Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ Yesterday a chef used $32$ eggs to make $4$ chocolate souffles and $6$ lemon meringue pies. The day before, he made $1$ chocolate souffle, which used $5$ eggs. How many eggs does each dessert require? $\newline$ A chocolate souffle requires $\_\_\_\_$ eggs and a lemon meringue pie requires $\_\_\_\_$ eggs.

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Grade 8

Solve a system of equations using elimination: 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

Yesterday a chef used

32

eggs to make

4

chocolate souffles and

6

lemon meringue pies. The day before, he made

1

chocolate souffle, which used

5

eggs. How many eggs does each dessert require?

\newline

A chocolate souffle requires

\_\_\_\_

eggs and a lemon meringue pie requires

\_\_\_\_

eggs.

Define Variables: Let $x$ be the number of eggs needed for a chocolate souffle and $y$ be the number of eggs needed for a lemon meringue pie. We can write two equations based on the information given:
Write Equations: Equation for yesterday: $4x + 6y = 32$ (since $4$ souffles and $6$ pies were made using $32$ eggs).
Convert to Matrix: Equation for the day before: $1x = 5$ (since $1$ souffle was made using $5$ eggs).
Eliminate Variables: Now we have the system of equations: $\newline$ $1$ . $4x + 6y = 32$ $\newline$ $2$ . $x = 5$
Solve for $y$ : Convert the system of equations into an augmented matrix: $\newline$ $\begin{bmatrix} 4 & 6 & | & 32 \ 1 & 0 & | & 5 \end{bmatrix}$
Substitute and Find $x$ : Use the second equation to eliminate $x$ from the first equation by multiplying the second equation by $-4$ and adding it to the first equation: $\newline$ $-4\times\begin{bmatrix} $1$ & $0$ | & $5$ \end{bmatrix} = \begin{bmatrix} $-4$ & $-0$ | & $-20$ \end{bmatrix}(\newline$\begin{bmatrix} $4$ & $6$ | & $32$ \end{bmatrix} + \begin{bmatrix} $-4$ & $-0$ | & $-20$ \end{bmatrix} = \begin{bmatrix} $0$ & $6$ | & $12$ \end{bmatrix}
Final Results: Now we have the new system in matrix form: $\begin{bmatrix} 4 & 6 & \vert & 32 \ 0 & 6 & \vert & 12 \end{bmatrix}$
Final Results: Now we have the new system in matrix form: $\newline$ $\begin{bmatrix}4 & 6 \ 0 & 6 \end{bmatrix}\begin{bmatrix}32 \ 12 \end{bmatrix}$ Divide the second row by $6$ to solve for $y$ : $\newline$ $\begin{bmatrix}0 & 6 \ 12 \end{bmatrix} \div 6 = \begin{bmatrix}0 & 1 \ 2 \end{bmatrix}$ $\newline$ So, $y = 2$ .
Final Results: Now we have the new system in matrix form: $\newline$ \begin{bmatrix}4 & 6 \ 0 & 6\end{bmatrix}\begin{bmatrix}32\12\end{bmatrix}Divide the second row by $6$ to solve for $y$ : $\newline$ $\begin{bmatrix}0 & 6 \ 12\end{bmatrix} \div 6 = \begin{bmatrix}0 & 1 \ 2\end{bmatrix}$ $\newline$ So, $y = 2$ .Substitute $y = 2$ into the second original equation, $x = 5$ , to find $x$ . But since $x$ is already given as $5$ , there's no need to substitute, and we can say $x = 5$ .
Final Results: Now we have the new system in matrix form: $\newline$ $\begin{matrix} 4 & 6 & | & 32 \ 0 & 6 & | & 12 \end{matrix}$ Divide the second row by $6$ to solve for $y$ : $\newline$ $\begin{matrix} 0 & 6 & | & 12 \end{matrix}$ $\div 6 = \begin{matrix} 0 & 1 & | & 2 \end{matrix}$ $\newline$ So, $y = 2$ .Substitute $y = 2$ into the second original equation, $x = 5$ , to find $x$ . But since $x$ is already given as $5$ , there's no need to substitute, and we can say $x = 5$ .Therefore, a chocolate souffle requires $5$ eggs and a lemon meringue pie requires $y$ $1$ eggs.