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EXPENSES: Camille has no more than $50.00 to spend each week for lunch and bus fare. Lunch costs $3.00 each day, and bus fare is $1.75 each ride. Write an inequality for this situation. Can Camille buy lunch 7 times and ride the bus 14 times in one​ week????

Answer :

MrRoyal

Answer:

Inequality: [tex]3.00 L + 1.75B \leq 50[/tex]

She can afford to buy the lunch and bus rides

Step-by-step explanation:

Given

[tex]Maximum\ Amount = \$50.00[/tex]

[tex]Lunch = \$3.00[/tex] daily

[tex]Bus\ Fare = \$1.75[/tex] daily

Solving (a) The inequality

Let bus ride be represented with B and Lunch be represented with L

If 1 bus ride costs $1.75, then

B bus ride would cost $1.75 * B

If 1 lunch costs $3.00, then

L lunch would cost $3.00 * L

Total Spending is then the sum of the above expression i.e.

$3.00 * L + $1.75 * B

The question says Camille can not spend more than $50.00.

This is represented by less than or equal to.

So, the expression is:

[tex]3.00 * L + 1.75 * B \leq 50[/tex]

[tex]3.00 L + 1.75B \leq 50[/tex]

Solving (b): Can she afford 7 lunch and 14 bus ride?

To do this, we simply substitute 7 for L and 14 for B

[tex]3.00 * 7 + 1.75 * 14 \leq 50[/tex]

[tex]21 + 24.5 \leq 50[/tex]

[tex]45.5 \leq 50[/tex]

The above inequality is true.

Hence, she can afford to buy the lunch and bus rides

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