Can someone please help me?​

9514 1404 393

Step-by-step explanation:

I like a graphing calculator for creating graphs. Here, each inequality is in "slope-intercept" form, so the y-intercepts are easily located (+8 and -4).

The slope (x-coefficient) is the "rise" divided by the "run". The boundary line of the first inequality has a rise of -1 for each run of 3 to the right. The second rises 5 units for each 3 to the right. Starting from the y-intercept, additional points to plot are easily found.

Of course, the boundary line is solid for y ≥ ( ), and is dashed for y < ( ). The shading will be above the line (for higher y-values) for the first inequality, and below the line (for lower y-values) for the second inequality. That puts the double-shaded solution area in the right-side quadrant of the X where the lines cross.

There are an infinite number of solution points to choose from. One of them is (12, 5). Points with an x-value that is a multiple of 3 will be easiest to substitute into the inequalities to check.

  5 ≥ (-1/3)(12) +8   ⇒   5 ≥ 4 . . . true

  5 < (5/3)(12) -4   ⇒   5 < 16 . . . true

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step-by-step explanation: