Solve using augmented matrices. –9x − 9y = 18 x = 2 (_____, _____)

Write Equations in Matrix Form: First, let's write the system of equations in matrix form. We have: –9x − 9y = 18 x = 2 This can be written as: [–9 –9 | 18] [ 1 0 | 2 ]Eliminate x-term: Now, we need to use the second equation to eliminate the x-term in the first equation. We can multiply the second row by 9 and add it to the first row. 9 * [ 1 0 | 2 ] = [ 9 0 | 18 ] Adding this to the first row: [–9 –9 | 18] + [ 9 0 | 18 ] = [ 0 –9 | 36 ] So the new matrix is: [ 0 –9 | 36 ] [ 1 0 | 2 ]Solve for y: Now we can solve for y by dividing the first row by –9. [ 0 –9 | 36 ] ÷ –9 = [ 0 1 | –4 ] So y = –4.Find Solution: We already know x from the second equation, which is x = 2. So the solution to the system is (x, y) = (2, –4).

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Solve using augmented matrices. $\newline$ $-9x - 9y = 18$ $\newline$ $x = 2$ $\newline$ (_, _)

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Precalculus

Solve a system of equations using elimination

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Q. Solve using augmented matrices.

\newline

-9x - 9y = 18

\newline

x = 2

\newline

(_____, _____)

Write Equations in Matrix Form: First, let's write the system of equations in matrix form. We have: $\newline$ $-9x - 9y = 18$ $\newline$ $x = 2$ $\newline$ This can be written as: $\newline$ $\begin{bmatrix} -9 & -9 & | & 18 \ 1 & 0 & | & 2 \end{bmatrix}$
Eliminate x-term: Now, we need to use the second equation to eliminate the x-term in the first equation. We can multiply the second row by $9$ and add it to the first row. $\newline$ $9 \times \left[ \begin{array}{cc|c} 1 & 0 & 2 \end{array} \right] = \left[ \begin{array}{cc|c} 9 & 0 & 18 \end{array} \right]$ $\newline$ Adding this to the first row: $\newline$ $\left[ \begin{array}{cc|c} -9 & -9 & 18 \end{array} \right] + \left[ \begin{array}{cc|c} 9 & 0 & 18 \end{array} \right] = \left[ \begin{array}{cc|c} 0 & -9 & 36 \end{array} \right]$ $\newline$ So the new matrix is: $\newline$ $\left[ \begin{array}{cc|c} 0 & -9 & 36 \end{array} \right]$ $\newline$ $\left[ \begin{array}{cc|c} 1 & 0 & 2 \end{array} \right]$
Solve for $y$ : Now we can solve for $y$ by dividing the first row by $–9$ . $\left[ 0 \quad –9 \mid 36 \right] \div –9 = \left[ 0 \quad 1 \mid –4 \right]$ So $y = –4$ .
Find Solution: We already know $x$ from the second equation, which is $x = 2$ . So the solution to the system is $(x, y) = (2, -4)$ .