Respuesta :

We have the following equation system:

[tex]\begin{cases}y=-\frac{1}{3}x-1 \\ 2x+y=-6\end{cases}[/tex]

Those are two lines, to plot them you just need two points. The intercepts are the easiest points to calculate. Let's calculate for the first line:

[tex]\begin{gathered} x=0\Rightarrow y=1 \\ y=0\Rightarrow0=-\frac{1}{3}x-1\Rightarrow x=-3 \end{gathered}[/tex]

Then, this set of points belong to the first line:

[tex]\lbrace(0,1),(-3,0)\rbrace[/tex]

By the same logic, this set of points belong to the second line:

[tex]\lbrace(0,-6),(-3,0)\rbrace[/tex]

Ploting your system, it should look like this

Coincidentally, we already know the solution because it is on the x-intercept(the point (-3,0)). But let's calculate using our equations.

First, let's rewrite the second equation in slope intercept form.

[tex]2x+y=-6\Leftrightarrow y=-2x-6[/tex]

Now we match our equations:

[tex]-\frac{1}{3}x-1=-2x-6\Rightarrow x=-3[/tex]

If you evaluate any of the functions at x = -

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