Can you please help?
At a store, you can buy three packs of heavy duty batteries and six boxes of cereal for $48.00. You can also buy two packs of heavy-duty batteries and eight boxes of cereal for $46.00. What is the cost of one pack of heavy duty batteries at this store.
3x +6y = $48.00
2x + 8y = $46.00
3 (3x + 7y )=144
9x + 21y = 144
2 (2x + 8y) =92
4x + 16y =92
is this all incorrect?
don't make it so complicated, all you want is the coefficients of either the x's or the y's to be the same.
3x +6y = $48.00 , divide by 3 --> x + 2y = 16
2x + 8y = $46.00 , divide by 2 --> x + 4y = 23
subtract them:
2y = 7
y = 3.5
then in x+2y = 16
x + 7 = 16
x = 9
To solve this problem, we will need to use a system of equations. However, the equations you provided seem to contain some errors. Let's try to correct them.
Let's use the variables x and y to represent the cost of one pack of heavy-duty batteries and one box of cereal, respectively.
From the given information, we can set up the following two equations:
Equation 1: 3x + 6y = $48.00
This equation represents the total cost of buying three packs of batteries and six boxes of cereal.
Equation 2: 2x + 8y = $46.00
This equation represents the total cost of buying two packs of batteries and eight boxes of cereal.
Now, we can solve this system of equations.
To eliminate one variable, we can multiply Equation 1 by 2 and Equation 2 by 3, which gives us:
Equation 3: 6x + 12y = $96.00
Equation 4: 6x + 24y = $138.00
Next, let's subtract Equation 3 from Equation 4:
(6x + 24y) - (6x + 12y) = $138.00 - $96.00
6x - 6x + 24y - 12y = $42.00
Simplifying:
12y = $42.00
y = $42.00 / 12
y = $3.50
Now, substitute y = $3.50 back into either Equation 1 or Equation 2 to solve for x.
Using Equation 1:
3x + 6($3.50) = $48.00
3x + $21.00 = $48.00
3x = $48.00 - $21.00
3x = $27.00
x = $27.00 / 3
x = $9.00
Therefore, the cost of one pack of heavy-duty batteries at this store is $9.00.