cost of one coffee

Question

Kevin picked up two donuts and four large coffees for the office staff. He paid $ 5.18 . On? Thursday, Connor picked up four donuts and three large coffees for the office staff. She paid $ 5.46. What is the cost of one? donut? What is the cost of one large? coffee?

Answer 1

For this word problem, use systems of linear equations.

We will first write some equations, or number sentences, describing what is happening in the story.

 

"Kevin picked up 2 donuts and 4 large coffees for the office staff ... He paid $5.18"    

(Think of how you would calculate the total cost, if you knew the price per donut and price per cup of coffee...)

 

2 *(cost per donut) + 4 *(cost per coffee) = total

Now some variables!!! 

Let's use d for donut and c for coffee!

 

2d + 4c = 5.18   

 

Now, notice that we have two different variables in our equation above ... that means we need two equations to solve for our variables! So let's make another.

 

"On Thursday, Connor picked up 4 donuts and 3 large coffees for the office staff ... She paid $5.46"

 

4 *(cost per donut) + 3 *(cost per coffee) = total

 

4d + 3c = 5.46

 

 

>>> Now we have two different equations.  We can solve using either substitution or elimination... or graphing.  

 

ELIMINATION: Try to get both d coefficients, or both c coefficients, to be opposites ... meaning +4 and -4 (You can do that by multiplying the WHOLE equation by a number)

 

2d + 4c = 5.18    > turn 2d into 4d  > multiply by 2   > 2(2d) + 2(4c) = 2(5.18) 

4d + 3c = 5.46    > turn 4d into -4d > multiply by -1 > -1(4d) + -1(3c) = -1(5.46) 

 

 

  4d + 8c = 10.36       Then add down!

- 4d - 3c = -5.46

 

 

0d + 5c = 4.90

 

Notice that our d is  now gone!  We have only one variable! Now we solve for c!

 

5c = 4.90        (divide both sides by 5)

 

c = 0.98           (so 1 large coffee costs $0.98)

 

Now that we know c ... we can plug that back into one of the original equations to find d.

 

2d + 4c = 5.18      (NOTE: it does not matter which equation you plug it back into)

2d + 4(0.98) = 5.18

 

Now we can solve for d!

 

2d + 3.92 = 5.18       (subtract 3.92 from both sides)

2d = 1.26                  (divide both sides by 2)

d = 0.63                    (So 1 donut costs $0.63)

 

 

SOLUTION: The cost of 1 donut is $0.63 and the cost of 1 large coffee is $0.98.

 

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

 

The other method is SUBSTITUTION ... where you solve for a variable and plug it into the other equation ... shown below:

 

2d + 4c = 5.18                

4d + 3c = 5.46

 

(NOTE: This method produces more fractions, and "unpretty" numbers, than the elimination method)

 

You can solve for whichever variable you like ... lets pick d.

 

You can start with whatever equation you like ... lets pick the first one.

 

2d + 4c = 5.18              (solve for d .... means to get it by itself)

2d = 5.18 - 4c               (subtract 4c from both sides)

 

d = (5.18)/2 - (4c)/2      (divide both sides by 2)

 

d = 2.59 - 2c                  (simplify)

 

Plug this above d ... in for the d in the other equation ....

 

4d + 3c = 5.46                  (Since d = 2.59 - 2c)

 

4(2.59 - 2c) + 3c = 5.46     (Then solve for c)     (distributive property)

 

10.36 - 8c + 3c = 5.46         (combine like terms ... -8c + 3c = -5c))

 

10.36 - 5c = 5.46                 (subtract both sides by 10.36)

 

-5c = -4.90                         (divide both sides by -5)

 

c = 0.98                             (so 1 large coffee costs $0.98)

 

 

Now that we know c ... we can plug that back into one of the original equations to find d.

2d + 4c = 5.18 (NOTE: it does not matter which equation you plug it back into)
2d + 4(0.98) = 5.18

Now we can solve for d!

2d + 3.92 = 5.18 (subtract 3.92 from both sides)
2d = 1.26 (divide both sides by 2)
d = 0.63 (So 1 donut costs $0.63)


SOLUTION: The cost of 1 donut is $0.63 and the cost of 1 large coffee is $0.98.