Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To determine which equation can pair with [tex]\( x - y = -2 \)[/tex] to create a consistent and dependent system, we need to understand the criteria for consistency and dependency.
1. Consistent system: A system of equations is consistent if there is at least one solution that satisfies all the equations simultaneously.
2. Dependent system: A system of equations is dependent if it has infinitely many solutions. This typically occurs when the equations describe the same line, or one equation is a multiple of the other.
Here are our given equations:
1. [tex]\( 6x + 2y = 15 \)[/tex]
2. [tex]\( -3x + 3y = 6 \)[/tex]
3. [tex]\( -8x - 3y = 2 \)[/tex]
4. [tex]\( 4x - 4y \pm 6 \)[/tex]
We need to find the pair of equations that form a consistent and dependent system with [tex]\( x - y = -2 \)[/tex].
First, let's evaluate each option.
1. Equation: [tex]\( 6x + 2y = 15 \)[/tex]:
- Simplify [tex]\( x - y = -2 \)[/tex] to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]: [tex]\( x = y - 2 \)[/tex].
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( 6x + 2y = 15 \)[/tex]:
[tex]\[ 6(y - 2) + 2y = 15 \implies 6y - 12 + 2y = 15 \implies 8y - 12 = 15 \implies 8y = 27 \implies y = \frac{27}{8}. \][/tex]
- This is a unique solution, making the system consistent but not dependent.
- Result: Not dependent.
2. Equation: [tex]\( -3x + 3y = 6 \)[/tex]:
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( -3x + 3y = 6 \)[/tex]:
[tex]\[ -3(y - 2) + 3y = 6 \implies -3y + 6 + 3y = 6 \implies 6 = 6. \][/tex]
- This is an identity, meaning the original equation [tex]\( -3x + 3y = 6 \)[/tex] is equivalent to [tex]\( x - y = -2 \)[/tex] when both sides are multiplied by [tex]\(-3\)[/tex].
- Result: Dependent.
3. Equation: [tex]\( -8x - 3y = 2 \)[/tex]:
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( -8x - 3y = 2 \)[/tex]:
[tex]\[ -8(y - 2) - 3y = 2 \implies -8y + 16 - 3y = 2 \implies -11y + 16 = 2 \implies -11y = -14 \implies y = \frac{14}{11}. \][/tex]
- This is a unique solution, making the system consistent but not dependent.
- Result: Not dependent.
4. Equation: [tex]\( 4x - 4y \pm 6 \)[/tex]:
- We will consider the equation [tex]\( 4x - 4y + 6 = 0 \)[/tex] as it was likely meant to represent an equation precisely.
- Simplify to [tex]\( 4(x - y) + 6 = 0 \)[/tex]:
[tex]\[ 4(x - y) + 6 = 0 \implies 4(-2) + 6 = 0 \implies -8 + 6 = -2 \ne 0. \][/tex]
- This indicates a different process is needed. Consider adjusting terms:
Let's denote [tex]\( 4(x - y) \pm 6 = 0 \)[/tex], where for consistency with [tex]\(x - y = -2\)[/tex], substituting [tex]\(x = y - 2\)[/tex]:
[tex]\[ 4(x - y) = -8 \rightarrow -2 + 6 = -2 \rightarrow -8 + 8 = 0 \rightarrow \rightarrow Different terms arise - Result: Impossible with need of specific term - Result: After evaluating the given equations, the equation that can pair with \( x - y = -2 \) to form a consistent and dependent system is: \[ -3x + 3y = 6 \][/tex]
1. Consistent system: A system of equations is consistent if there is at least one solution that satisfies all the equations simultaneously.
2. Dependent system: A system of equations is dependent if it has infinitely many solutions. This typically occurs when the equations describe the same line, or one equation is a multiple of the other.
Here are our given equations:
1. [tex]\( 6x + 2y = 15 \)[/tex]
2. [tex]\( -3x + 3y = 6 \)[/tex]
3. [tex]\( -8x - 3y = 2 \)[/tex]
4. [tex]\( 4x - 4y \pm 6 \)[/tex]
We need to find the pair of equations that form a consistent and dependent system with [tex]\( x - y = -2 \)[/tex].
First, let's evaluate each option.
1. Equation: [tex]\( 6x + 2y = 15 \)[/tex]:
- Simplify [tex]\( x - y = -2 \)[/tex] to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]: [tex]\( x = y - 2 \)[/tex].
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( 6x + 2y = 15 \)[/tex]:
[tex]\[ 6(y - 2) + 2y = 15 \implies 6y - 12 + 2y = 15 \implies 8y - 12 = 15 \implies 8y = 27 \implies y = \frac{27}{8}. \][/tex]
- This is a unique solution, making the system consistent but not dependent.
- Result: Not dependent.
2. Equation: [tex]\( -3x + 3y = 6 \)[/tex]:
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( -3x + 3y = 6 \)[/tex]:
[tex]\[ -3(y - 2) + 3y = 6 \implies -3y + 6 + 3y = 6 \implies 6 = 6. \][/tex]
- This is an identity, meaning the original equation [tex]\( -3x + 3y = 6 \)[/tex] is equivalent to [tex]\( x - y = -2 \)[/tex] when both sides are multiplied by [tex]\(-3\)[/tex].
- Result: Dependent.
3. Equation: [tex]\( -8x - 3y = 2 \)[/tex]:
- Substitute [tex]\( x = y - 2 \)[/tex] into [tex]\( -8x - 3y = 2 \)[/tex]:
[tex]\[ -8(y - 2) - 3y = 2 \implies -8y + 16 - 3y = 2 \implies -11y + 16 = 2 \implies -11y = -14 \implies y = \frac{14}{11}. \][/tex]
- This is a unique solution, making the system consistent but not dependent.
- Result: Not dependent.
4. Equation: [tex]\( 4x - 4y \pm 6 \)[/tex]:
- We will consider the equation [tex]\( 4x - 4y + 6 = 0 \)[/tex] as it was likely meant to represent an equation precisely.
- Simplify to [tex]\( 4(x - y) + 6 = 0 \)[/tex]:
[tex]\[ 4(x - y) + 6 = 0 \implies 4(-2) + 6 = 0 \implies -8 + 6 = -2 \ne 0. \][/tex]
- This indicates a different process is needed. Consider adjusting terms:
Let's denote [tex]\( 4(x - y) \pm 6 = 0 \)[/tex], where for consistency with [tex]\(x - y = -2\)[/tex], substituting [tex]\(x = y - 2\)[/tex]:
[tex]\[ 4(x - y) = -8 \rightarrow -2 + 6 = -2 \rightarrow -8 + 8 = 0 \rightarrow \rightarrow Different terms arise - Result: Impossible with need of specific term - Result: After evaluating the given equations, the equation that can pair with \( x - y = -2 \) to form a consistent and dependent system is: \[ -3x + 3y = 6 \][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
