originally, the cost was to be divided among all members
![(1400)/(m)=x](https://img.qammunity.org/2023/formulas/mathematics/college/b5bdn5b0g98p89c0jjfhoz64ruue6e3hhr.png)
Where m is the number of members and x was the cost per person originally
When 5 people decided not to go, x was increased $14 and m decreased 5...
![(1400)/(m-5)=x+14](https://img.qammunity.org/2023/formulas/mathematics/college/c1uv6luz934na0dhnzo1pbybbjwez90f17.png)
Now, we have two equations that we need to solve...
![(1400)/(m)=x,\: (1400)/(m-5)=x+14](https://img.qammunity.org/2023/formulas/mathematics/college/5arcz6a84v1vej9r0joxplbog078q85zfn.png)
![\begin{gathered} (1400)/(m)=x \\ - \\ \underline{(1400)/(m-5)=x+14} \\ (1400)/(m)-(1400)/(m-5)=x-\mleft(x+14\mright) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6abuihex6ebbfuwrb4w52tendnsmtu9hed.png)
if we subtract the two expressions, we get the above
Now, we simplify...
![\begin{gathered} (1400)/(m)-(1400)/(m-5)=x-\mleft(x+14\mright) \\ -(7000)/(m\left(m-5\right))=-14 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p6s4enynrtqlo2wzu7itexpg0qfjqigjf7.png)
Now, we solve for m
![\begin{gathered} -(7000)/(m\left(m-5\right))=-14 \\ -(7000)/(m\left(m-5\right))m\mleft(m-5\mright)=-14m\mleft(m-5\mright) \\ -7000=-14m\mleft(m-5\mright) \\ -7000=-14m^2+70m \\ -14m^2+70m=-7000 \\ -14m^2+70m+7000=-7000+7000 \\ -14m^2+70m+7000=0 \\ m_(1,\: 2)=(-70\pm√(70^2-4\left(-14\right)\cdot\:7000))/(2\left(-14\right)) \\ m_(1,\: 2)=(-70\pm\:630)/(2\left(-14\right)) \\ m_1=(-70+630)/(2\left(-14\right)),\: m_2=(-70-630)/(2\left(-14\right)) \\ m1=-20,\: m2=25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o3nf349jvodzpu93idb7y6wyn6ppcu2wo0.png)
we obtain that m=25, meaning there are 25 members, so 25-5=20, this is, 20 people went to the expedition