Question

On Monday Harold picked up six donuts and two large coffees for the office staff. He paid $5.80. On Tuesday, Melinda picked up four donuts and 5 large coffees for the office staff. She paid $7.02. What is the cost of one donut? What is the cost of one large coffee?

Answer 1

Cost of tjhe donut: D

Cost of the large coffee: C

On Monday Harold picked up six donuts and two large coffees for the office staff, he paid $5.80:

$6D+2C=5.80$

On Tuesday, Melinda picked up four donuts and 5 large coffees for the office staff. She paid $7.02:

$4D+5C=7.02$

Use the next system of linear equations to find the value of D and C:

$\begin{gathered} 6D+2C=5.80 \\ 4D+5C=7.02 \end{gathered}$

1. Solve D in the first equation:

$\begin{gathered} \text{Subtract 2C in both sides of the equation:} \\ 6D+2C-2C=5.80-2C \\ 6D=5.80-2C \\ \\ \text{Divide both sides of the equation into 6:} \\ \frac{6}{6}D=\frac{5.80}{6}-\frac{2}{6}C \\ \\ D=\frac{5.80}{6}-\frac{1}{3}C \end{gathered}$

2. Substitute the D in the second equation by the equation you get in step 1:

$4(\frac{5.80}{6}-\frac{1}{3}C)+5C=7.02$

3. Solve C:

$\begin{gathered} \frac{23.2}{6}-\frac{4}{3}C+5C=7.02 \\ \\ \frac{-4C+15C}{3}=7.02-\frac{23.2}{6} \\ \\ \frac{11}{3}C=\frac{42.12-23.2}{6} \\ \\ \frac{11}{3}C=\frac{18.92}{6} \\ \\ C=\frac{3}{11}\cdot\frac{18.92}{6} \\ \\ C=\frac{56.76}{66} \\ \\ C=0.86 \end{gathered}$

4. Use the value of C=0.86 to find D;

$\begin{gathered} D=\frac{5.80}{6}-\frac{1}{3}C \\ \\ D=\frac{5.80}{6}-\frac{1}{3}(0.86) \\ \\ D=\frac{5.80}{6}-\frac{0.86}{3} \\ \\ D=\frac{17.4-5.16}{18} \\ \\ D=\frac{12.24}{18} \\ \\ D=0.68 \end{gathered}$

The solution fot the system is:

$\begin{gathered} C=0.86 \\ D=0.68 \end{gathered}$

The cost of one dount is $0.68

The cost of one large coffee is $0.86