Answer:
The original duration of the tour = 20 days
Step-by-step explanation:
Solution:
Total expenses for the tour = $360
Let the original tour duration be for
days.
So, for
days the total expense = $360
<em>Thus the daily expense in dollars can be given by</em> = ![\frac{360}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B360%7D%7Bx%7D)
Tour extension and effect on daily expenses.
The tour is extended by 4 days.
<em>Tour duration now</em> =
days
On extension, his daily expense is cut by $3
<em>New daily expense in dollars </em>= ![(\frac{360}{x}-3)](https://tex.z-dn.net/?f=%28%5Cfrac%7B360%7D%7Bx%7D-3%29)
Total expense in dollars can now be given as: ![(x+4)(\frac{360}{x}-3)](https://tex.z-dn.net/?f=%28x%2B4%29%28%5Cfrac%7B360%7D%7Bx%7D-3%29)
Simplifying by using distribution (FOIL).
![(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))](https://tex.z-dn.net/?f=%28x.%5Cfrac%7B360%7D%7Bx%7D%29%2B%28x%28-3%29%2B%284.%5Cfrac%7B360%7D%7Bx%7D%29%2B%284%28-3%29%29)
![360-3x+\frac{1440}{x}-12](https://tex.z-dn.net/?f=360-3x%2B%5Cfrac%7B1440%7D%7Bx%7D-12)
![348-3x+\frac{1440}{x}](https://tex.z-dn.net/?f=348-3x%2B%5Cfrac%7B1440%7D%7Bx%7D)
We know total expense remains the same which is = $360.
So, we have the equation as:
![348-3x+\frac{1440}{x}=360](https://tex.z-dn.net/?f=348-3x%2B%5Cfrac%7B1440%7D%7Bx%7D%3D360)
Multiplying each term with
to remove fractions.
![348x-3x^2+1440=360x](https://tex.z-dn.net/?f=348x-3x%5E2%2B1440%3D360x)
Subtracting
both sides
![348x-348x-3x^2+1440=360x-348x](https://tex.z-dn.net/?f=348x-348x-3x%5E2%2B1440%3D360x-348x)
![-3x^2+1440=12x](https://tex.z-dn.net/?f=-3x%5E2%2B1440%3D12x)
Dividing each term with -3.
![\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}](https://tex.z-dn.net/?f=%5Cfrac%7B-3x%5E2%7D%7B-3%7D%2B%5Cfrac%7B1440%7D%7B-3%7D%3D%5Cfrac%7B12x%7D%7B-3%7D)
![x^2-480=-4x](https://tex.z-dn.net/?f=x%5E2-480%3D-4x)
Adding
both sides.
![x^2+4x-480=-4x+4x](https://tex.z-dn.net/?f=x%5E2%2B4x-480%3D-4x%2B4x)
![x^2+4x-480=0](https://tex.z-dn.net/?f=x%5E2%2B4x-480%3D0)
Solving using quadratic formula.
For a quadratic equation: ![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Plugging in values from the equation we got.
![x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-4%5Cpm%5Csqrt%7B%284%29%5E2-4%281%29%28-480%29%7D%7D%7B2%281%29%7D)
![x=\frac{-4\pm\sqrt{16+1920}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-4%5Cpm%5Csqrt%7B16%2B1920%7D%7D%7B2%7D)
![x=\frac{-4\pm\sqrt{1936}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-4%5Cpm%5Csqrt%7B1936%7D%7D%7B2%7D)
![x=\frac{-4\pm44}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-4%5Cpm44%7D%7B2%7D)
So, we have
and ![x=\frac{-4-44}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-4-44%7D%7B2%7D)
and ![x=\frac{-48}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-48%7D%7B2%7D)
∴
and ![x=-24](https://tex.z-dn.net/?f=x%3D-24)
Since number of days cannot be negative, so we take
as the solution for the equation.
Thus, the original duration of the tour = 20 days