Question

Find the value of $2da+d+a,$ if
$2ab+a+b=3$, $2bc+b+c=4$, $2cd+c+d=-5.$

Answer 1

$\Huge\text{3}$

$\large\underline{\large\underline{\text{Topic}}}$

Algebra- Simon's favorite factorization trick

$\large\underline{\large\underline{\text{Explanation}}}$

Given that,

$\cdots\longrightarrow\begin{cases} 2ab+a+b=3 \\ 2bc+b+c=4 \\ 2cd+c+d=-5. \end{cases}$

Now, it is time for Simon's favorite factorize factorization. We multiply two and add 1.

$\cdots\longrightarrow\begin{cases} 4ab+2a+2b+1=7 \\ 4bc+2b+2c+1=9 \\ 4cd+2c+2d+1=-9. \end{cases}$

By Simon's favorite factorization,

$\cdots\longrightarrow\begin{cases} (2a+1)(2b+1)=7\ &\cdots(1) \\ (2b+1)(2c+1)=9\ &\cdots(2) \\ (2c+1)(2d+1)=-9. &\cdots(3) \end{cases}$

By $(3)\times(1)\div(2),$

$\cdots\longrightarrow\underline{(2d+1)(2a+1)=-7.}$

We know how to rewrite $(2d+1)(2a+1)$ from the above. Reverse the calculations by subtracting and dividing.

Accordingly,

$\cdots\longrightarrow(2d+1)(2a+1)=7$

$\cdots\longrightarrow4da+2d+2a+1=7$

$\cdots\longrightarrow4da+2d+2a=6$

$\cdots\longrightarrow\underline{2da+d+a=3.}$

So,

$\cdots\longrightarrow\boxed{2da+d+a=3.}$