Find two numbers whose sum is 27 and product is 182.

Hint: First assume any two variables for the 2 required numbers. Now by using the first condition find a relation between these 2. Then by using the second condition find another relation between two variables. By using these 2 relations find their values. The values of these variables are the required result.

Complete step-by-step answer:
Given conditions in the question, can be written here as :
Sum of two random numbers is given to be 27.
Product of above two numbers is given to be as 182.
Let us assume the first number to be denoted by x.
Let us assume the second number to be denoted by y.
By writing the first condition, mathematically, we get it as:
\[x+y=27\] ……… (i)
By writing the second condition, mathematically, we get it as:
\[xy=182\] ……… (ii)
By Subtracting y on both sides of equation (i), we get it as:
\[x+y-y=27-y\]
By cancelling common term, we can write the equations in form of:
\[x=27-y\] ……… (iii)
By Substituting equation (iii) into the equation (ii), we get it as.
\[\left( 27-y \right)y=182\]
Distributive law is defined as product of 2 terms, given by:
\[a\left( b\text{ }+\text{ }c \right)\text{ }=\text{ }ab+ac\]
By using above law we can write the above equation as:
\[27y-{{y}^{2}}=\text{ }182\]
By adding ${{y}^{2}}$ and subtracting \[27y\] on both sides, we get
\[{{y}^{2}}-27y+182=0\]
Factorizations: For factoring a quadratic follow the steps:
Allot ${{x}^{2}}$ coefficient as "a", x co-efficient as "b", Constant as "c”.
Find the product of the 2 numbers a ,c. Let it be K.
Find 2 numbers such that their product is K, Sum is b.
Rewrite the term \[bx\]in terms of those 2 numbers.
Now factor the first two terms.
Next factor the last two terms.
If we do it correctly our 2 new terms will have one more common factor.
Now take that common to get quadratic as a product of 2 terms, for which you get the roots.
By comparing it to $a{{x}^{2}}+bx+c=0$, we get values of a, b, c, as:
a=1, b=-27,c=182.
Two numbers whose product is 182, sum is 27, are -13, -14.
By writing -27y as -13y-14y, we get the equation as:
${{y}^{2}}-13y-14y+182=0$
By factoring, we get the equation as product given by:
$(y-13)(y-14)=0$
By the above equation, we can say roots of y are given by.
y=13, 14.
By substituting y=13 in equation (iii), we get x as:
$x=27-13=14$
By substituting y=14, in equation (iii), we get x as:
$x=27-14=13$
So, the 2 roots satisfying the condition all given by:
(13, 14); (14,13). Therefore these are solutions.

Note: Alternate method is to divide with x on both sides of equation (ii), then substitute y in terms of x into equation (i), any ways you get the same quadratic in y. Also you can write -27y as -14y-13y instead of -13y-14y any ways you will have the same roots for y.