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Solve the system of linear equations using

6x+2y=14
augmented matrices.

2x-y=13

Solve the system of linear equations using $\newline$ $6 x+2 y=14$ $\newline$ augmented matrices. $\newline$ $2 x-y=13$

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Convert between exponential and logarithmic form: rational bases

Full solution

Q. Solve the system of linear equations using

\newline

6 x+2 y=14

\newline

augmented matrices.

\newline

2 x-y=13

Write Equations as Matrix: Write the system of equations as an augmented matrix. $\newline$ $\begin{bmatrix} 6 & 2 & \vert & 14 \ 2 & -1 & \vert & 13 \end{bmatrix}$
Get Leading $1$ in First Row: Use row operations to get a leading $1$ in the first row, first column. $\newline$ Divide the first row by $6$ . $\newline$ $\begin{bmatrix}1 & 1/3 & | & 7/3 \ 2 & -1 & | & 13\end{bmatrix}$
Eliminate $x$ -term in Second Row: Subtract $2$ times the first row from the second row to eliminate the $x$ -term in the second row. $\newline$ $\begin{bmatrix} 1 & \frac{1}{3} | & \frac{7}{3} \ 0 & -\frac{5}{3} | & \frac{11}{3} \end{bmatrix}$
Get Leading $1$ in Second Column: Multiply the second row by $-\frac{3}{5}$ to get a leading $1$ in the second row, second column. $\newline$ $\begin{bmatrix} 1 & \frac{1}{3} | & \frac{7}{3} \ 0 & 1 | & -\frac{11}{5} \end{bmatrix}$
Eliminate $0$ Above Leading $1$ : Subtract $\frac{1}{3}$ times the second row from the first row to get a $0$ above the leading $1$ in the second row. $\newline$ $\begin{bmatrix} 1 & 0 & | & \frac{26}{5} \ 0 & 1 & | & -\frac{11}{5} \end{bmatrix}$
Read Solutions: Read off the solutions from the augmented matrix. $\newline$ $x = \frac{26}{5}$ $\newline$ $y = -\frac{11}{5}$

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