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Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ At an ice cream store, a family ordered $1$ banana split, paying a total of $\$6$ for their order. The next customer ordered $4$ banana splits and $6$ hot fudge sundaes and paid $\$54$ . How much does each item cost? $\newline$ A banana split costs $\$\_\_\_\_$ and a hot fudge sundae costs $\$\_\_\_\_\_$ .

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Solve a system of equations using elimination: word problems

Full solution

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

\newline

At an ice cream store, a family ordered

1

banana split, paying a total of

\$6

for their order. The next customer ordered

4

banana splits and

6

hot fudge sundaes and paid

\$54

. How much does each item cost?

\newline

A banana split costs

\$\_\_\_\_

and a hot fudge sundae costs

\$\_\_\_\_\_

.

Define Variables: Let $x$ be the cost of a banana split and $y$ be the cost of a hot fudge sundae. $\newline$ First family's order: $1$ banana split ( $x$ ) for $\$6$ . $\newline$ $1x + 0y = 6$
First Family's Order: Next customer's order: $4$ banana splits $(x)$ and $6$ hot fudge sundaes $(y)$ for $\$54$ . $4x + 6y = 54$
Next Customer's Order: Create an augmented matrix to represent the system of equations: $\newline$ \begin{array}{cc|c} 1 & 0 & 6 \ 4 & 6 & 54 \ \end{array}
Create Augmented Matrix: Use row operations to find the cost of each item. $\newline$ First, multiply the first row by $-4$ and add it to the second row to eliminate $x$ from the second equation. $\newline$ $-4 \times \left| \begin{array}{cc} 1 & 0 \ 4 & 6 \end{array} \right| \begin{array}{c} 6 \ 54 \end{array} + \left| \begin{array}{cc} 4 & 6 \end{array} \right| \begin{array}{c} 54 \end{array} = \left| \begin{array}{cc} 4 & 6 \end{array} \right| \begin{array}{c} 54 \end{array} + \left| \begin{array}{cc} -4 & 0 \end{array} \right| \begin{array}{c} -24 \end{array} = \left| \begin{array}{cc} 0 & 6 \end{array} \right| \begin{array}{c} 30 \end{array}$
Use Row Operations: Now we have the new system of equations: $\newline$ $1x + 0y = 6$ $\newline$ $0x + 6y = 30$
New System of Equations: Solve for $y$ by dividing the second equation by $6$ . $\frac{6y}{6} = \frac{30}{6}$ $y = 5$
Solve for $y$ : Substitute $y = 5$ into the first equation to solve for $x$ . $1x + 0(5) = 6$ $x = 6$

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