin{array}{l}4x+3y=-2\\ 8x-2y=12\end{array}$
Multiply the first equation by $-2$ and add the result to the second equation.
$\begin{array}{l}-8x-6y=4\\ \underset{\_}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8x-2y=12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-8y=16\end{array}$
Solve for $y$ .
$y=-2$
Substitute for $y$ in either of the original equations and solve for $x$ .
$\begin{array}{l}4x+3(-2)=-2\\ 4x-6=-2\\ 4x=4\\ x=1\end{array}$
The solution is $(1,-2)$ .