Ordered Pairs In Linear Inequality Solution Sets

When we talk about systems of linear inequalities, we're essentially looking for a region on a graph where all the inequalities in the system are true simultaneously. This is different from solving a single inequality, where the solution is a region. In a system, we're interested in specific points, or ordered pairs, that fall within that common solution region. Today, we're going to dive deep into how to determine if a given ordered pair is part of the solution set for a system of linear inequalities, using the example:

$$\begin{array}{l} y \geq-\frac{1}{3} x+2 \\ y\ \textless \ 2 x+3 \end{array}$$

To identify which ordered pairs satisfy this system, we need to test each pair against both inequalities. An ordered pair $(x, y)$ is in the solution set only if it makes both inequalities true. If it fails even one, it's out! We'll be examining two sets of potential solutions: A. (2,2), (3,1), (4,2) and B. (2,2), (3,-1), (4,1). Let's break down the process step-by-step to ensure we leave no stone unturned in finding the correct solutions.

Understanding Systems of Linear Inequalities

A system of linear inequalities is a collection of two or more linear inequalities that are considered together. Each inequality in the system defines a half-plane on the coordinate plane. The solution set of the system is the intersection of all these half-planes. Graphically, this intersection is often a region, but when we're asked about specific ordered pairs, we're checking if those individual points lie within this common region. The inequalities we're working with are $y \geq-\frac{1}{3} x+2$ and $y\ \textless \ 2 x+3$. The first inequality, $y \geq-\frac{1}{3} x+2$, includes all points on or above the line $y = -\frac{1}{3} x+2$. The second inequality, $y\ \textless \ 2 x+3$, includes all points strictly below the line $y = 2 x+3$. Finding the ordered pairs that satisfy both means finding points that are simultaneously on or above the first line and below the second line. This graphical interpretation is crucial for understanding why we test the ordered pairs the way we do. Each ordered pair represents a specific coordinate $(x, y)$ that we substitute into the inequalities. If the substitution results in a true statement for both inequalities, then that ordered pair is a valid solution. This process is fundamental to understanding how algebraic conditions translate into geometric locations on a graph. The elegance of this method lies in its simplicity and directness: plug in the values and check the truthfulness of the statements. It’s like a key fitting into multiple locks; only the correct key (the ordered pair) will open all the doors (satisfy all the inequalities).

Testing Ordered Pairs for the First Inequality

Let's begin by testing the ordered pairs against the first inequality: $y \geq-\frac{1}{3} x+2$. Remember, an ordered pair $(x, y)$ satisfies this inequality if the y-coordinate is greater than or equal to the value of $-\frac{1}{3} x+2$ for the given x-coordinate. We'll go through each pair in option A and then option B.

Option A: (2,2), (3,1), (4,2)

• For (2,2): Substitute $x=2$ and $y=2$ into the inequality. $2 \geq-\frac{1}{3}(2)+2$ $2 \geq-\frac{2}{3}+2$ $2 \geq\frac{-2+6}{3}$ $2 \geq\frac{4}{3}$ This statement is true. So, (2,2) satisfies the first inequality.

• For (3,1): Substitute $x=3$ and $y=1$ into the inequality. $1 \geq-\frac{1}{3}(3)+2$ $1 \geq-1+2$ $1 \geq 1$ This statement is true. So, (3,1) satisfies the first inequality.

• For (4,2): Substitute $x=4$ and $y=2$ into the inequality. $2 \geq-\frac{1}{3}(4)+2$ $2 \geq-\frac{4}{3}+2$ $2 \geq\frac{-4+6}{3}$ $2 \geq\frac{2}{3}$ This statement is true. So, (4,2) satisfies the first inequality.

All ordered pairs in Option A satisfy the first inequality. This is a good start, but they must also satisfy the second inequality to be part of the solution set.

Option B: (2,2), (3,-1), (4,1)

• For (2,2): We already tested this pair and found it to be true for the first inequality.

• For (3,-1): Substitute $x=3$ and $y=-1$ into the inequality. $-1 \geq-\frac{1}{3}(3)+2$ $-1 \geq-1+2$ $-1 \geq 1$ This statement is false. So, (3,-1) does not satisfy the first inequality. This means Option B cannot be the correct solution set, as one of its pairs fails the first test.

• For (4,1): Substitute $x=4$ and $y=1$ into the inequality. $1 \geq-\frac{1}{3}(4)+2$ $1 \geq-\frac{4}{3}+2$ $1 \geq\frac{-4+6}{3}$ $1 \geq\frac{2}{3}$ This statement is true. So, (4,1) satisfies the first inequality.

From this first round of testing, we can already see that Option B has a pair (3,-1) that doesn't work. Therefore, we can tentatively rule out Option B and focus on Option A. However, we must still confirm that all pairs in Option A also satisfy the second inequality.

Testing Ordered Pairs for the Second Inequality

Now, let's test the ordered pairs from Option A against the second inequality: $y\ \textless \ 2 x+3$. An ordered pair $(x, y)$ satisfies this inequality if the y-coordinate is strictly less than the value of $2 x+3$ for the given x-coordinate.

Option A: (2,2), (3,1), (4,2)

• For (2,2): Substitute $x=2$ and $y=2$ into the inequality. $2\ \textless \ 2(2)+3$ $2\ \textless \ 4+3$ $2\ \textless \ 7$ This statement is true. So, (2,2) satisfies the second inequality.

• For (3,1): Substitute $x=3$ and $y=1$ into the inequality. $1\ \textless \ 2(3)+3$ $1\ \textless \ 6+3$ $1\ \textless \ 9$ This statement is true. So, (3,1) satisfies the second inequality.

• For (4,2): Substitute $x=4$ and $y=2$ into the inequality. $2\ \textless \ 2(4)+3$ $2\ \textless \ 8+3$ $2\ \textless \ 11$ This statement is true. So, (4,2) satisfies the second inequality.

Since all ordered pairs in Option A, namely (2,2), (3,1), and (4,2), have satisfied both the first inequality ($y \geq-\frac{1}{3} x+2$) and the second inequality ($y\ \textless \ 2 x+3$), Option A represents the correct solution set.

Conclusion: Identifying the Solution Set

To reiterate, for an ordered pair to be in the solution set of a system of linear inequalities, it must satisfy every single inequality within that system. We systematically tested each ordered pair from both options against both inequalities. Option A's pairs (2,2), (3,1), and (4,2) all proved to be true for $y \geq-\frac{1}{3} x+2$. They also all proved to be true for $y\ \textless \ 2 x+3$. Therefore, A. (2,2), (3,1), (4,2) is the correct solution set. Option B contained the ordered pair (3,-1), which failed the first inequality, immediately disqualifying the entire set.

This methodical approach ensures accuracy. It's not just about guessing or looking at a graph (though graphing can be a helpful visual tool!); it's about rigorous algebraic verification. Each ordered pair is a potential candidate, and we put each candidate through the necessary tests. The ones that pass all tests are the members of the solution set. This process is fundamental in various areas of mathematics, including linear programming and optimization problems, where finding feasible regions and specific points within them is critical.

If you're interested in learning more about graphing linear inequalities and systems of inequalities, exploring resources on Khan Academy can provide excellent visual explanations and practice problems.