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Laura is planning a party for her son. She has \[tex]$50 remaining in her budget and wants to provide one party favor per person to at least 10 guests. She found some miniature stuffed animals for \$[/tex]6.00 each and some toy trucks for \$4.00 each.

Which system of inequalities represents this situation, where [tex]\(x\)[/tex] is the number of stuffed animals and [tex]\(y\)[/tex] is the number of toy trucks?

A.
[tex]\[ 6x + 4y \leq 50 \][/tex]
[tex]\[ x + y \leq 10 \][/tex]

B.
[tex]\[ 6x + 4y \leq 50 \][/tex]
[tex]\[ x + y \geq 10 \][/tex]

C.
[tex]\[ 6x + 4y \geq 50 \][/tex]
[tex]\[ x + y \leq 10 \][/tex]

D.
[tex]\[ 6x + 4y \geq 50 \][/tex]
[tex]\[ x + y \geq 10 \][/tex]


Sagot :

Let's analyze the given problem and break it down into the necessary inequalities:

1. Budget Constraint:
Laura has a budget of [tex]$50. The cost of one stuffed animal is $[/tex]6, and the cost of one toy truck is $4. If [tex]\(x\)[/tex] represents the number of stuffed animals and [tex]\(y\)[/tex] represents the number of toy trucks, then the total cost of party favors can be expressed as [tex]\(6x + 4y\)[/tex]. This total cost should not exceed Laura's budget:
[tex]\[ 6x + 4y \leq 50 \][/tex]

2. Guest Constraint:
Laura wants to provide one party favor per person to at least 10 guests. Therefore, the total number of party favors, represented by [tex]\(x + y\)[/tex], must be at least 10:
[tex]\[ x + y \geq 10 \][/tex]

Combining these constraints, we get the following system of inequalities:

[tex]\[ \begin{cases} 6x + 4y \leq 50 \\ x + y \geq 10 \end{cases} \][/tex]

This matches with option B:
[tex]\[ \begin{cases} 6x + 4y \leq 50 \\ x + y \geq 10 \end{cases} \][/tex]

Therefore, the correct answer is B.
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