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The $a, b, c$ and $d$ satisfy the equations$$\begin{matrix} a & + & 7b & + & 3c & + & 5d & = &16 \\ 8a & + & 4b & + & 6c & + & 2d & = &-16 \\ 2a & + & 6b & + & 4c & + & 8d & = &16 \\ 5a & + & 7b & + & 3c & + & 5d & = &-16 \end{matrix}$$Then $(a+d)(b+c)$ equals

1. $-4$
2. $0$
3. $16$
4. $-16$

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+1 vote

Let the given equations be 1,2,3,4 in the order they are given.

Adding equation 2 and 3 -

10a + 10b + 10c + 10d = 0;    Let this be equation 5

Adding equation 1 and 4 -

6a + 10b  + 10c + 6d = 0;       Let this be equation 6

Equation 5 minus 6 :

we get : 4a + 4d =0

so (a+d) =0

if (a+d) =0 then (a+d)(b+c)=0

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