A fruit stand has to decide what to charge for their produce. They need $5 for 1 apple and 1 orange. They also need $15 for 3 apples and 3 oranges. Put this information into a system of linear equations. Can we find a unique price for an apple and an orange? Choose 1 answer: A Yes; they should charge $2.00 for an apple and $3.00 for an orange. (B) Yes; they should charge $1.00 for an apple and $4.00 for an orange. (C) No; the system has many solutions. (D) No; the system has no solution.

Translate Equations: Let's denote the price of an apple as A and the price of an orange as O. The information given can be translated into two equations based on the cost of the fruits: 1 apple + 1 orange = $5 3 apples + 3 oranges = $15 In terms of A and O, these equations are: A + O = 5 ...(1) 3A + 3O = 15 ...(2)Simplify Equation (2): Now, let's simplify equation (2) by dividing each term by 3 to see if it provides any new information: 3A/3 + 3O/3 = 15/3 A + O = 5 This is the same as equation (1), which means that the two equations are actually the same line when graphed. Therefore, they do not provide unique solutions for A and O.Infinite Solutions: Since both equations are the same, there are infinitely many solutions that satisfy both equations. This means that we cannot determine a unique price for an apple and an orange based on the given information alone. The system of equations is dependent and consistent with many solutions.

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A fruit stand has to decide what to charge for their produce. They need $\$5$ for $1$ apple and $1$ orange. They also need $\$15$ for $3$ apples and $3$ oranges. Put this information into a system of linear equations. $\newline$ Can we find a unique price for an apple and an orange? $\newline$ Choose $1$ answer: $\newline$ (A) Yes; they should charge $\$2.00$ for an apple and $\$3.00$ for an orange. $\newline$ (B) Yes; they should charge $\$1.00$ for an apple and $\$4.00$ for an orange. $\newline$ (C) No; the system has many solutions. $\newline$ (D) No; the system has no solution.

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Q. A fruit stand has to decide what to charge for their produce. They need

\$5

for

1

apple and

1

orange. They also need

\$15

for

3

apples and

3

oranges. Put this information into a system of linear equations.

\newline

Can we find a unique price for an apple and an orange?

\newline

Choose

1

answer:

\newline

(A) Yes; they should charge

\$2.00

for an apple and

\$3.00

for an orange.

\newline

(B) Yes; they should charge

\$1.00

for an apple and

\$4.00

for an orange.

\newline

\newline

(D) No; the system has no solution.

Translate Equations: Let's denote the price of an apple as $A$ and the price of an orange as $O$ . The information given can be translated into two equations based on the cost of the fruits: $\newline$ $1$ apple + $1$ orange = $\$(5)$ $\newline$ $3$ apples + $3$ oranges = $\$(15)$ $\newline$ In terms of $A$ and $O$ , these equations are: $\newline$ $A + O = 5$ ...( $1$ ) $\newline$ $3A + 3O = 15$ ...( $2$ )
Simplify Equation ( $2$ ): Now, let's simplify equation ( $2$ ) by dividing each term by $3$ to see if it provides any new information: $\newline$ $\frac{3A}{3} + \frac{3O}{3} = \frac{15}{3}$ $\newline$ $A + O = 5$ $\newline$ This is the same as equation ( $1$ ), which means that the two equations are actually the same line when graphed. Therefore, they do not provide unique solutions for $A$ and $O$ .
Infinite Solutions: Since both equations are the same, there are infinitely many solutions that satisfy both equations. This means that we cannot determine a unique price for an apple and an orange based on the given information alone. The system of equations is dependent and consistent with many solutions.