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At the first store the price for one burger was dollars, where "n" was the number of burgers in the box.
At the second store there were (n+8) burgers in their box for the same cost of 35 dollars per box,
so the cost of one single burger was dollars.
The difference in price for one single burger is 50 cents = 0.5 dollar, which gives you an equation
- = 0.5 dollars. (1)
To solve this equation, multiply both sides by 2*n*(n+8). You will get
70(n+8) - 70n = n*(n+8).
Simplify and then solve this quadratic equation
70n + 70*8 - 70n = n^2 + 8n
n^2 + 8n - 560 = 0.
Factor left side
(n+28)*(n-20) = 0.
Only positive root n = 20 is meaningful - so it is the solution.
Answer. There were 20 burgers in the box at the first store at the price = $1.75 per one burger.
CHECK. I will check if the equation (1) is satisfied.
The price for 1 burger at the first store is = cents = 175 cents.
The price for 1 burger at the second store is dollars = = 125 cents, or exactly 0.5 dollars less.
So the problem is solved correctly (!)
Solved.
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I like this approach: it is short and straightforward, and at every step of building the base equation (1) you follow
exactly to the problem's description. It prevents you of making errors - as much as possible.
In this site, there are several lessons, where you can find similar problems explained ans solved
- Challenging word problems solved using quadratic equations
- Had they sold . . .
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Be careful: The post (the solution) by @swincher4391 has an error !