Math Problem Statement

Use Gaussian elimination and back-substitution to solve the following system of equations. If there is a solution, write your answer in the format (x,y) 2x + 5y = 23; - 4x - 10y = - 46

Solution

We are given the system of linear equations:

  1. 2x+5y=232x + 5y = 23
  2. 4x10y=46-4x - 10y = -46

Step 1: Write the system in matrix form.

We write the system as an augmented matrix:

2 & 5 & | & 23 \\ -4 & -10 & | & -46 \end{pmatrix}$$ ### Step 2: Perform Gaussian elimination. To eliminate one of the variables, let's start by eliminating $$x$$. To do this, we multiply the first row by 2 and add it to the second row: $$R_2 \rightarrow R_2 + 2R_1$$ This gives: $$\begin{pmatrix} 2 & 5 & | & 23 \\ 0 & 0 & | & 0 \end{pmatrix}$$ ### Step 3: Interpret the matrix. The second row becomes $$0 = 0$$, which is a true statement and indicates that the second equation does not add any new information. Therefore, the system has infinitely many solutions. ### Step 4: Solve for one variable. From the first row, we can solve for $$x$$ in terms of $$y$$: $$2x + 5y = 23$$ Solve for $$x$$: $$x = \frac{23 - 5y}{2}$$ ### Step 5: Write the solution. The system has infinitely many solutions and can be written in parametric form. Let $$y = t$$ (where $$t$$ is any real number), then the solution is: $$x = \frac{23 - 5t}{2}, \quad y = t$$ Thus, the solution is: $$(x, y) = \left( \frac{23 - 5t}{2}, t \right)$$ ### Conclusion: The system has infinitely many solutions, parameterized by $$t$$, where $$t$$ can be any real number. --- Do you want further details or have any questions? Here are five related questions you might consider: 1. How can Gaussian elimination handle systems with no solutions? 2. What are the implications of having a row of zeros in a matrix? 3. How can you determine if a system has a unique solution, no solution, or infinitely many solutions? 4. How can you express the general solution of a system with infinitely many solutions? 5. What are the differences between Gaussian elimination and other methods of solving linear systems? **Tip:** When performing Gaussian elimination, always aim to create zeros below the pivot (leading) entries to simplify back-substitution.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Gaussian Elimination

Formulas

Gaussian Elimination Steps: Row operations to convert a system to row echelon form.
Back-Substitution: Solving from the bottom row upwards in triangular form.

Theorems

Consistency Theorem: A system of linear equations is consistent if there is at least one solution.

Suitable Grade Level

Grades 10-12