How can you factor this portion of the equation

Question

This is a part of another person's question that was answered in the first question . I was confused on how they took $3x+4y+xy=2012$ and factored it out to $(y+3)(x+4)=2024$. If someone could explain this to me that would be great.

Answer 1

In general, suppose $xy + ax + by = c$. Since $(x+b)(y+a) =xy+ax+by+ab $,

$\begin{array}\\ (x+b)(y+a) &=xy + ax + by+ab \\ &= c+ab\\ \end{array} $.

Therefore, $(x+b)$ and $(y+a)$ are divisors of $c+ab $. By looking at all the divisors of $c+ab$, you can find all possible values of $x$ and $y$.

In your problem, $a=3$, $b=4$, and $c = 2012$, so that $c+ab = 2012+3\cdot 4 =2024 $.

Look at all the factors of $2024$ and, for each $m$ and $n$ such that $m\cdot n = 2024$, set $x+4 = n$ and $y+3 = m$.

Answer 2

$$\begin{array}{l} 3x + 4y +xy = 2012\\ 3x + 4y + xy + 12 = 2012 + 12\quad\quad\quad{\text{added 12 to both sides}}\\ \color{blue}{3x}+ \color{red}{4y}+\color{blue}{xy}+\color{red}{12} = 2024\quad\quad\quad\quad\quad{\text{factoring by grouping}}\\ x\cdot(3+y)+4\cdot(y+3) = 2024\quad\quad\quad\;\text{common factor $(y+3)$}\\ (x+4)(y+3) = 2024 \end{array} $$