Select the best answer for the question. Use Gauss-Jordan elimination to solve the following system of equations. [tex]\begin{array}{l} 3 x+5 y=7 \\ 6 x-y=-8 \end{array}[/tex] A. [tex]$x=2, y=1$[/tex] B. [tex]$x=3, y=-1$[/tex] C. [tex]$x=-1, y=2$[/tex] D. [tex]$x=5, y=6$[/tex]
Answer (2)
Write the augmented matrix for the system.
Perform row operations to get a leading 1 in the first row and a 0 below it.
Perform row operations to get a leading 1 in the second row and a 0 above it.
Read the values of x and y from the reduced row echelon form: x = − 1 , y = 2 . The final answer is x = − 1 , y = 2 .
Explanation
Problem Setup We are given a system of two linear equations with two variables x and y :
{ 3 x + 5 y = 7 6 x − y = − 8
We will solve this system using Gauss-Jordan elimination.
Augmented Matrix First, we write the augmented matrix for the system:
[ 3 6 5 − 1 7 − 8 ]
Leading 1 in First Row We want to get a leading 1 in the first row, first column. We can divide the first row by 3:
R 1 → 3 1 R 1
[ 1 6 3 5 − 1 3 7 − 8 ]
Zero Below Leading 1 Next, we want to get a 0 below the leading 1 in the first column. We can subtract 6 times the first row from the second row:
R 2 → R 2 − 6 R 1
[ 1 0 3 5 − 11 3 7 − 22 ]
Leading 1 in Second Row Now, we want to get a leading 1 in the second row, second column. We can divide the second row by -11:
R 2 → − 11 1 R 2
[ 1 0 3 5 1 3 7 2 ]
Zero Above Leading 1 Finally, we want to get a 0 above the leading 1 in the second column. We can subtract 3 5 times the second row from the first row:
R 1 → R 1 − 3 5 R 2
[ 1 0 0 1 − 1 2 ]
Solution The matrix is now in reduced row echelon form. We can read the values of x and y directly from the matrix:
x = − 1 , y = 2
Therefore, the solution to the system of equations is x = − 1 and y = 2 .
Verification with Options Alternatively, we can substitute each of the given options into the equations and check which one satisfies both equations:
A. x = 2 , y = 1 : 3 ( 2 ) + 5 ( 1 ) = 6 + 5 = 11 = 7 , so this is not the solution. B. x = 3 , y = − 1 : 3 ( 3 ) + 5 ( − 1 ) = 9 − 5 = 4 = 7 , so this is not the solution. C. x = − 1 , y = 2 : 3 ( − 1 ) + 5 ( 2 ) = − 3 + 10 = 7 and 6 ( − 1 ) − 2 = − 6 − 2 = − 8 , so this is the solution. D. x = 5 , y = 6 : 3 ( 5 ) + 5 ( 6 ) = 15 + 30 = 45 = 7 , so this is not the solution.
Thus, the correct answer is C.
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling supply and demand in economics. In engineering, systems of equations can be used to analyze circuits or structural designs. Understanding how to solve these systems is crucial for making informed decisions in many fields.
Using Gauss-Jordan elimination, we found that the solution to the system of equations is x = − 1 and y = 2 . Therefore, the correct answer is C: x = − 1 , y = 2 .
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