Problem 33
Question
Write a system of equations that does not have a unique solution.
Step-by-Step Solution
Verified Answer
A system with infinitely many solutions: \( 2x + 3y = 6 \) and \( 4x + 6y = 12 \). Parallel lines with no solution: \( 2x + 3y = 6 \) and \( 2x + 3y = 8 \).
1Step 1: Identify Types of Solutions
A system of equations can have three types of solutions: a unique solution (one intersection point), no solution (parallel lines), or infinitely many solutions (coinciding lines). To ensure the system does not have a unique solution, we can either have no solution or infinitely many solutions.
2Step 2: Create a System with Infinitely Many Solutions
To create a system with infinitely many solutions, we use two dependent equations that represent the same line. For example, consider the equations: \( 2x + 3y = 6 \) and \( 4x + 6y = 12 \). Notice that the second equation is just a multiple of the first. This implies that both equations represent the same line, hence there are infinitely many solutions.
3Step 3: Create a System with No Solution
To create a system with no solution, we need two parallel lines. This occurs if the lines have the same slope but different y-intercepts. For example, consider the equations: \( 2x + 3y = 6 \) and \( 2x + 3y = 8 \). Since both have the same slope but different intercepts, they are parallel and will never intersect.
Key Concepts
Types of SolutionsInfinitely Many SolutionsNo SolutionDependent Equations
Types of Solutions
When solving a system of equations, it's important to understand the possible outcomes, which are known as the 'types of solutions.' These outcomes determine how many solutions (if any) a system has. Here are the three key types:
- Unique Solution: This occurs when there is exactly one solution to the system. In a graph, this corresponds to two lines intersecting at a single point.
- No Solution: A system will have no solutions if the lines are parallel and never meet. These lines have the same slope but different y-intercepts.
- Infinitely Many Solutions: This is the case when the lines overlap entirely, meaning they represent the same equation. This results in an infinite number of intersection points, as every point on the line is a solution.
Infinitely Many Solutions
A system of equations has infinitely many solutions when the equations are dependent, meaning they represent the same line. In simple terms, both equations describe the same line repeated in different forms.
To create such a system, make one equation a scalar multiple of the other. For instance:
To create such a system, make one equation a scalar multiple of the other. For instance:
- First equation: \(2x + 3y = 6\)
- Second equation: \(4x + 6y = 12\)
No Solution
A system of equations will have no solution if the equations form parallel lines. Parallel lines have the same slope but will never intersect, meaning there is no common point that satisfies both equations.
To construct a system with no solution, ensure that both equations share the same slope but have different y-intercepts. For example:
To construct a system with no solution, ensure that both equations share the same slope but have different y-intercepts. For example:
- First equation: \(2x + 3y = 6\)
- Second equation: \(2x + 3y = 8\)
Dependent Equations
Dependent equations are key to understanding systems with infinitely many solutions. When you have dependent equations, it means one equation is essentially a rearranged version of the other.
Such equations render the term 'dependent' because their solutions depend on being identical lines. To spot dependent equations, check if one is a consistent multiple of the other.
For instance, take:
Such equations render the term 'dependent' because their solutions depend on being identical lines. To spot dependent equations, check if one is a consistent multiple of the other.
For instance, take:
- First equation: \(2x + 3y = 6\)
- Second equation: \(4x + 6y = 12\)
Other exercises in this chapter
Problem 32
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View solution Problem 32
Use \(A=\left[\begin{array}{cc}{1} & {-2} \\ {4} & {3}\end{array}\right], B=\left[\begin{array}{cc}{-5} & {2} \\ {4} & {3}\end{array}\right], C=\left[\begin{arr
View solution Problem 33
Find the inverse of each matrix, if it exists. $$ \left[\begin{array}{rr}{\frac{1}{2}} & {-\frac{3}{4}} \\ {\frac{1}{6}} & {\frac{1}{4}}\end{array}\right] $$
View solution Problem 33
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