Solve using augmented matrices. 3x − 3y = –15 x = –2 (_____, _____)

Write Equations in Matrix Form: Step 1: Write the system of equations in matrix form. We have the equations: 3x - 3y = -15 x = -2 This can be rewritten as: 3x - 3y = -15 1x + 0y = -2 The corresponding augmented matrix is: [3 -3 | -15] [1 0 | -2]Eliminate x from Equations: Step 2: Use the second equation to eliminate x from the first equation. We'll multiply the second row by -3 and add it to the first row to eliminate x: -3 * [1 0 | -2] = [-3 0 | 6] Adding this to the first row: [3 -3 | -15] + [-3 0 | 6] = [0 -3 | -9] Now the matrix looks like: [0 -3 | -9] [1 0 | -2]Solve for y: Step 3: Solve for y from the first row of the transformed matrix. The first row [0 -3 | -9] simplifies to y = 3.

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Solve using augmented matrices. $\newline$ $3x - 3y = -15$ $\newline$ $x = -2$ $\newline$ (_, _)

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Grade 8

Solve a system of equations using any method

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Q. Solve using augmented matrices.

\newline

3x - 3y = -15

\newline

x = -2

\newline

(_____, _____)

Write Equations in Matrix Form: Step $1$ : Write the system of equations in matrix form. $\newline$ We have the equations: $\newline$ $3x - 3y = -15$ $\newline$ $x = -2$ $\newline$ This can be rewritten as: $\newline$ $3x - 3y = -15$ $\newline$ $1x + 0y = -2$ $\newline$ The corresponding augmented matrix is: $\newline$ \begin{bmatrix}3 & -3 & | & -15\1 & 0 & | & -2\end{bmatrix}
Eliminate $x$ from Equations: Step $2$ : Use the second equation to eliminate $x$ from the first equation. $\newline$ We'll multiply the second row by $-3$ and add it to the first row to eliminate $x$ : $\newline$ $-3 \times [1 0 | -2] = [-3 0 | 6]$ $\newline$ Adding this to the first row: $\newline$ $[3 -3 | -15] + [-3 0 | 6] = [0 -3 | -9]$ $\newline$ Now the matrix looks like: $\newline$ $[0 -3 | -9]$ $\newline$ $[1 0 | -2]$
Solve for y: Step $3$ : Solve for $y$ from the first row of the transformed matrix. $\newline$ The first row $[0 -3 | -9]$ simplifies to $y = 3$ .