Solve the first equation for [latex]y[/latex].

[latex]\begin{array}{c}2x+y=-8\\ y=-2x - 8\end{array}[/latex]

Solve the second equation for [latex]y[/latex].

[latex]\begin{array}{c}x-y=-1\\ y=x+1\end{array}[/latex]

Graph both equations on the same set of axes as seen below.

The lines appear to intersect at the point [latex]\left(-3,-2\right)[/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations

[latex]\begin{array}{ll}2\left(-3\right)+\left(-2\right)=-8\hfill & \hfill \\ \text{ }-8=-8\hfill & \text{True}\hfill \\ \text{ }\left(-3\right)-\left(-2\right)=-1\hfill & \hfill \\ \text{ }-1=-1\hfill & \text{True}\hfill \end{array}[/latex]

The solution to the system is the ordered pair [latex]\left(-3,-2\right)[/latex], so the system is independent.

First, graph [latex]y=2x+1[/latex] using the slope [latex]m = 2[/latex] and the [latex]y[/latex]-intercept [latex](0,1)[/latex]

Next, add [latex]y=2x-3[/latex] using the slope [latex]m = 2[/latex], and the [latex]y[/latex]-intercept [latex](0,-3)[/latex]

Notice how these are parallel lines, and they don’t cross.  In the previous section, we discussed how there are no solutions to a system of equations that are parallel lines.  We classified this type of system as an inconsistent system.

First, find the [latex]x[/latex]– and [latex]y[/latex]– intercepts of [latex]y=\frac{1}{2}x+2[/latex]

The [latex]x[/latex]-intercept will have a value of [latex]0[/latex] for [latex]y[/latex], so substitute [latex]y=0[/latex] into the equation, and isolate the variable [latex]x[/latex].

[latex]\begin{array}{c}0=\frac{1}{2}x+2\\\underline{\,\,\,\,\,\,\,\,-2\,\,\,\,\,\,-2}\\-2=\frac{1}{2}x\\\left(2\right)\left(-2\right)=\left(2\right)\frac{1}{2}x\\-4=x\end{array}[/latex]

The [latex]x[/latex]-intercept of [latex]y=\frac{1}{2}x+2[/latex] is [latex]\left(-4,0\right)[/latex].

The [latex]y[/latex]-intercept is easier to find since this equation is in slope-intercept form.  The [latex]y[/latex]-intercept is [latex](2,0)[/latex].

Now we can plot [latex]y=\frac{1}{2}x+2[/latex] using the intercepts

Now find the intercepts of [latex]2y-x=4[/latex]

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 the same intercepts as [latex]y=\frac{1}{2}x+2[/latex]!  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 you were to write them both in slope-intercept form you would see that they are the same equation.

When you graph them, they are the same line. In the previous section, we learned that systems with two of the same equations in them have an infinite number of solutions and are classified as a dependent system.