Solving Systems Of Linear Inequalities

When you're tackling problems involving systems of linear inequalities, like the one presented with $\begin{array}{l} y \leq 2 x-5 \\ y>-3 x+1 \end{array}$, the goal is to find the region on a graph where all the inequalities are true simultaneously. Think of it like finding a sweet spot on a map where multiple conditions are met. The first inequality, $y \leq 2x-5$, tells us we're looking for all the points on or below the line represented by $y = 2x-5$. This line itself is a boundary, and the "less than or equal to" sign means the points directly on the line are included in the solution. To visualize this, you'd first graph the line $y = 2x-5$. It has a y-intercept of -5 (where it crosses the y-axis) and a slope of 2, meaning for every 1 unit you move to the right on the x-axis, you move 2 units up on the y-axis. Once the line is drawn, you'd shade the region below it. Now, let's consider the second inequality, $y > -3x+1$. This one is a bit different because it uses a "greater than" sign. This means we are interested in all the points above the line $y = -3x+1$. Importantly, the points on this line are not included in the solution because it's strictly "greater than." So, when we graph $y = -3x+1$, we'd draw it with a dashed line to signify that the line itself is not part of the solution set. This line has a y-intercept of 1 and a slope of -3, meaning for every 1 unit you move right on the x-axis, you move 3 units down on the y-axis. After drawing this dashed line, you'd shade the region above it. The real solution to the system of inequalities is the area where these two shaded regions overlap. This overlapping region represents all the coordinate pairs $(x, y)$ that satisfy both $y \leq 2x-5$ AND $y > -3x+1$ at the same time. When you're asked to identify the correct graph showing this solution, you're essentially looking for a graph that displays these two boundary lines, correctly indicates whether each line is solid or dashed, and has the shading in the appropriate regions that overlap. The intersection point of the two lines is also a critical feature to observe, as it often marks a vertex of the solution region. Understanding these components—boundary lines, shading, and line types (solid vs. dashed)—is fundamental to accurately representing and interpreting solutions to systems of linear inequalities.

Understanding the Boundaries and Shading

To truly master solving systems of linear inequalities and identifying their graphical representation, a deep understanding of how each inequality dictates the boundary and shading is paramount. Let's break down the inequality $y \leq 2x-5$. The core of this is the line $y = 2x-5$. This is a linear equation, and its graph is a straight line. The slope-intercept form, $y = mx+b$, is incredibly useful here, where $m$ is the slope and $b$ is the y-intercept. In $y = 2x-5$, the slope $m$ is 2, and the y-intercept $b$ is -5. This means the line crosses the y-axis at the point (0, -5). The slope of 2 tells us that for every unit increase in $x$, $y$ increases by 2 units. Graphically, this means you can find another point by moving one unit to the right and two units up from any point on the line. Since the inequality is $y \leq 2x-5$ (less than or equal to), the boundary line $y = 2x-5$ is included in the solution. This is represented on a graph by drawing a solid line. The "less than" part ($y \leq$) means we are interested in all the points whose y-coordinates are less than or equal to the y-coordinates on the line for a given x. On a standard Cartesian plane, this corresponds to the region below the line. So, for $y \leq 2x-5$, we draw a solid line for $y = 2x-5$ and shade the entire area underneath it. Now, let's move to the second inequality: $y > -3x+1$. Again, the foundation is the line $y = -3x+1$. Here, the slope $m$ is -3, and the y-intercept $b$ is 1. The line crosses the y-axis at (0, 1). The slope of -3 means that for every unit increase in $x$, $y$ decreases by 3 units. You can find other points by moving one unit to the right and three units down from any point on the line. Because the inequality is $y > -3x+1$ (strictly greater than), the boundary line $y = -3x+1$ is not included in the solution. This is indicated on a graph by drawing a dashed line. The "greater than" part ($y >$) means we are interested in all the points whose y-coordinates are greater than the y-coordinates on the line for a given x. In the Cartesian plane, this translates to the region above the line. Therefore, for $y > -3x+1$, we draw a dashed line for $y = -3x+1$ and shade the entire area above it. The final solution to the system of inequalities is the region where these two shaded areas overlap. This overlap is the set of all points $(x, y)$ that satisfy both conditions simultaneously. When evaluating graphs for this system, you must check that the boundary lines are drawn correctly (solid for $\leq$ or $\geq$, dashed for $<$ or $>$), that they are in the correct positions based on their equations, and that the shading accurately represents the "less than" or "greater than" conditions, with the final solution being the intersection of these shaded regions.

