Which Ordered Pairs Make Both Inequalities True

Which Ordered Pairs Make Both Inequalities True?

When you’re presented with two inequalities that must hold simultaneously, the challenge is to find all the ordered pairs ((x, y)) that satisfy both conditions at the same time. This is a common problem in algebra and precalculus, and mastering it helps you understand how to work with systems of inequalities, sketch feasible regions, and apply the concepts to real‑world situations such as optimization or feasibility studies.

No fluff here — just what actually works.

Below we walk through the entire process, from interpreting the inequalities to graphing the solution set, and finish with a worked example that ties everything together That alone is useful..

1. Understanding the Problem

Suppose you’re given two inequalities:

1. (2x + 3y \leq 12)
2. (x - y \geq 1)

You’re asked: Which ordered pairs ((x, y)) satisfy both inequalities?

The answer will be a region on the coordinate plane—a set of infinite points that meet both conditions. Our goal is to describe that region precisely, either algebraically or graphically, and list a few sample points that belong to it Simple as that..

2. Rewriting Inequalities in Slope–Intercept Form

The first step is to rewrite each inequality so you can graph it easily.

2.1. Inequality 1

(2x + 3y \leq 12)

Solve for (y):

[ 3y \leq 12 - 2x \quad \Rightarrow \quad y \leq \frac{12 - 2x}{3} = 4 - \frac{2}{3}x ]

So the boundary line is (y = 4 - \frac{2}{3}x), and the inequality tells us to shade below this line (including the line itself because of the “(\leq)” symbol).

2.2. Inequality 2

(x - y \geq 1)

Solve for (y):

[ -x + y \leq -1 \quad \Rightarrow \quad y \geq x - 1 ]

Here the boundary line is (y = x - 1), and the inequality requires shading above this line (including the line because of the “(\geq)” symbol) And that's really what it comes down to. But it adds up..

3. Plotting the Boundary Lines

Tips for plotting:

• For each line, pick two easy (x) values, compute the corresponding (y) values, and plot the points.
• Draw a straight line through the two points.
• Use a solid line because the inequalities are inclusive (≤ or ≥).

4. Determining the Feasible Region

The feasible region is the intersection of the two shaded areas. To find it:

1. Identify the intersection points of the boundary lines.
   Solve the equations simultaneously:

   [ \begin{cases} y = 4 - \frac{2}{3}x \ y = x - 1 \end{cases} ]

   Set them equal:

   [ 4 - \frac{2}{3}x = x - 1 \quad \Rightarrow \quad 4 + 1 = x + \frac{2}{3}x \quad \Rightarrow \quad 5 = \frac{5}{3}x \quad \Rightarrow \quad x = 3 ]

   Plug (x = 3) back into (y = x - 1):

   [ y = 3 - 1 = 2 ]

   So the lines intersect at ((3, 2)).

2. Test a point in one of the shaded areas to see if it also satisfies the other inequality.
   Take this case: choose the point ((0, 0)):

   • Inequality 1: (2(0) + 3(0) = 0 \leq 12) ✔️
   • Inequality 2: (0 - 0 = 0 \geq 1) ❌

   Since ((0,0)) fails the second inequality, the feasible region lies above the line (y = x - 1) but below the line (y = 4 - \frac{2}{3}x) Turns out it matters..

3. Sketch the intersection.
   The region will be a convex polygon (in this case, a triangle) bounded by the two lines and the (x)-axis or (y)-axis if one of the inequalities forces a coordinate to be non‑negative. In our example, the region is the infinite strip between the two lines, extending indefinitely in the direction where both inequalities hold.

5. Describing the Solution Set Algebraically

The solution set can be expressed as:

[ {(x, y) \in \mathbb{R}^2 \mid 4 - \frac{2}{3}x \geq y \geq x - 1} ]

Or, if you prefer inequalities involving only (x) and (y) without fractions:

[ {(x, y) \mid 3y \leq 12 - 2x,; x - y \geq 1} ]

These two conditions are equivalent and capture all ordered pairs that satisfy both inequalities Which is the point..

6. Generating Sample Points

To illustrate, here are five sample ordered pairs that lie inside the feasible region:

Notice that points 3 and 5 fail the first inequality, so they’re not part of the solution set. Only the points that satisfy both conditions remain.

7. Common Pitfalls and How to Avoid Them

8. Extending to More Complex Systems

When you have more than two inequalities, the same principles apply:

1. Rewrite each inequality in slope–intercept form (if linear).
2. Graph all boundary lines.
3. Shade the appropriate side for each inequality.
4. Find the intersection of all shaded regions.
5. Describe the resulting feasible region algebraically or graphically.

For non‑linear inequalities (e.Which means g. , involving (x^2) or (\sqrt{y})), you’ll plot curves instead of straight lines, but the intersection logic remains the same Simple as that..

9. Real‑World Applications

• Resource Allocation: Determining how many units of two products can be produced given constraints on labor and materials.
• Budget Planning: Figuring out feasible spending combinations when there are upper and lower limits on expenses.
• Engineering Design: Ensuring that dimensions meet both safety (upper bound) and cost (lower bound) requirements.

In each scenario, the feasible region represents all viable solutions; picking a specific ordered pair corresponds to a concrete decision.

10. Conclusion

Finding all ordered pairs that satisfy two (or more) inequalities is a matter of:

1. Translating each inequality into a form that’s easy to graph.
2. Visualizing the shaded regions corresponding to each condition.
3. Identifying the intersection of these regions—this intersection is the set of all valid ordered pairs.

By mastering this process, you gain a powerful tool for tackling linear programming problems, optimizing resources, and making informed decisions based on multiple constraints. The key is to approach each step methodically, verify your calculations, and remember that the solution is a region, not just a handful of points.