At a school cafeteria a cold lunch costs $1.80, and a hot lunch costs $3.00. During one school year a school year a teacher spent a total of $288.60 on cold lunches and hot lunches. The number of cold lunches the teacher bought was 1 fewer than twice the number of hot lunches the teacher bought. Which system of equations can be used to find c, the number of cold lunches, and h, the number of hot lunches, the teacher taught during the school year?

The system of equations that can be used to calculate the number of hot and cold lunches is and

Let h represents hot lunch, and c represents cold lunch.

So, we have:

Substitute in

Expand

Collect like terms

Divide both sides by 6.60

Substitute 44 for h in

Hence, the system of equations that can be used to calculate the number of hot and cold lunches is and

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System of Equations is given as:

c = 2h - 1 Equation 1

1.8c + 3h = 288.60Equation 2

Step-by-step explanation:

Let us represent

The number of Cold lunch = c

The number of Hot lunch = h

The number of cold lunches the teacher bought was 1 fewer than twice the number of hot lunches the teacher bought.

Hence:

c = 2h - 1

At a school cafeteria a cold lunch costs $1.80, and a hot lunch costs $3.00. During one school year a school year a teacher spent a total of $288.60 on cold lunches and hot lunches.

Hence:

$1.80 × c + $3.00 × h = $288.60

1.8c + 3h = 288.60Equation 2

The system of equations can be used to find c, the number of cold lunches, and h, the number of hot lunches, the teacher taught during the school year is given as:

c = 2h - 1 Equation 1

1.8c + 3h = 288.60Equation 2

step-by-step explanation:

b and c