Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ Dr. Kelly, a pediatrician, has $1$ annual checkup scheduled next Tuesday, which will fill a total of $52$ minutes on her schedule. Next Wednesday, she has $2$ annual checkups and $2$ sick visits on the schedule, which should take $160$ minutes. How much time is allotted for each type of appointment? $\newline$ The time allotted is $\_\_\_\_$ minutes for an annual checkup and $\_\_\_\_$ minutes for a sick visit.

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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

Dr. Kelly, a pediatrician, has

1

annual checkup scheduled next Tuesday, which will fill a total of

52

minutes on her schedule. Next Wednesday, she has

2

annual checkups and

2

sick visits on the schedule, which should take

160

minutes. How much time is allotted for each type of appointment?

\newline

The time allotted is

\_\_\_\_

minutes for an annual checkup and

\_\_\_\_

minutes for a sick visit.

Define Variables: question_prompt: How much time is allotted for each type of appointment in Dr. Kelly's schedule?
Form Equations: Let $x$ be the time for an annual checkup and $y$ be the time for a sick visit.
Convert to Matrix: Equation for Tuesday: $1x = 52$ .
Row Operations: Equation for Wednesday: $2x + 2y = 160$ .
Eliminate Variables: System of equations: $\newline$ $1x + 0y = 52$ $\newline$ $2x + 2y = 160$ .
Solve for $y$ : Convert the system of equations into an augmented matrix: $\newline$ $\begin{bmatrix}1 & 0 | & 52\ 2 & 2 | & 160\end{bmatrix}$ .
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\newline$ \begin{array}{cc|c} -2 & 0 & -104 \ 2 & 2 & 160 \end{array}.
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\newline$ \begin{array}{cc|c} -2 & 0 & -104 \ 2 & 2 & 160 \end{array}. Add the rows to eliminate $x$ from the second row: $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 2 & 56 \end{array}.
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\newline$ \begin{array}{cc|c} -2 & 0 & -104 \ 2 & 2 & 160 \end{array}. Add the rows to eliminate $x$ from the second row: $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 2 & 56 \end{array}. Divide the second row by $2$ to solve for $y$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 1 & 28 \end{array}.
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\newline$ \begin{array}{cc|c} -2 & 0 & -104 \ 2 & 2 & 160 \end{array}. Add the rows to eliminate $x$ from the second row: $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 2 & 56 \end{array}. Divide the second row by $2$ to solve for $y$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 1 & 28 \end{array}. $y = 28$ .
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\begin{bmatrix} -2 & 0 & | & -104 \ 2 & 2 & | & 160 \end{bmatrix}$ . Add the rows to eliminate $x$ from the second row: $\begin{bmatrix} 1 & 0 & | & 52 \ 0 & 2 & | & 56 \end{bmatrix}$ . Divide the second row by $2$ to solve for $y$ : $\begin{bmatrix} 1 & 0 & | & 52 \ 0 & 1 & | & 28 \end{bmatrix}$ . $y = 28$ . Substitute $y$ back into the first equation to solve for $x$ : $-2$ $1$ .
Substitute and Solve: Use row operations to find the value of $x$ . Multiply the first row by $-2$ and add it to the second row: $\newline$ \begin{array}{cc|c} -2 & 0 & -104 \ 2 & 2 & 160 \end{array}. Add the rows to eliminate $x$ from the second row: $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 2 & 56 \end{array}. Divide the second row by $2$ to solve for $y$ : $\newline$ \begin{array}{cc|c} 1 & 0 & 52 \ 0 & 1 & 28 \end{array}. $y = 28$ . Substitute $y$ back into the first equation to solve for $x$ : $\newline$ $1x + 0(28) = 52$ . $x = 52$ .