Solving 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.)
Zero solutions:
$\begin{array}{l}y=-2x+4\\ y=-2x-3\end{array}$
One solution:
$\begin{array}{l}y=0.5x+2\\ y=-2x-3\end{array}$
Infinitely many solutions:
$\begin{array}{l}y=-2x-4\\ y+4=-2x\end{array}$
There are a few different methods of solving systems of linear equations:
- The Graphing Method . This is useful when you just need a rough answer, or you're pretty sure the intersection happens at integer coordinates. Just graph the two lines, and see where they intersect!
- The Substitution Method . First, solve one linear equation for $y$ in terms of $x$ . Then substitute that expression for $y$ in the other linear equation. You'll get an equation in $x$ . Solve this, and you have the $x$ -coordinate of the intersection. Then plug in $x$ to either equation to find the corresponding $y$ -coordinate. (If it's easier, you can start by solving an equation for $x$ in terms of $y$ , also – same difference!)
- The Linear Combination Method , aka The Addition Method , aka The Elimination Method. 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. Then solve for $x$ (or $y$ , whichever's left) and substitute back to get the other coordinate.
- The Matrix Method . This is really just the Linear Combination method, made simpler by shorthand notation.
See the second graph above. The solution is where the two lines intersect, the point $(-2,1)$ .
Example 1:
Solve the system $\{\begin{array}{l}3x+2y=16\\ 7x+y=19\end{array}$
Solve the second equation for $y$ .
$y=19-7x$
Substitute $19-7x$ for $y$ in the first equation and solve for $x$ .
$\begin{array}{l}3x+2(19-7x)=16\\ 3x+38-14x=16\\ -11x=-22\\ x=2\end{array}$
Substitute $2$ for $x$ in $y=19-7x$ and solve for $y$ .
$\begin{array}{l}y=19-7\left(2\right)\\ y=5\end{array}$
The solution is $(2,5)$ .
Example 2:
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 $(1,-2)$ .