Answer:
y =
![(1)/(5)(x^(2)-35x)](https://img.qammunity.org/2020/formulas/mathematics/high-school/aye3jvefj471sn3l9qtzvqx5xsrnqp8ell.png)
Explanation:
Let the total numbers are n.
If the average of y numbers is x then we can form an equation
![(y)/(n)=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/n5y9qduacz56rcnhkrktcdtaxkqw9rrv22.png)
⇒
![(n)/(y)=(1)/(x)](https://img.qammunity.org/2020/formulas/mathematics/high-school/89ly7pena0vtw2mftlm5lqjq43lcy0jnxr.png)
⇒ n =
--------(1)
Now 30 is added to the set of numbers then average becomes (x - 5)
![(y+30)/(n+1)=(x-5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/mjxcwjn643hqcdw20i9u7upkcs63ayuvhh.png)
⇒
![((n+1))/((y+30))=(1)/((x-5))](https://img.qammunity.org/2020/formulas/mathematics/high-school/ol7ibhdvuwwfzeqtwbcier90k7bdgpvoxy.png)
⇒ (n + 1) =
![(y+30)/(x-5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/usam59nxia5lfy56t9edliqomvkd8r75r2.png)
⇒ n =
- 1 ----- (2)
Now we equate the values of n from equation 1 and 2
=
- 1
y(x - 5) = x(y + 30) - x(x - 5) [ By cross multiplication ]
xy - 5y = xy + 30x - x² + 5x
xy - xy - 5y = 35x - x²
-5y = 35x - x²
x² - 35x = 5y
y =
![(1)/(5)(x^(2)-35x)](https://img.qammunity.org/2020/formulas/mathematics/high-school/aye3jvefj471sn3l9qtzvqx5xsrnqp8ell.png)