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Sagot :
A linear equation can have one or more variables.
- The constraints on the formula are the costs of plain coffee and cost of flavored coffee
- The equation of f is: [tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex].
- The maximum number of packages of flavored coffee she can buy is 3
The function is given as:
[tex]\mathbf{T = 5.50p + 7f}[/tex]
(a) The constraints when T = 40
Substitute 40 for T in [tex]\mathbf{T = 5.50p + 7f}[/tex]
[tex]\mathbf{5.50p + 7f = 40}[/tex]
The variables in the above equation are p and f
Hence, the constraints are the cost of a package of plain coffee and the cost of a package of flavored coffee
(b) Solve for f
In (a), we have:
[tex]\mathbf{5.50p + 7f = 40}[/tex]
Subtract both sides by 5.50p
[tex]\mathbf{7f = 40 -5.50p }[/tex]
Divide both sides by 7
[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]
Hence, the equation of f is:
[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]
(c) The maximum number of flavored coffee when she buys 3 packages of plain coffee
In (b), we have:
[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]
Substitute 3 for p
[tex]\mathbf{f = \frac{40 -5.50 \times 3}7 }[/tex]
[tex]\mathbf{f = \frac{40 -16.50}7 }[/tex]
[tex]\mathbf{f = \frac{23.5}7 }[/tex]
Divide
[tex]\mathbf{f = 3.36 }[/tex]
Remove the decimal part (do not approximate)
[tex]\mathbf{f = 3 }[/tex]
Hence, the maximum number of packages of flavored coffee she can buy is 3
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