Identifying the Correct Graph

When presented with multiple-choice options for a system of linear inequalities like $\begin{array}{l} y \leq 2 x-5 \\ y>-3 x+1 \end{array}$, the key is to systematically eliminate incorrect graphs by checking the characteristics of each inequality. The first step is always to analyze the boundary lines. For $y \leq 2x-5$, the boundary line is $y = 2x-5$. You need to verify that the graph shows a solid line passing through the point (0, -5) with a slope of +2. If a graph shows a dashed line for this inequality, or if the line doesn't match the equation, it's incorrect. Similarly, for $y > -3x+1$, the boundary line is $y = -3x+1$. The graph must display a dashed line passing through the point (0, 1) with a slope of -3. Any graph that uses a solid line here, or has the wrong intercept or slope, can be immediately discarded. After confirming the boundary lines, the next crucial step is to examine the shading. For $y \leq 2x-5$, the shading must be below the solid line $y = 2x-5$. This indicates that y-values are less than or equal to those on the line. If the shading is above this line, the graph is incorrect. For $y > -3x+1$, the shading must be above the dashed line $y = -3x+1$. This indicates that y-values are greater than those on the line. If the shading is below this line, it’s also incorrect. The ultimate solution to the system is the region where the shading from both inequalities overlaps. Therefore, the correct graph will show the region that is simultaneously below the solid line $y = 2x-5$ AND above the dashed line $y = -3x+1$. It's often helpful to pick a test point within the potential solution region and plug its coordinates into both original inequalities. For example, if you identify a region as the potential solution, pick a point within it (say, (3, -2)). Substitute these values into the inequalities: Is $-2 \leq 2(3)-5$? This simplifies to $-2 \leq 6-5$, or $-2 \leq 1$, which is true. Now check the second inequality: Is $-2 > -3(3)+1$? This simplifies to $-2 > -9+1$, or $-2 > -8$, which is also true. Since the test point satisfies both inequalities, the region containing this point is likely the correct solution. Conversely, if a point does not satisfy even one of the inequalities, it cannot be in the solution set. By carefully checking the line types, intercepts, slopes, and shading for each inequality, and then confirming the overlap, you can confidently identify the correct graphical solution to the system.

Visualizing the Intersection of Solution Regions

To solidify your understanding of solving systems of linear inequalities, let's visualize the process of finding the intersection of the solution regions for $\begin{array}{l} y \leq 2 x-5 \\ y>-3 x+1 \end{array}$. We've established that the first inequality, $y \leq 2x-5$, requires us to consider the solid line $y = 2x-5$ and shade everything on or below it. Imagine this as painting a territory on a map. The second inequality, $y > -3x+1$, demands we focus on the dashed line $y = -3x+1$ and shade everything strictly above it. This is like painting a second, different territory. The solution to the system is where these two painted territories overlap. This overlapping area is the graphical representation of all points $(x, y)$ that simultaneously satisfy both conditions. Let's think about the characteristics of this overlap. The boundary of the solution region will be formed by segments of the two lines we've graphed. Specifically, the upper boundary of the solution region will be part of the dashed line $y = -3x+1$ (since we need points below $y=2x-5$ and above $y=-3x+1$), and the lower boundary will be part of the solid line $y = 2x-5$. The point where these two lines intersect is a crucial feature. To find this intersection point algebraically, we can set the expressions for $y$ equal to each other: $2x - 5 = -3x + 1$. Adding $3x$ to both sides gives $5x - 5 = 1$. Adding 5 to both sides yields $5x = 6$. Dividing by 5, we get $x = \frac{6}{5}$. Now, we can substitute this value of $x$ back into either of the original line equations to find the corresponding $y$-coordinate. Using $y = 2x - 5$: $y = 2\left(\frac{6}{5}\right) - 5 = \frac{12}{5} - \frac{25}{5} = -\frac{13}{5}$. So, the intersection point is $\left(\frac{6}{5}, -\frac{13}{5}\right)$ or $(1.2, -2.6)$. This intersection point itself is not part of the solution because the inequality $y > -3x+1$ uses a dashed line, meaning points on that line are excluded. However, the solution region will extend infinitely in the direction indicated by the shading, forming a wedge or an unbounded region. The shape of the solution region is determined by the slopes of the lines. Since the slopes are different (2 and -3), the lines will intersect at exactly one point. If the slopes were the same, the lines would be parallel, leading to either no solution (if the inequalities were contradictory) or infinitely many solutions (if one inequality's region completely contained the other). For this specific system, the solution region is the area that is below the solid line $y=2x-5$ and above the dashed line $y=-3x+1$. When looking at a graph, you're seeking this specific geometric shape, bounded by segments of the two lines and extending outwards. The vertex of this region is the intersection point, and the nature of the boundary lines (solid vs. dashed) dictates whether this vertex is included in the overall solution set.

Conclusion: Mastering Graphical Solutions

In conclusion, accurately graphing the solution to a system of linear inequalities like $\begin{array}{l} y \leq 2 x-5 \\ y>-3 x+1 \end{array}$ hinges on understanding the distinct roles of each inequality. Each inequality defines a half-plane, and the system's solution is the intersection of these half-planes. The boundary line, determined by the equality part of the inequality, dictates the edge of this half-plane. A solid line signifies that the boundary is included in the solution set (for $\leq$ and $\geq$), while a dashed line indicates the boundary is excluded (for $<$ and $>$). The direction of shading—below the line for "less than" and above for "greater than"—is equally crucial. The final solution is the region where all shaded areas overlap. For this particular system, the solution is the area below the solid line $y = 2x-5$ and above the dashed line $y = -3x+1$. The intersection point of the boundary lines, $\left(\frac{6}{5}, -\frac{13}{5}\right)$, serves as a critical reference, marking the vertex of the solution region, although it is not included in the solution due to the strict inequality in the second condition. By systematically analyzing the line type, intercept, slope, and shading for each inequality, and then identifying the overlapping region, you can confidently determine the correct graphical representation of the solution. For further exploration into graphical methods in mathematics, you can refer to resources like Khan Academy for detailed explanations and practice exercises.