Mastering Graphing: $x+3y > -3$ & $y < \frac{1}{2}x+1$

Understanding Systems of Inequalities

Systems of inequalities might sound a bit intimidating at first, but don't worry, we're going to break it down together! Imagine you have a few rules or conditions, and you need to find all the possible scenarios that fit all those rules simultaneously. That's essentially what a system of inequalities is all about in mathematics. Instead of finding a single point that works (like in a system of equations), we're looking for an entire region on a graph where all the conditions are met. This solution region represents every single point $(x,y)$ that makes every inequality in the system true. It's a powerful tool used in many real-world situations, from figuring out the best budget for a project to optimizing production schedules or even planning a healthy diet within certain nutritional constraints. Think of it as finding the 'sweet spot' that satisfies multiple demands at once. Our goal today is to tackle a specific system: $x+3y > -3$ and $y < \frac{1}{2}x + 1$. We'll learn how to graph inequalities individually and then combine them to pinpoint that magical solution region. By the end of this article, you'll feel much more confident in visualizing mathematical conditions and understanding what they represent. So, grab your imaginary graph paper and let's dive into the fascinating world of linear inequalities and their graphical representations. It's more intuitive than you might think!

Graphing the First Inequality: $x+3y > -3$

Let's kick things off by graphing the first inequality: $x+3y > -3$. The first step to graphing any inequality is to pretend it's an equation for a moment. This helps us draw the boundary line that separates the points that satisfy the inequality from those that don't. So, we'll convert $x+3y > -3$ into its boundary line equation: $x+3y = -3$. Now, how do we draw this line? There are a couple of popular ways. One easy method is to find the x- and y-intercepts. To find the x-intercept, we set $y=0$: $x + 3(0) = -3$, which simplifies to $x = -3$. So, our x-intercept is $(-3, 0)$. To find the y-intercept, we set $x=0$: $0 + 3y = -3$, which gives us $3y = -3$, meaning $y = -1$. Our y-intercept is $(0, -1)$. Now that we have two points, $(-3,0)$ and $(0,-1)$, we can draw our line.

Here's a crucial detail: look at the inequality symbol. Since it's ' $>$ ' (greater than) and not ' $\ge$ ' (greater than or equal to), the points on the line $x+3y = -3$ are not part of our solution. This means we need to draw a dashed line. A dashed line tells us, "Hey, these points on the line itself aren't included!" If it were $\ge$ or $\le$, we'd draw a solid line. Once the dashed line is drawn, we need to figure out which side of the line to shade. This shaded area will be the solution region for this single inequality. The easiest way to do this is to pick a test point that is not on the line. The origin $(0,0)$ is usually the simplest choice, unless the line passes through it. Let's test $(0,0)$ in our original inequality: $0 + 3(0) > -3$. This simplifies to $0 > -3$, which is true! Since the test point $(0,0)$ makes the inequality true, it means all the points on the same side of the line as $(0,0)$ are solutions. So, you would shade the region that contains the origin. If our test point had resulted in a false statement, we would shade the opposite side. By following these steps, you've successfully identified the entire region of the coordinate plane where $x+3y$ is indeed greater than $-3$. Pretty neat, right?

Graphing the Second Inequality: $y < \frac{1}{2}x + 1$

Now that we've mastered the first one, let's move on to graphing the second inequality: $y < \frac{1}{2}x + 1$. Just like before, our initial move is to determine the boundary line by converting the inequality into an equation. So, for $y < \frac{1}{2}x + 1$, our boundary line equation is $y = \frac{1}{2}x + 1$. This form is super convenient because it's already in slope-intercept form ($y = mx + b$), where 'm' is the slope and 'b' is the y-intercept. From $y = \frac{1}{2}x + 1$, we can immediately see that the y-intercept is $(0, 1)$. This gives us a great starting point on our graph! The slope 'm' is $\frac{1}{2}$. Remember, slope is