Question 48121This question is from textbook Algebra 2
: This is a Value Word Problems lesson. I learned it in chapter 19 but it doesn't help for my problem. Here it is
The expensive ones cost $7 each, whereas the worthless ones sold for only $2 each. Monongahela spent $111 and bought three more expensive ones than worthless ones. How may of each kind did she buy.
THE ANSWER IS 10 WORTHLESS ONES AND 13 EXPENSIVE ONES. I WANT TO KNOW HOW DID THEY GET THAT.
PLEASE AND THANK YOU
This question is from textbook Algebra 2
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! You are working with two different variables, x is one variable and y is the other variable. Since you have two variables, use the following formula:
Ax + By = C
.
.
A = rate, cost or percent of A = $7
x = the amount or number of A = x
B = rate, cost or percent of B = $2
y = the amount or number of B = y
C = the total amount = $111
.
Expensive DISABLED_event_ones= $7x [We do not know how many expensive items are available]
Worthless DISABLED_event_ones= $2y [We do not know how many worthless items are available]
Total amount spent = C = $111
.
Ax + By = C [Formula for linear equations]
7x + 2y = 111 [Plug-in the values]
.
But, there is one more piece of information -- "...and bought three more expensive ones than worthless ones".
This translate into:
Number of expensive ones (x) = Number of Worthless ones (y) + 3 more
x = y + 3 [Substitute this (y + 3) in the equation]
.
7(y+3) + 2y = 111 [Simplify]
7y + 21 + 2y = 111 [Solve for y]
9y + 21 = 111
9y + 21 - 21 = 111 - 21
9y = 90
9y/9 = 90/9
y = 10
.
Plug (y=10) back into the original equation and solve for x]
7x + 2y = 111
7x + 2(10) = 111
7x + 20 = 111
7x + 20 - 20 = 11 - 20
7x = 91
7x/7 = 91/7
x = 13
.
Check by plugging all of the values back into the original equation.
7x + 2y = 111
7(13) + 2(10) = 111
7(13) + 2(10) = 111
111 = 111 [Checks out]
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