For one Fredburger, one Milk Shake and two order of Fries, we pay $4.34
Explanation:
Let us assume the cost of Fredburger as B, cost of Milk Shake as M, and cost of Order of Fries as Y.
As mentioned in the Question:
[tex]B + M + Y = $3.72[/tex] ----------- Equation 1
[tex]B = M + 2 \times Y[/tex], ----------- Equation 2
[tex]3 \times M = B + Y[/tex], ----------- Equation 3
Let us assume that we have to pay $X for one Fredburger, one Milk Shake and two order of Fries.
[tex]B+M+2 \times Y = $X[/tex], consider this as Equation 4
Substituting Equation 3 into Equation 1:
[tex]B + M + Y = $3.72[/tex]
[tex] 3 \times M + M = $3.72[/tex]
[tex] 4 \times M = $3.72[/tex]
[tex]M = \frac{3.72}{4}[/tex]
M = $ 0.93
Substituting value of M in Equation 2,
[tex]B = M + 2 \times Y\\\\B = $0.93 + 2 \times Y ------ Equation\ 5[/tex]
Substituting value of M in Equation 3,
[tex]3 \times M = B + Y[/tex]
[tex]3 \times $0.93 = B + Y [/tex]
[tex]B + Y = $2.79[/tex] ------------ Equation 6
Substituting Equation 5 in Equation 6,
[tex]B + Y = $2.79 \\ \\$0.93 + 2 \times Y + Y = $2.79[/tex]
[tex]3 \times Y = $2.79 - $0.93[/tex]
[tex]3 \times Y = $1.86[/tex]
[tex]Y = \frac{1.86}{3}[/tex]
Y = $0.62
Substituting value of F in Equation 6,
[tex]B + Y = $2.79[/tex]
[tex]B + $0.62 = $2.79[/tex]
[tex]B = $2.79 - $0.62[/tex]
B = $2.17
Substituting values of B, Y, and M in Equation 4,
[tex]B+M+2 \times Y = $X[/tex]
[tex] $2.17 + $ 0.93 +2 \times $ 0.62 = $X[/tex]
[tex] $2.17 + $ 0.93 + $1.24 = $X[/tex]
$X = $4.34
