Chapter 12: Problem 16
Graph each equation of the system. Then solve the system to find the points of intersection. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=10 \\ y &=x+2 \end{aligned}\right. $$
Short Answer
Expert verified
The points of intersection are \((-3, -1)\) and \((1, 3)\).
Step by step solution
01
Understand the Equations
There are two equations in the system. The first equation is a circle: \(x^2 + y^2 = 10\). The second equation is a straight line: \(y = x + 2\).
02
Graph the Circle
The equation \(x^2 + y^2 = 10\) represents a circle with its center at (0,0) and a radius of \(\sqrt{10}\). Plot the circle on the graph.
03
Graph the Line
The equation \(y = x + 2\) is a straight line with a slope of 1 and a y-intercept at 2. Plot the line on the same graph.
04
Find the Points of Intersection
To find the points of intersection, solve the system algebraically. Substitute \(y = x + 2\) into \(x^2 + y^2 = 10\): \[ x^2 + (x + 2)^2 = 10 \] Solve for \(x\): \[ x^2 + x^2 + 4x + 4 = 10 \] Simplify: \[ 2x^2 + 4x + 4 = 10 \] \[ 2x^2 + 4x - 6 = 0 \] \[ x^2 + 2x - 3 = 0 \] Factor the quadratic equation: \[ (x + 3)(x - 1) = 0 \] Solve for \(x\): \[ x = -3 \] and \[ x = 1 \]. Plug these \(x\) values back into the line equation to get \(y\): \( y = -3 + 2 = -1 \) \( y = 1 + 2 = 3 \). Thus, the points of intersection are \((-3, -1)\) and \((1, 3)\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Equations
Graphing equations is a visual method of finding the solutions to a system of equations. Start by understanding the type of equations you're working with. In this example, we have a circle and a line.
For the circle, the equation is given by \(x^2 + y^2 = 10\). This indicates all points that are at a constant distance (radius) from the center. Here, the center is at (0,0), with a radius of \(\sqrt{10}\). Drawing this circle involves placing the center at the origin and drawing all points that are at a distance of \(\sqrt{10}\).
For the line, the equation is \(y = x + 2\). This is a straight line with a slope of 1 and a y-intercept at 2. To graph this line, you start at the y-intercept (0, 2) and use the slope to determine the steepness and direction. By plotting these on the same graph, you can visually see where the circle and line intersect.
For the circle, the equation is given by \(x^2 + y^2 = 10\). This indicates all points that are at a constant distance (radius) from the center. Here, the center is at (0,0), with a radius of \(\sqrt{10}\). Drawing this circle involves placing the center at the origin and drawing all points that are at a distance of \(\sqrt{10}\).
For the line, the equation is \(y = x + 2\). This is a straight line with a slope of 1 and a y-intercept at 2. To graph this line, you start at the y-intercept (0, 2) and use the slope to determine the steepness and direction. By plotting these on the same graph, you can visually see where the circle and line intersect.
Algebraic Substitution
Algebraic substitution is a method of solving systems of equations. It involves replacing one variable with an equivalent expression from another equation. This reduces the system to a single equation with one variable, making it easier to solve.
In our example, the second equation is \(y = x + 2\). We substitute this expression for \(y\) in the first equation \(x^2 + y^2 = 10\). After substitution, the equation becomes \[x^2 + (x + 2)^2 = 10\].
Solving this involves squaring \(x + 2\), combining like terms, and simplifying the equation to find the values of \(x\). These solutions for \(x\) can then be used to find corresponding \(y\) values by substituting back into the linear equation.
In our example, the second equation is \(y = x + 2\). We substitute this expression for \(y\) in the first equation \(x^2 + y^2 = 10\). After substitution, the equation becomes \[x^2 + (x + 2)^2 = 10\].
Solving this involves squaring \(x + 2\), combining like terms, and simplifying the equation to find the values of \(x\). These solutions for \(x\) can then be used to find corresponding \(y\) values by substituting back into the linear equation.
Quadratic Equations
Quadratic equations, commonly written in the form \(ax^2 + bx + c = 0\), can be solved using various methods like factoring, completing the square, or using the quadratic formula.
In our exercise, after substituting and simplifying, we arrive at the quadratic equation: \(x^2 + 2x - 3\). To solve this, we factor the quadratic expression into \((x + 3)(x - 1) = 0\).
By setting each factor to zero, we get the solutions \(x = -3\) and \(x = 1\). These are the \(x\)-values where the circle and the line intersect.
In our exercise, after substituting and simplifying, we arrive at the quadratic equation: \(x^2 + 2x - 3\). To solve this, we factor the quadratic expression into \((x + 3)(x - 1) = 0\).
By setting each factor to zero, we get the solutions \(x = -3\) and \(x = 1\). These are the \(x\)-values where the circle and the line intersect.
Points of Intersection
The points of intersection between two graphs represent the solutions to the system of equations. These are the points where both equations hold true simultaneously.
In our example, the solutions found from the quadratic equation are \(x = -3\) and \(x = 1\). To find the corresponding \(y\)-values, substitute these back into the linear equation \(y = x + 2\).
For \(x = -3\), \(y = -3 + 2 = -1\). For \(x = 1\), \(y = 1 + 2 = 3\). Hence, the points of intersection are \((-3, -1)\) and \((1, 3)\). These points indicate where the circle and line cross each other on the graph.
In our example, the solutions found from the quadratic equation are \(x = -3\) and \(x = 1\). To find the corresponding \(y\)-values, substitute these back into the linear equation \(y = x + 2\).
For \(x = -3\), \(y = -3 + 2 = -1\). For \(x = 1\), \(y = 1 + 2 = 3\). Hence, the points of intersection are \((-3, -1)\) and \((1, 3)\). These points indicate where the circle and line cross each other on the graph.