Solve Linear Inequalities: Find Points That Satisfy Y < 0.5x + 2

Linear inequalities are a fundamental concept in algebra, often appearing in various mathematical and real-world applications. Understanding how to determine if a point is a solution to a linear inequality is crucial for graphing these inequalities and solving systems of inequalities. When we're presented with an inequality like $y < 0.5x + 2$, we're essentially looking for all the coordinate pairs $(x, y)$ that make this statement true. This means that for any given $x$-value, the corresponding $y$-value must be strictly less than the value obtained by plugging $x$ into the expression $0.5x + 2$. This isn't just about finding one or two points; it's about identifying an entire region on the coordinate plane where all these points reside. The line $y = 0.5x + 2$ acts as a boundary, and the inequality < tells us which side of this boundary contains our solutions. Points on the line itself are not included because the inequality is strict (less than, not less than or equal to). To find these solution points, we can adopt a systematic approach. For each given point, we substitute its $x$ and $y$ coordinates into the inequality and check if the resulting statement is true. If it is, then that point is a solution. If it's false, then it's not. This process is straightforward but requires careful arithmetic. Let's dive into testing the provided points to see which ones satisfy $y < 0.5x + 2$.

Testing Potential Solutions for $y < 0.5x + 2$

To rigorously determine which of the given points satisfy the linear inequality $y < 0.5x + 2$, we will substitute the $x$ and $y$ coordinates of each point into the inequality and evaluate the result. A point is a solution if, after substitution, the inequality holds true. This method ensures accuracy and helps us build confidence in our understanding of linear inequalities.

Evaluating Point $(-3, -2)$

Let's begin with the point $(-3, -2)$. Here, $x = -3$ and $y = -2$. We substitute these values into the inequality $y < 0.5x + 2$:

$-2 < 0.5(-3) + 2$

First, calculate the right side of the inequality:

$0.5(-3) = -1.5$

Now, add 2:

$-1.5 + 2 = 0.5$

So the inequality becomes:

$-2 < 0.5$

This statement is true. Since $-2$ is indeed less than $0.5$, the point $(-3, -2)$ is a solution to the linear inequality $y < 0.5x + 2$. This confirms that points in certain regions of the coordinate plane will satisfy the inequality, and this point falls within that region.

Evaluating Point $(-2, 1)$

Next, we examine the point $(-2, 1)$. Here, $x = -2$ and $y = 1$. Substitute these into the inequality:

$1 < 0.5(-2) + 2$

Calculate the right side:

$0.5(-2) = -1$

Now, add 2:

$-1 + 2 = 1$

So the inequality becomes:

$1 < 1$

This statement is false. Since $1$ is not strictly less than $1$ (it is equal), the point $(-2, 1)$ is not a solution to the linear inequality $y < 0.5x + 2$. This highlights the importance of the strict inequality sign; if it were $\leq$, this point would have been a solution. The point lies on the boundary line, but the inequality excludes points on the boundary.

Evaluating Point $(-1, -2)$

Let's test the point $(-1, -2)$. Here, $x = -1$ and $y = -2$. Substitute these values into the inequality:

$-2 < 0.5(-1) + 2$

Calculate the right side:

$0.5(-1) = -0.5$

Now, add 2:

$-0.5 + 2 = 1.5$

So the inequality becomes:

$-2 < 1.5$

This statement is true. Since $-2$ is indeed less than $1.5$, the point $(-1, -2)$ is a solution to the linear inequality $y < 0.5x + 2$. This confirms that this point lies in the solution region defined by the inequality.

Evaluating Point $(-1, 2)$

Now, let's consider the point $(-1, 2)$. Here, $x = -1$ and $y = 2$. Substitute these into the inequality:

$2 < 0.5(-1) + 2$

Calculate the right side:

$0.5(-1) = -0.5$

Now, add 2:

$-0.5 + 2 = 1.5$

So the inequality becomes:

$2 < 1.5$

This statement is false. Since $2$ is not less than $1.5$, the point $(-1, 2)$ is not a solution to the linear inequality $y < 0.5x + 2$. This point lies above the boundary line and does not satisfy the condition.

Evaluating Point $(1, -2)$

Finally, we check the point $(1, -2)$. Here, $x = 1$ and $y = -2$. Substitute these values into the inequality:

$-2 < 0.5(1) + 2$

Calculate the right side:

$0.5(1) = 0.5$

Now, add 2:

$0.5 + 2 = 2.5$

So the inequality becomes:

$-2 < 2.5$

This statement is true. Since $-2$ is indeed less than $2.5$, the point $(1, -2)$ is a solution to the linear inequality $y < 0.5x + 2$. This point also lies within the solution region.

Conclusion: Identifying the Three Solutions

After meticulously testing each point by substituting its coordinates into the linear inequality $y < 0.5x + 2$, we have identified the points that satisfy the condition. The process involves simple arithmetic but requires careful attention to detail, especially with the strict inequality sign. The points that resulted in a true statement are the solutions.

• For $(-3, -2)$, we found $-2 < 0.5$, which is true. So, $(-3, -2)$ is a solution.
• For $(-2, 1)$, we found $1 < 1$, which is false. So, $(-2, 1)$ is not a solution.
• For $(-1, -2)$, we found $-2 < 1.5$, which is true. So, $(-1, -2)$ is a solution.
• For $(-1, 2)$, we found $2 < 1.5$, which is false. So, $(-1, 2)$ is not a solution.
• For $(1, -2)$, we found $-2 < 2.5$, which is true. So, $(1, -2)$ is a solution.

Therefore, the three points that are solutions to the linear inequality $y < 0.5x + 2$ are $(-3, -2)$, $(-1, -2)$, and $(1, -2)$. These points lie in the region below the line $y = 0.5x + 2$ on the coordinate plane, representing all the possible pairs of $(x, y)$ that satisfy the given inequality.

For further exploration into linear inequalities and graphing, you can visit resources like Khan Academy's section on linear inequalities, which offers comprehensive explanations and practice problems.