Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ Two coworkers picked up some writing instruments at the office supply store. Olivia selected $1$ box of pencils, paying $\$2$ . Next, Grayson spent $\$18$ on $6$ boxes of pencils and $2$ boxes of ballpoint pens. How much does a box of each cost? $\newline$ A box of pencils costs $\$\_\_\_\_$ , and a box of pens costs $\$\_\_\_\_$ .

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Algebra 1

Solve a system of equations using any method: word problems

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Q. Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks.

\newline

Two coworkers picked up some writing instruments at the office supply store. Olivia selected

1

box of pencils, paying

\$2

. Next, Grayson spent

\$18

6

boxes of pencils and

2

boxes of ballpoint pens. How much does a box of each cost?

\newline

A box of pencils costs

\$\_\_\_\_

, and a box of pens costs

\$\_\_\_\_

Define Variables: Let $x$ be the cost of a box of pencils and $y$ be the cost of a box of pens. Olivia's purchase gives us the equation: $1x = 2$ .
Olivia's Equation: Grayson's purchase gives us the equation: $6x + 2y = 18$ .
Grayson's Equation: We now have the system of equations: $\newline$ $1x + 0y = 2$ $\newline$ $6x + 2y = 18$
Create Augmented Matrix: To solve using an augmented matrix, write the coefficients in matrix form: $\begin{bmatrix} 1 & 0 & \vert & 2 \ 6 & 2 & \vert & 18 \end{bmatrix}$
Row Operations: Perform row operations to get the leading coefficient of the first row to be $1$ . It's already $1$ , so no operation is needed on the first row.
Make Leading Coefficient $1$ : Next, make the first coefficient of the second row $0$ by replacing $R_2$ with $R_2 - 6\cdot R_1$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 2 \ 0 & 2 & 12 \end{array}
Divide Second Row: Now, divide the second row by $2$ to get the leading coefficient to be $1$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 2 \ 0 & 1 & 6 \end{array}
Read Solutions: We can now read the solutions from the matrix: $x = 2$ and $y = 6$ . But wait, let's check if this is correct by plugging the values back into the original equations.
Check First Equation: Check with the first equation: $1x = 2$ . Plugging in $x = 2$ gives us $1(2) = 2$ , which is true.
Check Second Equation: Check with the second equation: $6x + 2y = 18$ . Plugging in $x = 2$ and $y = 6$ gives us $6(2) + 2(6) = 12 + 12 = 24$ , which is not equal to $18$ . There's a mistake.