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2x + 2/3y= -2 x, y intercept

Sagot :

We need to find the points at which the expression below intercept the axis of the coordinate plane:

[tex]2x+\frac{2}{3}y=-2[/tex]

To find the "x" intercept we need to find the value of "x" that results in a value of "y" equal to 0. We have:

[tex]\begin{gathered} 2x+\frac{2}{3}\cdot0=-2 \\ 2x+0=-2 \\ 2x=-2 \\ x=\frac{-2}{2}=-1 \end{gathered}[/tex]

To find the "y" intercept we need to find which value of "y" the function outputs when we make x equal to 0.

[tex]\begin{gathered} 2\cdot0+\frac{2}{3}y=-2 \\ \frac{2}{3}y=-2 \\ 2y=-6 \\ y=\frac{-6}{2}=-3 \end{gathered}[/tex]

The x intercept is -1 and the y intercept is -3.