
To write: the solution of the given system of linear equation.

Answer to Problem 58E
The solution is x=7,y=3 .
Explanation of Solution
Given information:
A system of linear equation is given as
{2x+6y=162x+3y=7
Concept used:
Co-efficient matrix of a system of linear equation of the form {a1x+b1y=c1a2x+b2y=c2 is given by
[a1b1a2b2]
The augmented matrix of the system of linear equation is given by
[a1b1⋮c1a2b2⋮c2]
A matrix is in row-echelon form is
- If any row containing all elements as zero is in the bottom of the matrix.
- Each row that does not consist entirely of zeros has the first nonzero entry as 1.
- For successive nonzero rows the leading 1 in the higher row is further than the leading in the lower row.
And a matrix is in reduced row echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1.
To solve a system of linear equation, the corresponding augmented matrix need to reduce in row echelon form using Gaussian elimination method.
Consider the given system of equation.
{2x+6y=162x+3y=7
Now, the augmented matrix is given by
[26⋮1623⋮7]
Now, to solve the system of equation the corresponding augmented matrix need to reduce in row echelon form using Gaussian elimination method.
Multiply row 1 of the matrix by (−1) and add with row 2.
Apply R2→(−1)R1+R2
[26⋮160−3⋮−9]
Multiply row 1 by (12) and row 2 by (−13) .
Apply R1→(12)R1,R2→(−13)R2
[13⋮1601⋮3]
Multiply row 2 by −3 and add with row 1.
Apply R1→(−3)R2+R1
[10⋮701⋮3]
So, the required row echelon formis [10⋮701⋮3] .
The matrix is row reduced augmented matrix by Gauss Elimination method.
Now, the corresponding solution will be
x=7y=3
Variables x , y have been used with the coefficients.
Therefore, solution of the system of linear equation is
x=7,y=3
Chapter 7 Solutions
Precalculus with Limits: A Graphing Approach
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