Problem 26
Question
In Exercises 19-26, solve the system by graphing. $$ \left\\{\begin{array}{l} 2 x+3 y=6 \\ 4 x+6 y=12 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The system of equations has an infinite number of solutions since both equations represent the same line on the graph.
1Step 1: Graph the First Equation
To graph the equation \(2x + 3y = 6\), first, it can be rewritten in the slope-intercept form, \(y = mx + c\), where \(m\) is the slope of the line and \(c\) is the y-intercept. Isolating \(y\), gives \(y = -(2/3)x + 2\). Then, on the x-y plane, plot the y-intercept (\(0, 2\)), and use the slope (\(-2/3\)) to find another point on the line by moving two units down and three units to the right from the y-intercept.
2Step 2: Graph the Second Equation
Next, graph the equation \(4x + 6y = 12\) in the same manner. This time, isolated \(y\) gives \(y = -(4/6)x + 2\), which simplifies to \(y = -(2/3)x + 2\). This is the exact same line as in step 1.
3Step 3: Identify the Solution
Both equations yield the exact same line, meaning that every point on this line is a solution to the system of equations. Therefore, the system of equations has an infinite number of solutions, instead of a single intersecting point.
Key Concepts
System of Linear EquationsSlope-Intercept FormGraphing Linear EquationsInfinite Solutions to System of Equations
System of Linear Equations
When we talk about a system of linear equations, we're referring to a set of two or more linear equations that we're working to solve at the same time. Each equation in the system represents a line when graphed on a coordinate plane. The solution to the system is the point or points where the lines intersect, indicating the values of the variables that satisfy all equations simultaneously.
For example, in the exercise given, we are dealing with a system of two equations:
For example, in the exercise given, we are dealing with a system of two equations:
- \(2x + 3y = 6\)
- \(4x + 6y = 12\)
Slope-Intercept Form
The slope-intercept form is an equation of a line written as \(y = mx + c\), where \(m\) is the slope of the line and \(c\) is the y-intercept, the point where the line crosses the y-axis. This form makes it particularly easy to graph a line, as you can readily identify the steepness and direction of the line (via the slope) and the starting point on the graph (via the y-intercept).
To convert a linear equation to the slope-intercept form, you solve for \(y\). Both equations from the exercise convert to \(y = -(2/3)x + 2\). In this case, the slope \(m\) is \(-2/3\), indicating a downward slope of 2 units for every 3 units moved to the right on the graph, and the y-intercept \(c\) is 2, the point on the graph at \((0, 2)\).
To convert a linear equation to the slope-intercept form, you solve for \(y\). Both equations from the exercise convert to \(y = -(2/3)x + 2\). In this case, the slope \(m\) is \(-2/3\), indicating a downward slope of 2 units for every 3 units moved to the right on the graph, and the y-intercept \(c\) is 2, the point on the graph at \((0, 2)\).
Graphing Linear Equations
To graph linear equations, you start by plotting the y-intercept on the y-axis. From there, you use the slope to determine the direction and steepness of the line, plotting another point to help you draw the line accurately. If the slope is positive, the line rises to the right; if it's negative, it falls to the right.
The slope, represented as a fraction \(\frac{rise}{run}\), gives you a pair of instructions for how to move from the y-intercept to another point on the line. For example, a slope of \(-2/3\) means from the y-intercept you go down 2 units (the rise) and then 3 units to the right (the run). Once at least two points are plotted, a straight line can be drawn through them, extending to the edges of the graph. This extended line represents all possible solutions for the given linear equation.
The slope, represented as a fraction \(\frac{rise}{run}\), gives you a pair of instructions for how to move from the y-intercept to another point on the line. For example, a slope of \(-2/3\) means from the y-intercept you go down 2 units (the rise) and then 3 units to the right (the run). Once at least two points are plotted, a straight line can be drawn through them, extending to the edges of the graph. This extended line represents all possible solutions for the given linear equation.
Infinite Solutions to System of Equations
Infinite solutions to a system of equations occur when the system's equations represent the exact same line on a graph. This means that instead of crossing at just one point, the lines lie on top of each other, and every point along that line is a solution to the system.
In the provided exercise, both equations resolved into the same line, \(y =(2/3)x + 2\). This reveals that the two equations are actually the same line expressed differently, and thus, they intersect at not just one point but at an infinite number of points. It's essential for students to recognize this scenario as not just a mere coincidence but a significant result indicating that the system doesn't have a single solution, but rather an infinite number of them.
In the provided exercise, both equations resolved into the same line, \(y =(2/3)x + 2\). This reveals that the two equations are actually the same line expressed differently, and thus, they intersect at not just one point but at an infinite number of points. It's essential for students to recognize this scenario as not just a mere coincidence but a significant result indicating that the system doesn't have a single solution, but rather an infinite number of them.
Other exercises in this chapter
Problem 26
In Exercises 25-28, solve the system by the method of elimination. $$ \left\\{\begin{array}{r} 7 x+10 y=0 \\ 21 x+30 y=0 \end{array}\right. $$
View solution Problem 26
In Exercises 21-26, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} 0.3 x-0.3 y=0 \\ x-\quad y=4 \end{array}\right. $$
View solution Problem 27
In Exercises 23-28, sketch the graph of the system of linear inequalities. $$ \left\\{\begin{array}{rr} x-y & \leq 8 \\ 2 x+5 y & \leq 25 \\ x & \geq 0 \\ y & \
View solution Problem 27
In Exercises 23-28, use a system of linear equations to determine the number of each type of coin. 44 Nickels and dimes \(\$ 3.00\)
View solution