lesson 16-2 practice

Question

Identify a possible solution to the linear inequality below:

y<-2x-1

Answer 1

y < -2x - 1 is a linear inequality that can be read "y is less than -2 times x minus one."

A solution is any point, (x, y) that satisfies this equation. We can pick any point and see if it works by plugging in the x and y values into the equation and seeing if it is true.

For example, let's select something easy to try like (0, 0), where x = 0 and y = 0.

When we plug it into the equation, we get:

y < -2x - 1

0 < -2 * 0 - 1

0 < 0 - 1

0 < -1

Note that by order of operations (PEMDAS), we multiply -2 times 0 before we subtract -1 to get the inequality:

0 < -1

This statement is false, because 0 is not less than negative one; it is greater. However, we can see that if we had chosen a smaller number for y or a larger number for x, it could have worked.

For example, let's try (0, -3), where x = 0 and y = -3:

y < -2x - 1

-3 < -2 * 0 - 1

-3 < 0 - 1

-3 < -1

Because -3 is to the left of -1 on the number line, -3 < -1 is true and therefore we have shown that (0, -3) or x = 0, y = -3 is one possible solution.

There are many possible solutions to this inequality. For example, (-1, -1) also works. If we graph the line

y = -2x - 1, any points that lie below that line would satisfy the inequality.

Answer 2

I believe the best way to solve this problem is to first graph the linear inequality on the coordinate plane. So you have a y-intercept at -1 and a slope of -2/1. Because this is a < symbol, after graphing several points, you should connect your points with a dotted line. Then you need to shade below the line you just graphed (because it says less than). You can then find tons of solutions once you have correctly graphed your inequality. These solutions will lie in the shaded region of your graph (just not on the line since it is dotted). So for example if, once I graphed and shaded, the point (2, 3) is in the shaded region, that would be a solution. If the point (-2, 0) was not in the shaded region that can NOT be a solution. So you should have many solutions to choose from.

Does that help?