# Question #73ee7

Nov 14, 2017

Are the equations $y < 3 x + 4$ and $- 3 x - 2 y \ge 6$? If so, the intersection lies at $\left(- 2 , 0\right)$, though the solution set is all the points where the blue and red areas meet on the graph here.

#### Explanation:

The first inequality is ready to graph. In the attached picture, it is red. To begin, graph the line y=3x + 4. It has a y-intercept of $4$ and a slope of $3$.
The line should be dotted, since the sign is "less than", not "less than or equal to". When there is no "equal to" part in an inequality, you always use a dashed line in place of a solid.
Then, since it is "less than", shade the graph under the line.

Next, you have to reformat $- 3 x - 2 y \ge 6$ to graph it. You want it in slope-intercept form ($y = m x + b$).
Add $3 x$ to both sides to get $- 2 y \ge 6 + 3 x$.
Divide through by $- 2$ to get $y$ by itself, and remember that when you multiply or divide an inequality by a negative number, you have to flip the sign. So $\ge$ becomes $\le$.
$- 2 \frac{y}{-} 2 \ge \frac{6}{-} 2 + 3 \frac{x}{-} 2$
$\implies y \le - 3 - 3 \frac{x}{2}$
$\implies y \le - 3 \frac{x}{2} - 3$

Graph this inequality as a solid line, since "equal to" is part of the expression. $y = - 3 \frac{x}{2} - 3$ is in blue on the graph here. Shade down to include all the answers (values that make it true) for the expression.

There is not one point that is the solution for an inequality. Here, there are an infinite number of points that satisfy the set of inequalities. They are where the blue and red areas meet in the graph. 