0 added
0 removed
Original
2026-01-01
Modified
2026-02-28
1
<p>In solving a system of linear equations, first find the values of the variables that make all the equations true at the same time. For example, solve the system:</p>
1
<p>In solving a system of linear equations, first find the values of the variables that make all the equations true at the same time. For example, solve the system:</p>
2
<p>\(2x + y = 8\)</p>
2
<p>\(2x + y = 8\)</p>
3
<p>\(3x -2y = -1\)</p>
3
<p>\(3x -2y = -1\)</p>
4
<p><strong>Step 1:</strong>Write it as augmented matrix</p>
4
<p><strong>Step 1:</strong>Write it as augmented matrix</p>
5
<p>\( \begin{bmatrix} 2 & 1 & | & 8 \\ 3 & -2 & | & -1 \end{bmatrix} \)</p>
5
<p>\( \begin{bmatrix} 2 & 1 & | & 8 \\ 3 & -2 & | & -1 \end{bmatrix} \)</p>
6
<p><strong>Step 2:</strong>Replace the first element of the matrix with 1</p>
6
<p><strong>Step 2:</strong>Replace the first element of the matrix with 1</p>
7
<p>To do that, we divide the first row by 2:</p>
7
<p>To do that, we divide the first row by 2:</p>
8
<p>\( R_1 = \tfrac{1}{2} R_1 \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right] \)</p>
8
<p>\( R_1 = \tfrac{1}{2} R_1 \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right] \)</p>
9
<p><strong>Step 3:</strong>Eliminate the 3 below the leading 1 Using row operation:</p>
9
<p><strong>Step 3:</strong>Eliminate the 3 below the leading 1 Using row operation:</p>
10
<p>\(R2 = R2 - 3 × R1\) </p>
10
<p>\(R2 = R2 - 3 × R1\) </p>
11
<p>\(R2 = 3,-2,-1 - 3 - 1, 0.5, 4 = 3-3,-2-1.5,-1-12 = 0,-3.5,-13 \)</p>
11
<p>\(R2 = 3,-2,-1 - 3 - 1, 0.5, 4 = 3-3,-2-1.5,-1-12 = 0,-3.5,-13 \)</p>
12
<p>So now, the new matrix is: \(\left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right]\) </p>
12
<p>So now, the new matrix is: \(\left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right]\) </p>
13
<p><strong>Step 4:</strong>Make the second row’s middle into 1 Divide row 2 by -3.5</p>
13
<p><strong>Step 4:</strong>Make the second row’s middle into 1 Divide row 2 by -3.5</p>
14
<p>\(R2 = 1-3.5 \) </p>
14
<p>\(R2 = 1-3.5 \) </p>
15
<p>\( R_2 = \frac{(0,\;1,\;-13)}{-3.5} = (0,\;1,\;3.714\ldots) \)</p>
15
<p>\( R_2 = \frac{(0,\;1,\;-13)}{-3.5} = (0,\;1,\;3.714\ldots) \)</p>
16
<p>Or as<a>fraction</a>: \(133.5 = 13035 = 267\)</p>
16
<p>Or as<a>fraction</a>: \(133.5 = 13035 = 267\)</p>
17
<p>\( \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 0 & 1 & 267 \end{array}\right] \)</p>
17
<p>\( \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 0 & 1 & 267 \end{array}\right] \)</p>
18
<p><strong>Step 5:</strong>make 0.5 into 0 </p>
18
<p><strong>Step 5:</strong>make 0.5 into 0 </p>
19
<p>\(R1= R1 -0.5 × R2\)</p>
19
<p>\(R1= R1 -0.5 × R2\)</p>
20
<p>\( \left[\begin{array}{cc|c} 1 & 0 & 157 \\ 0 & 1 & 267 \end{array}\right] \)</p>
20
<p>\( \left[\begin{array}{cc|c} 1 & 0 & 157 \\ 0 & 1 & 267 \end{array}\right] \)</p>
21
<p>\(x = 157\) and \(y = 267\) </p>
21
<p>\(x = 157\) and \(y = 267\) </p>
22
22