Answer:
System of Equations is given as:
c = 2h - 1...... Equation 1
1.8c + 3h = 288.60....Equation 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 = 288.60....Equation 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 = 288.60....Equation 2
The system of equations that can be used to calculate the number of hot and cold lunches is [tex]\mathbf{1.80c +3.00h = 288.60}[/tex] and [tex]\mathbf{c = 2h - 1}[/tex]
Let h represents hot lunch, and c represents cold lunch.
So, we have:
[tex]\mathbf{1.80c +3.00h = 288.60}[/tex]
[tex]\mathbf{c = 2h - 1}[/tex]
Substitute [tex]\mathbf{c = 2h - 1}[/tex] in [tex]\mathbf{1.80c +3.00h = 288.60}[/tex]
[tex]\mathbf{1.80(2h -1) +3.00h = 288.60}[/tex]
Expand
[tex]\mathbf{3.60h -1.80 +3.00h = 288.60}[/tex]
Collect like terms
[tex]\mathbf{3.60h +3.00h = 288.60 + 1.80}[/tex]
[tex]\mathbf{6.60h = 290.4}[/tex]
Divide both sides by 6.60
[tex]\mathbf{h = 44}[/tex]
Substitute 44 for h in [tex]\mathbf{c = 2h - 1}[/tex]
[tex]\mathbf{c = 2(44) - 1}[/tex]
[tex]\mathbf{c = 83}[/tex]
Hence, the system of equations that can be used to calculate the number of hot and cold lunches is [tex]\mathbf{1.80c +3.00h = 288.60}[/tex] and [tex]\mathbf{c = 2h - 1}[/tex]
Read more about simultaneous equations at:
https://brainly.com/question/16763389
