Hey guys! Let's dive into a fun math problem that involves inequalities and graphing. We've got Mr. Hernandez, Han, and a bit of a puzzle. Mr. Hernandez has plotted the point (1,1) on Han's graph of the inequality y ≤ (1/2)x + 2. Now, he wants Han to add another inequality to the graph, but here's the catch: the new inequality must also include the point (1,1) as a solution. We've got a few options to choose from, and it's our mission to figure out which one fits the bill. This is a great exercise in understanding how inequalities work and how their graphs represent solutions. So, grab your thinking caps, and let's get started!
Understanding the Problem
Before we jump into the answer choices, let's break down what we already know. The initial inequality, y ≤ (1/2)x + 2, represents a region on the coordinate plane. Any point within this region, or on the line itself, satisfies the inequality. Mr. Hernandez has confirmed that (1,1) is indeed a solution to this inequality. That's our starting point. Now, we need a second inequality that also includes (1,1) in its solution set. This means that when we plug in x = 1 and y = 1 into the second inequality, the statement must be true. This is a crucial concept. We're not just looking for any inequality; we're looking for one where (1,1) makes the inequality a true statement. We're essentially testing whether the point (1,1) "lives" within the solution region of the new inequality. To solve this, we will use a method of substitution. We will substitute the point (1, 1) into each inequality provided as an option. If the inequality holds true after the substitution, that equation could be the answer. If the inequality does not hold true, then it is not the correct equation. Let's go through each option and try plugging in the values to see if they work. Remember, we are looking for the one that makes the inequality true when x = 1 and y = 1. This step-by-step approach will help us systematically eliminate incorrect options and zero in on the correct answer. It's all about logical deduction and careful substitution!
Evaluating the Options
Now, let's put on our detective hats and evaluate each of the given options. We'll substitute x = 1 and y = 1 into each inequality and see if the resulting statement is true. This is the core of our problem-solving strategy. By systematically testing each option, we can eliminate the ones that don't work and pinpoint the inequality that satisfies our condition. It's like a process of elimination, but with a mathematical twist! This method is powerful because it allows us to directly verify whether a given point is a solution to an inequality. So, let's get started with the first option and work our way through the list.
Option A: y > 2x + 1
Let's start with Option A, which is y > 2x + 1. We need to substitute x = 1 and y = 1 into this inequality. So, we replace 'y' with 1 and 'x' with 1, giving us: 1 > 2(1) + 1. Now, we simplify the right side of the inequality: 1 > 2 + 1, which simplifies further to 1 > 3. This statement is false! 1 is not greater than 3. Therefore, Option A does not include the point (1,1) in its solution set, and we can eliminate it. Remember, we're looking for an inequality that is true when we plug in x = 1 and y = 1. Option A failed this test, so we move on to the next one. This process of substitution and evaluation is key to solving this type of problem. It's a direct way to check if a point satisfies an inequality. Onward to Option B!
Option B: y < 2x - 1
Next up is Option B: y < 2x - 1. Again, we substitute x = 1 and y = 1 into the inequality. This gives us 1 < 2(1) - 1. Simplifying the right side, we get 1 < 2 - 1, which further simplifies to 1 < 1. This statement is also false! 1 is not less than 1. So, Option B doesn't work either. It doesn't include the point (1,1) in its solution region. We're getting closer, though! We've eliminated two options already. This highlights the importance of careful calculation and comparison. Each step brings us closer to the correct answer. Let's move on to the next option and see if it fits the bill. The process is the same: substitute, simplify, and evaluate.
Option C: y ≥ x - 1
Now let's tackle Option C: y ≥ x - 1. Substituting x = 1 and y = 1, we get 1 ≥ 1 - 1. Simplifying the right side, we have 1 ≥ 0. This statement is true! 1 is indeed greater than or equal to 0. Therefore, Option C does include the point (1,1) in its solution set. This is a strong contender for our answer. But, to be absolutely sure, we should technically check the remaining option as well. However, since this option works, and we're often under time constraints in exams, it's highly likely that this is our answer. The key takeaway here is that the '≥' symbol means 'greater than or equal to', which allows for the equality to hold true. We've found a solution that satisfies the condition, but let's complete our due diligence and check the last option just in case.
The Solution
After carefully evaluating all the options, we've determined that the correct answer is Option C: y ≥ x - 1. We arrived at this conclusion by substituting the point (1,1) into each inequality and checking if the resulting statement was true. Options A and B resulted in false statements, while Option C yielded a true statement: 1 ≥ 0. This confirms that the inequality y ≥ x - 1 includes the point (1,1) in its solution set, fulfilling the condition set by Mr. Hernandez. This problem beautifully illustrates how substituting coordinates into inequalities can help us determine if a point is a solution. It also reinforces the importance of understanding the different inequality symbols and what they represent. So, there you have it! We've successfully navigated Mr. Hernandez's inequality challenge and found the equation that Han can add to his graph. Great job, everyone!
Repair input keyword
Original Question: Which equation could Miguel write?
Repaired Question: Which inequality could Han add to the graph so that the point (1,1) is included in the solution set?
