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Here is one more practice problem to solve.

Graph this system of linear inequalities. Shade in the correct areas.

3y < x + 6
2y < 6x + 7

Click through the slideshow to reveal the answer steps.

First rearrange both equations into slope-intercept form:

3y < x + 6
\(\small\mathsf{ \frac{3}{3} }\)y < \(\small\mathsf{ \frac{1}{3} }\)x + \(\small\mathsf{ \frac{6}{3} }\)
y < \(\small\mathsf{ \frac{1}{3} }\)x + 2

2y < 6x + 7
\(\small\mathsf{ \frac{2}{2} }\)y < \(\small\mathsf{ \frac{6}{2} }\)x + \(\small\mathsf{ \frac{7}{2} }\)
y < 3x + \(\small\mathsf{ \frac{7}{2} }\)

Draw the lines.Both lines will be dotted, and we will shade below both lines, since both equations are "less than" (<).

Now shade in the area below equation 1 because it is less than. (<) Here is the graph with just the shading for equation 1.

Now shade in the area below equation 2 becuase it is less than. (<) Here is the graph with just the shading for equation 2.

Here is the graph with shading for both equations:

Now, choose a point in the overlapping shaded area to check your answer. Let's go with (0,0). If we shaded the graph correctly, the solution should be true for both equations.

3y < x + 6
3(0) < 0 + 6
0 < 6
True		 2y < 6x + 7
2(0) < 6(0) + 7
0 < 7
True

This shows that the inequality is correctly shaded. You can also check by choosing another point outside any shaded area and the solution should be false for both equations.