Solve using augmented matrices. 7x − 10y = 2 x = 6 (_____, _____)

Write Augmented Matrix: First, let's write the system of equations as an augmented matrix. We have the equations: 7x − 10y = 2 x = 6 This can be written as: [7 -10 | 2] [1 0 | 6]Swap Rows: Now, let's use the second equation to make the first element of the first row a 1 by swapping rows. Swap R1 with R2: [1 0 | 6] [7 -10 | 2]Make First Element 0: Next, we need to make the first element of the second row a 0 by using row operations. We can multiply the first row by -7 and add it to the second row. -7 * R1 + R2 -> R2: [1 0 | 6] [0 -10 | -40]Make Second Element 1: Now, let's divide the second row by -10 to make the second element of the second row a 1. R2 / -10 -> R2: [1 0 | 6] [0 1 | 4]Final System of Equations: We have the matrix in reduced row echelon form, which corresponds to the system: x = 6 y = 4

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Solve using augmented matrices. $\newline$ $7x - 10y = 2$ $\newline$ $x = 6$ $\newline$ (_, _)

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Grade 8

Solve a system of equations using any method

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Q. Solve using augmented matrices.

\newline

7x - 10y = 2

\newline

x = 6

\newline

(_____, _____)

Write Augmented Matrix: First, let's write the system of equations as an augmented matrix. $\newline$ We have the equations: $\newline$ $7x − 10y = 2$ $\newline$ $x = 6$ $\newline$ This can be written as: $\newline$ \begin{bmatrix}7 & -10 & | & 2\1 & 0 & | & 6\end{bmatrix}
Swap Rows: Now, let's use the second equation to make the first element of the first row a $1$ by swapping rows. $\newline$ Swap $R_1$ with $R_2$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 6 \ 7 & -10 & 2 \end{array}
Make First Element $0$ : Next, we need to make the first element of the second row a $0$ by using row operations. $\newline$ We can multiply the first row by $-7$ and add it to the second row. $\newline$ $-7 \times R1 + R2 \rightarrow R2$ : $\newline$ $\begin{bmatrix} 1 & 0 | & 6 \ 0 & -10 | & -40 \end{bmatrix}$
Make Second Element $1$ : Now, let's divide the second row by $-10$ to make the second element of the second row a $1$ . $\newline$ $R2 / -10 \rightarrow R2$ : $\newline$ $\begin{bmatrix} 1 & 0 & | & 6 \ 0 & 1 & | & 4 \end{bmatrix}$
Final System of Equations: We have the matrix in reduced row echelon form, which corresponds to the system: $\newline$ $x = 6$ $\newline$ $y = 4$