Example

Graph the system [latex]\begin{array}{c}y=\frac{1}{2}x+2\\2y-x=4\end{array}[/latex].  Identify the type of system.

Solution

Equation 1:  [latex]y=\frac{1}{2}x+2[/latex] has a slope of [latex]m=\frac{1}{2}[/latex] and a [latex]y[/latex]-intercept at [latex](0,2)[/latex]. We plot the point [latex]\left ( 0,2\right )[/latex], then use the slope with a riuse of [latex]1[/latex] and a run of [latex]2[/latex] to find another point on the line [latex]\left (2,3\right )[/latex].

Equation 2, [latex]2y-x=4[/latex]is written in the form [latex]ax+by=c[/latex] so it is easiest to find the the intercepts.

Substitute [latex]y = 0[/latex] in to the equation to find the [latex]x[/latex]-intercept.

[latex]\begin{array}{c}2y-x=4\\2\left(0\right)-x=4\\x=-4\end{array}[/latex]

The [latex]x[/latex]-intercept of [latex]2y-x=4[/latex] is [latex]\left(-4,0\right)[/latex].

Now substitute [latex]x = 0[/latex] into the equation to find the y-intercept.

[latex]\begin{array}{c}2y-x=4\\2y-0=4\\2y=4\\y=2\end{array}[/latex]

The [latex]y[/latex]-intercept of [latex]2y-x=4[/latex] is [latex]\left(0,2\right)[/latex].

WAIT, these are exactly the same lines!  In fact, [latex]y=\frac{1}{2}x+2[/latex] and [latex]2y-x=4[/latex] are really the same equation, expressed in different ways.  If we were to write them both in slope-intercept form we would see that they are the same equation.

Systems with identical equations have an infinite number of solutions that lie on the line and are classified as a dependent system.

The solutions can be written in set builder notation: [latex]\{\;\left (x, y\right )\;\large | \normalsize \;y=\frac{1}{2}x+2\;\}[/latex].

They can also be written as a set of ordered pairs where [latex]y[/latex] is written in term of [latex]x[/latex]: [latex]\left (x,\frac{1}{2}x+2\right )[/latex].