Problem 12
Question
Determine whether each system of linear equations has (a) one and only one solution, (b) infinitely many solutions, or (c) no solution. Find all solutions whenever they exist. \(\begin{aligned} \frac{2}{3} x+y &=5 \\ \frac{1}{2} x+\frac{3}{4} y &=\frac{15}{4} \end{aligned}\)
Step-by-Step Solution
Verified Answer
The given system of linear equations has infinitely many solutions, and they can be represented by the equation \(y = \frac{15 - 2x}{3}\).
1Step 1: Rewrite the Equations in Standard Form
To rewrite our two equations in the standard form, we just need to eliminate the fractions.
System of linear equations:
\[
\begin{aligned}
\frac{2}{3} x+y &=5 \\
\frac{1}{2} x+\frac{3}{4} y &=\frac{15}{4}
\end{aligned}
\]
Multiplying the first equation by 3 to eliminate the fraction in the coefficients:
\(2x + 3y = 15\)
Multiplying the second equation by 4 to eliminate the fractions in coefficients:
\(2x + 3y = 15\)
Now, the system of linear equations in standard form is:
\[
\begin{aligned}
2x + 3y &= 15 \\
2x + 3y &= 15
\end{aligned}
\]
2Step 2: Apply the Elimination Method on the Standard Form
To apply the elimination method, subtract the second equation from the first equation.
The subtraction will look like this:
\((2x + 3y) - (2x + 3y) = 15 - 15\)
After subtracting, we get:
\(0 = 0\)
3Step 3: Find the Nature of the System of Linear Equations
Since 0 = 0 is a true statement, and the given system of equations led us to this identity, it indicates that both equations are actually the same line (also called coincident lines). As a result, the system has infinitely many solutions.
To represent all the solutions in the form of a parametric equation, we can solve one of the equations for x or y. Here, we will solve for y:
From Equation 1, we can write as:
\(2x + 3y = 15 \implies y = \frac{15 - 2x}{3}\)
Final answer: The given system of linear equations has infinitely many solutions, and they can be represented by the equation \(y = \frac{15 - 2x}{3}\).
Key Concepts
System of EquationsElimination MethodCoincident LinesParametric Equation
System of Equations
A system of equations involves two or more equations that share the same set of unknowns. In our example, we have linear equations involving variables, typically "x" and "y." These systems are crucial in mathematics as they represent multiple conditions or constraints that must be satisfied simultaneously.
An equation system can be visually represented using graphs, where each equation corresponds to a line. The solution to the system represents a point or points on the graph where the lines intersect.
Types of solutions for systems of linear equations include:
An equation system can be visually represented using graphs, where each equation corresponds to a line. The solution to the system represents a point or points on the graph where the lines intersect.
Types of solutions for systems of linear equations include:
- One solution: The lines intersect at a single point.
- Infinitely many solutions: The lines overlap entirely, called coincident lines.
- No solution: The lines are parallel and never intersect.
Elimination Method
The elimination method is a mathematical procedure used to find the solutions of a system of equations. This method involves eliminating one of the variables by adding or subtracting the equations.
Here's a simple breakdown of how it works:
Here's a simple breakdown of how it works:
- Modify each equation, if necessary, to align them for elimination (e.g., removing fractions).
- Add or subtract the equations to eliminate one variable.
- Solve the resulting equation to find the value of one variable.
- Substitute back into one of the original equations to find the other variable.
Coincident Lines
The concept of coincident lines emerges when two lines lie exactly on top of each other in a graph. This situation happens in a system of linear equations when each equation represents the same line.
In a coincident line scenario:
In a coincident line scenario:
- Every point on one line lies on the other line.
- This results in infinitely many solutions, as all points satisfy both equations.
- The graph may appear as a single line because overlap conceals the other line.
Parametric Equation
A parametric equation expresses the coordinates of the points that make up a geometric object by using a parameter, typically denoted as "t." In relation to linear systems, parametric equations are useful for representing solutions, especially for coincident lines.
When coincident lines produce infinitely many solutions, we can express these solutions in terms of a parametric equation by solving for one variable in terms of the other.
In the given problem, solving the equation for "y" in terms of "x" yielded:\[ y = \frac{15 - 2x}{3} \]This parametric form allows us to express all points (x, y) on the line. By choosing any value for "x," we can find the corresponding "y," illustrating the multitude of solutions existing on coincident lines.
When coincident lines produce infinitely many solutions, we can express these solutions in terms of a parametric equation by solving for one variable in terms of the other.
In the given problem, solving the equation for "y" in terms of "x" yielded:\[ y = \frac{15 - 2x}{3} \]This parametric form allows us to express all points (x, y) on the line. By choosing any value for "x," we can find the corresponding "y," illustrating the multitude of solutions existing on coincident lines.
Other exercises in this chapter
Problem 12
Given that the augmented matrix in row-reduced form is equivalent to the augmented matrix of a system of linear equations, (a) determine whether the system has
View solution Problem 12
Indicate whether the matrix is in rowreduced form. \(\left[\begin{array}{ll|l}0 & 1 & 3 \\ 0 & 0 & 5\end{array}\right]\)
View solution Problem 13
Find the inverse of the matrix, if it exists. Verify your answer. \(\left[\begin{array}{rrr}1 & 4 & -1 \\ 2 & 3 & -2 \\ -1 & 2 & 3\end{array}\right]\)
View solution Problem 13
Compute the indicated products. \(\left[\begin{array}{lll}2 & 1 & 2 \\ 3 & 2 & 4\end{array}\right]\left[\begin{array}{rr}-1 & 2 \\ 4 & 3 \\ 0 & 1\end{array}\rig
View solution