Let $a$ , $b$ , $c$ and $d$ satisfy the following equations
$a+7b+3c+5d=0$
$8a+4b+6c+2d=-16$
$2a+6b+4c+8d=16$
$5a+3b+7c+d=-16$
Then $(a+d)(b+c)$ equals?
Options: $16$ , $-16$ , $0$ or none of these
I know it can be solved by adjusting the equations through elimination or substitution to find the four variables. But I was wondering if there is a less tedious method?
On
You can solve this by inspection with clever combinations of the equations, as Key Flex did.
Another way is to define the intermediate varibles $m:=a+d,n:=b+c$, which are what we are looking for, and eliminate $c, d$. We have
$$\begin{cases}-4a&+4b&+3n&+5m&=0, \\6a&-2b&+6n&+2m&=-16, \\-6a&+2b&+4n&+8m&=16, \\4a&-4b&+7n&+m&=-16.\end{cases}$$
Now we eliminate $a$ and $b$,
$$\begin{cases}10n&+6m&=-16, \\10n&+10m&=0.\end{cases}$$
and the solution is $(-4)\cdot 4$.
I think finding the values of $a,b,c,d$ through elimination and substitution is only the easy and less tedious way which require basic calculation skills. Or else you need to do gauss elimination in order to find the variables $a,b,c,d$ which is a tedious process.
$$a + 7b + 3c + 5d = 0 ………… (1)$$$$8a + 4b + 6c + 2d = -16 ……. (2)$$$$2a + 6b + 4c + 8d = 16 ……… (3)$$$$5a + 3b + 7c + d = -16 ……….. (4)$$Adding equations $(2)$ and $(3)$ we get, $10a + 10b + 10c + 10d = 0$$$a + b + c + d = 0$$$$(a + d) = -(b + c) ……………. (5)$$Adding equations $(1)$ and $(4)$ we get, $6a + 10b + 10c + 6d = -16$ $$6(a + d) + 10(b + c) = -16$$$$ -6(b + c) + 10(b + c) = -16 (\mbox{ from} (5))$$$$4(b + c) = -16$$$$(b + c) = -4$$$$(a + d) = 4$$$$(a + d)(b + c) = 4(-4) = -16$$ $$\mbox{Therefore, } (a + d)(b + c) =-16$$