Question #0e32e

Let #c=# number of ceramic cups
Let #p=# number of plastic cups
#c+p=400#
#4c+3.5c=1456#
Multiply #c+p=400# by 3.5, so we get #3.5c+3.5p= 1400#
Use the process of elimination, so
#4c+3.5c=1456# #-#
#3.5c+3.5p= 1400#
which will equate to
#0.5c= 56# Simply divide both sides by 0.5 to get
#c=112#, which means 112 ceramic cups were sold. Now, go back to our earlier equation of #c+p=400# and substitute in our new value of #c#, so it becomes
#112+p=400# Subtract 112 from both sides to get
#p=288#, showing that 288 plastic cups were sold.

For the last bit, multiply the number of plastic cups by their price, and do the same with the ceramic cups to find out the dollar value of each type of cup sold.
#4*112= $448#,
#3.5*288= $1008#

To solve this problem we will have to lay some ground work. First, we identify the variables.

Let x = the number of plastic cups sold
Let y = the number of ceramic cups sold

Now we need 2 equations using these variables.

Because we know that there were 400 total cups sold, we know that

x + y = 400

We also know that the total cost of both types of cups is $1456. Because each plastic cup costs $3.50 and each ceramic cup costs $4.00, we can say that

3.5x + 4y = 1456

Now that we have these things, we can start solving.

1. Subtract y from both sides of the first equation to solve for x.
   x = 400 - y

2. Substitute this for x in the second equation.
   3.5 ( 400 - y ) + 4y = 1456

3. Distribute.
   1400 - 3.5y + 4y = 1456

4. Combine like terms and subtract 1400 from both sides.
   0.5y = 56

5. Multiply both sides by 2
   y = 112

Now that you have the number of ceramic cups, you can subtract it from 400 via the first equation to find that the number of plastic cups is 288.

These numbers can be used in turn to find the revenue generated by each type of cup.

$3.50 multiplied by 288 is $1008.
$4.00 multiplied by 112 is $448.

I do hope you have found this to be helpful. Let me know if you need more or if I was mistaken somehow.

The store sold #P# plastic cups and #C# ceramic cups to a total of 400 cups altogether. We can write that as:

#P+C=400#

Then: #C=400-P to 1)#

The price of the cups was #$3.5# for the plastic cups and #$4.# for the ceramic cups to a total of #$1456#. We can write that as:

#cancel$3.5P+cancel$4C = cancel$1456 to# Substitute #1)# for #C# into here:

#3.5P + 4(400-P) = 1456#

#3.5P + 1600-4P = 1456#

#cancel-0.5P = cancel-144#

#P = 288#

#P+C=400 to# substitute value for #P# here:

#288+C=400#

#C=400-288#

#C=112#

Dollar value of #P# cups sold is:

#$3.5P = $3.5(288) = $1008#

Dollar value of #C# cups sold is:

#$4C = $4(112) = $448#

Total dollar value of #P+C# cups sold is:

#$1008 + $448 = $1456#

#$1456 = $1456#

And because that addition equalized, then we know the answers above to both quantity sold and dollar value are correct.