1.6 System of Equations: The Intersection of Two Lines

Consider the following two linear equations:

$\begin{array}{rl}& ax+by=c\\ & dx+ey=f\end{array}$

A system of equations is any collection of two or more equations. The solution set of a system of two linear equations with two variables is the set of all ordered pairs that satisfy both equations in the system. There are three possibilities for the relative positions of the lines in a system of equations. First, the lines can be identical. We call this a dependent system, which has infinitely many solutions. Second, the lines could be parallel, which means that they do not cross each other. This is called an inconsistent system, which has no solution. Third, the lines could intersect at exactly one point. This is called an independent system, which has exactly one solution, or one intersection.

We will learn two methods to determine which type of the system is formed by a set of equations and to find the exact solution if the system is independent.

Substitution Method

The substitution method is the first mathematical method to find the solution of a system of equations. To use the substitution method, first solve one of the equations for $x$ or $y$, isolating the variable on the left-hand side of the equation. The equation will look like $x=$ or $y=$ . Then, plug in $x$ or $y$, whichever variable you have isolated on the left-hand side of the equation, into the other equation. In other words, if we solved the first equation for $y$ on the left-hand side, substitute the right-hand side of the equation for $y$ in the second equation; if we solved the first equation for $x$ on the left-hand side, substitute the right-hand side of the equation for $x$ in the second equation. This way, we can eliminate a variable in one equation by substituting from the other equation.

Suppose we have a system of two linear equations with two variables:

$\begin{array}{rl}2x+3y& =4\\ -2x+y& =6\end{array}$

We want to find the point at which the two equations intersect, which means that we are finding a distinct point. The second equation looks easier to manipulate into the form $y=$  by adding $2x$ to both sides of the equation. The result is as follows:

$y=2x+6$

Since $y$ is equal to $2x+6$, we can plug this in to replace $y$ in the first equation, and then solve for $x$:

$\begin{array}{rl}2x+6x+18& =4\\ 8x& =4-18\\ 8x& =-14\\ x& =-\frac{14}{8}=-\frac{7}{4}\end{array}$

Now we know the exact value of $x$. We can plug this value for $x$ into $y=2x+6$:

$y=2(-\frac{7}{4})+6=\frac{5}{2}$

We now have the solution to this system of linear equations, which indicates the only point at which the two lines intersect: $(-\frac{7}{4}$ , $\frac{5}{2})$.

How can we use the substitution method to identify whether a system is independent, dependent, or inconsistent? Consider the following system of two linear equations:

$\begin{array}{rl}x+y& =2\\ 4x+4y& =8\end{array}$

Let’s rewrite the first equation, then plug it into the second equation.

$\begin{array}{rl}y& =-x+2\\ & \\ 4x+4(-x+2)& =8\\ 4x-4x+8& =8\\ 8& =8\end{array}$

When you see a result like this, where a constant equals a constant and both constants are the same value, the system is dependent.

Next, consider the following system of two equations:

$\begin{array}{rl}x+y& =3\\ 4x+4y& =8\end{array}$

As we did before, we will rewrite the first equation, and then plug it into the second equation.

$\begin{array}{rl}y& =-x+2\\ & \\ 4x+4(-x+2)& =8\\ 4x-4x+8& =8\\ 8& =8\end{array}$

While both sides of the equations are constants, they are not the same value. This indicates that the system is inconsistent.

Addition Method

In the substitution method, we substituted one variable from one equation into the other equation to eliminate a variable. In the addition method, we will eliminate one variable by adding the two equations together. In other words, we will add the right-hand side of the two equations together and the left-hand side of the two equations together to solve the system.

Suppose we have a system of two equations:

$\begin{array}{rl}x+3y& =4\\ -2x+2y& =8\end{array}$

We cannot eliminate one of the variables simply by adding the two equations together. So, we will first multiply both sides of the first equation by 2:

$2x+6y=8$

Then, we will add the two equations together:

$2x+6y=8$
$\underset{\_}{-2x+2y=8}$
$8y=16$
$y=2$

Finally, plug the value of $y$ back into one of the equations in the system to solve for $x$:

$\begin{array}{rl}x+3(2)& =4\\ x+6& =4\\ x& =4-6=-2\end{array}$

Examples

Let’s run a few examples of identifying the type of system and finding the solution of an independent system of equations.

Example 1

Solve the following system using the addition method.

$\begin{array}{rl}& 3x-y=9\\ & 2x+y=1\end{array}$

Since the coefficient of $y$ has the same magnitude in both equations but with opposite signs, we can simply add the two equations together to eliminate $y$. Then we can solve for $x$.

$\begin{array}{rl}3x+2x& =9+1\\ 5x& =10\\ x& =2\end{array}$

Plug the value of $x$ into one of the equations in the system to find $y$:

$\begin{array}{rl}3(2)-y& =9\\ 6-y& =9\\ y& =6-9=-3\end{array}$

Since there is a unique solution, we know that the two equations are an independent system.

Example 2

Solve the following system using the substitution method.

$\begin{array}{rl}& 2x-3y=-2\\ & 3x-2y=12\end{array}$

We will rewrite the first equation as follows:

$\begin{array}{rl}2x-3y& =-2\\ 2x& =3y-2\\ x& =\frac{3}{2}y-1\end{array}$

Then, substitute for $x$ in the second equation:

$\begin{array}{rl}3(\frac{3}{2}y-1)-2y& =12\\ \frac{9}{2}y-3-2y& =12\\ \frac{5}{2}y& =15\\ y& =6\end{array}$

Then, plug the $y$ value into one of the equations in the system.

$\begin{array}{rl}2x-3(6)& =-2\\ 2x-18& =-2\\ 2x& =-2+18=16\\ x& =8\end{array}$

We know that the system is independent since there is a unique solution.

Example 3

Solve the following system using the addition method.

$\begin{array}{rl}& 0.2x-0.4y=0.5\\ & x-2y=1.3\end{array}$

We will first multiply the first equation by -5:

$\begin{array}{rl}& -x+2y=-2.5\\ & x-2y=1.3\end{array}$

Then, add the two equations together:

$-x+2y=-2.5$
$\underset{\_}{x-2y=1.3}$
$0=-1.2$

This result means that the system is inconsistent.