Question

a coffee shop sells regular coffee in three sizes. the table shows the sizes of each regular coffee sold throughout the day. write and solve a system of equations to determine the price of each regular coffee.

bajo: $1.60; mediano: $1.82; grande: $2.00
bajo: $1.61; mediano: $1.77; grande: $2.03
bajo: $1.62; mediano: $1.80; grande: $2.03
bajo: $1.65; mediano: $1.79; grande: $2.07

Explanation:

Step1: Set up equations

Let \(x\) be the price of Bajo, \(y\) be the price of Mediano, and \(z\) be the price of Grande.
From the table:
\(55x + 53y+54z=294.12\) (Equation 1)
\(17x + 22y+25z=117.89\) (Equation 2)
\(30x + 23y+29z=148.87\) (Equation 3)

Step2: Eliminate one variable

Multiply Equation 2 by \(3\): \(51x + 66y+75z = 353.67\) (Equation 4)
Subtract Equation 1 from Equation 4: \((51x + 66y+75z)-(55x + 53y+54z)=353.67 - 294.12\)
\(- 4x+13y + 21z=59.55\) (Equation 5)

Multiply Equation 2 by \(2\): \(34x + 44y+50z=235.78\) (Equation 6)
Subtract Equation 3 from Equation 6: \((34x + 44y+50z)-(30x + 23y+29z)=235.78 - 148.87\)
\(4x+21y+21z = 86.91\) (Equation 7)

Step3: Solve the new system

Add Equation 5 and Equation 7: \((-4x + 13y+21z)+(4x+21y + 21z)=59.55 + 86.91\)
\(34y+42z=146.46\), simplify to \(17y + 21z=73.23\) (Equation 8)

Multiply Equation 2 by \(5\): \(85x+110y + 125z=589.45\) (Equation 9)
Multiply Equation 3 by \( \frac{17}{30}\): \(17x+\frac{391}{30}y+\frac{493}{30}z=\frac{148.87\times17}{30}\approx84.27\) (Equation 10)
Subtract Equation 10 from Equation 9 (a more complex elimination, but we can also test the options)

Step4: Test the options

For option C: \(x = 1.62\), \(y=1.80\), \(z = 2.03\)
\(55\times1.62+53\times1.80 + 54\times2.03=55\times1.62+53\times1.80+54\times2.03\)
\(=89.1+95.4+109.62=294.12\)
\(17\times1.62+22\times1.80+25\times2.03=27.54+39.6+50.75 = 117.89\)
\(30\times1.62+23\times1.80+29\times2.03=48.6+41.4+58.87=148.87\)

Answer:

bajo: \(\$1.62\); mediano: \(\$1.80\); grande: \(\$2.03\) (Option C)