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Determine whether or not the given pair of values is a solution of the given system of linear equations. $$\begin{aligned} &6 i_{1}+i_{2}=5 \quad i_{1}=1, i_{2}=-1\\\ &3 i_{1}-4 i_{2}=-1 \end{aligned}$$

Short Answer

Expert verified
The pair \((1, -1)\) is not a solution to the system.

Step by step solution

01

Substitute Values into First Equation

To determine if \( (i_1, i_2) = (1, -1) \) is a solution to the system, substitute \( i_1 = 1 \) and \( i_2 = -1 \) into the first equation: \( 6i_1 + i_2 = 5 \). This becomes \( 6(1) + (-1) = 5 \). Simplifying, we get \( 6 - 1 = 5 \), which is true.
02

Substitute Values into Second Equation

Next, substitute \( i_1 = 1 \) and \( i_2 = -1 \) into the second equation: \( 3i_1 - 4i_2 = -1 \). This becomes \( 3(1) - 4(-1) = -1 \). Simplifying, we get \( 3 + 4 = -1 \), which is false since \( 7 eq -1 \).
03

Conclusion: Determine if the Pair is a Solution

Since the pair \( (1, -1) \) satisfies the first equation but not the second, it is not a solution to the system of equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Verification
Solution verification is an essential step in solving a system of linear equations. It involves checking if a proposed solution satisfies all equations in the system. For example, given the system:
  • Equation 1: \(6 i_{1} + i_{2} = 5\)
  • Equation 2: \(3 i_{1} - 4 i_{2} = -1\)
To verify if \((i_1, i_2) = (1, -1)\) is a valid solution, substitute these values into both equations. If both computations hold true, the pair is a solution. However, if even one of the equations is not satisfied, the pair is not a solution.
In this exercise, the pair satisfies the first equation but fails the second. Therefore, it is not a solution to the system. Solution verification ensures accuracy in linear algebra problem-solving by confirming that the proposed solution truly fits all given equations.
Substitution Method
The substitution method is a strategic approach to solving systems of linear equations. This method involves solving one of the equations for one variable and substituting this solution into the other equation.
In our exercise, we could solve the first equation for \(i_2\), yielding \(i_2 = 5 - 6i_1\). We could then substitute this expression for \(i_2\) in the second equation to solve for \(i_1\). This process transforms two-variable equations into simpler, single-variable ones.
This process can sometimes make solving complex systems simpler and more intuitive, offering clarity without the need for graphing or complex algebraic manipulations. It allows for step-by-step solutions which can be particularly helpful when verifying potential solutions.
Linear Algebra Concepts
Linear algebra involves studying vectors, matrices, and systems of linear equations. Understanding these concepts forms the foundation of solving equations like the ones in the exercise.
When dealing with systems of equations, one must grasp the idea of solutions being points in a multidimensional space pre-defined by the variables. Each linear equation defines a plane, and solutions are points where these planes intersect.
  • Consistent systems have at least one solution.
  • Inconsistent systems have no solution.
  • Dependent systems have infinitely many solutions.
Understanding these concepts not only helps in solving systems of equations but also builds a strong foundation for more advanced topics, such as vector spaces and transformations. Linear algebra's techniques and methods, like substitution, provide powerful tools for mathematical problem-solving across various domains.

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Most popular questions from this chapter

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