Problem 22
Question
Graph each system. $$ \left\\{\begin{array}{r} 3 x-4 y \leq 12 \\ x^{2}+y^{2}<16 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
Shade the region below the line and within the dashed circle.
1Step 1: Graph the Linear Inequality
To graph the inequality \(3x - 4y \leq 12\), first convert it to equality: \(3x - 4y = 12\). This represents a line. Find two points on this line by substituting values for \(x\) and solving for \(y\). For example, let \(x = 0\), then \(3(0) - 4y = 12\), so \(y = -3\). If \(x = 4\), then \(3(4) - 4y = 12\) results in \(y = 0\). Plot these points and draw the line. Shade below the line for the inequality \(\leq\).
2Step 2: Graph the Circle Equation
The inequality \(x^2 + y^2 < 16\) represents a circle centered at the origin \((0,0)\) with a radius of 4 (since \(\sqrt{16} = 4\)). Graph the circle by plotting points that satisfy \(x^2 + y^2 = 16\). For example, points like \((4, 0), (0, 4), (-4, 0), \) and \((0, -4)\) lie on the circle. Use a dashed circle since the inequality is strict (< rather than \(\leq\)).
3Step 3: Determine the Area of Overlap
The solution to the system is the region where the shaded area of the linear inequality overlaps with the inside of the circle; the solution must satisfy both inequalities. Thus, identify and shade the region that both lies under the line and inside the dashed circle.
Key Concepts
Graphing Linear InequalitiesGraphing a CircleInequality Solution Regions
Graphing Linear Inequalities
When solving a system of inequalities, it's essential to grasp how to graph linear inequalities. Start by converting the inequality into an equation. In our example, we transformed the inequality \(3x - 4y \leq 12\) into the equation \(3x - 4y = 12\). This equation represents a straight line.
To graph the line, find at least two points that lie on it. Choose easy values for \(x\). For instance, if \(x = 0\), then \(y = -3\), giving us the point \((0, -3)\). Alternatively, if \(x = 4\), then \(y = 0\), with the point \((4, 0)\). Plot these points on a graph and draw the line through them.
Since the inequality is \(\leq\), shade the area below this line. This shading indicates all the solutions for the inequality, meaning any \((x, y)\) pair in the shaded region satisfies \(3x - 4y \leq 12\). Remember that the line itself is also part of the solution set, as indicated by the \(\leq\) sign.
To graph the line, find at least two points that lie on it. Choose easy values for \(x\). For instance, if \(x = 0\), then \(y = -3\), giving us the point \((0, -3)\). Alternatively, if \(x = 4\), then \(y = 0\), with the point \((4, 0)\). Plot these points on a graph and draw the line through them.
Since the inequality is \(\leq\), shade the area below this line. This shading indicates all the solutions for the inequality, meaning any \((x, y)\) pair in the shaded region satisfies \(3x - 4y \leq 12\). Remember that the line itself is also part of the solution set, as indicated by the \(\leq\) sign.
Graphing a Circle
Understanding how to graph a circle is crucial when dealing with inequalities like \(x^2 + y^2 < 16\). This inequality describes a circle centered at the origin \((0, 0)\), with a radius of 4. The formula \(x^2 + y^2 = r^2\) gives us the circle's equation, where \(r\) is the radius. In our case, since \(r = \sqrt{16}\), the radius is 4.
To graph this, you plot several key points where \(x^2 + y^2 = 16\) is true, such as \((4, 0), (0, 4), (-4, 0),\) and \((0, -4)\). These points form the circular outline.
Since the inequality is strict \(<\), we use a dashed line to show the circle’s boundary. Within this dashed line is the solution area, implying that solutions lie inside the circle but not on the boundary itself.
To graph this, you plot several key points where \(x^2 + y^2 = 16\) is true, such as \((4, 0), (0, 4), (-4, 0),\) and \((0, -4)\). These points form the circular outline.
Since the inequality is strict \(<\), we use a dashed line to show the circle’s boundary. Within this dashed line is the solution area, implying that solutions lie inside the circle but not on the boundary itself.
Inequality Solution Regions
Finding the solution region for a system of inequalities means identifying where the individual solution areas overlap. Consider our two inequalities: \(3x - 4y \leq 12\) and \(x^2 + y^2 < 16\). After plotting both on the same graph, we observe their separate shaded areas.
The overlap is where all conditions of the system are met. This solution region is part of the plane where the shading from the linear inequality intersects with the interior region of the circle.
It is essential to only shade the section that satisfies both inequalities. This involves combining both concepts—being below the dashed circle and beneath the line—which together define the system's true solution in the graph.
The overlap is where all conditions of the system are met. This solution region is part of the plane where the shading from the linear inequality intersects with the interior region of the circle.
It is essential to only shade the section that satisfies both inequalities. This involves combining both concepts—being below the dashed circle and beneath the line—which together define the system's true solution in the graph.
Other exercises in this chapter
Problem 21
Solve each nonlinear system of equations. $$ \left\\{\begin{array}{r} x^{2}+y^{2}=1 \\ x^{2}+(y+3)^{2}=4 \end{array}\right. $$
View solution Problem 22
Graph each equation. $$ \frac{x^{2}}{36}=1-y^{2} $$
View solution Problem 22
Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} x^{2}+2 y^{2}=4 \\ x^{2}-y^{2}=4 \end{array}\right. $$
View solution Problem 23
Graph each equation. $$ 4(x-1)^{2}+9(y+2)^{2}=36 $$
View solution