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$$ \begin{cases} \begin{align} &\text{1) }a + 7b + 3c + 5d = 16 \\ &\text{2) }8a + 4b + 6c + 2d = -16 \\ &\text{3) }2a + 6b + 4c + 8d = 16 \\ &\text{4) }5a + 3b + 7c + d = -16 \\ \end{align} \end{cases} $$

Then $(a+d)(b+c)$ equals

(A) -4 (B) 0 (C) 16 (D) -16

Please note that this is a 1 mark MCQ sum.Solving for the linear equations will give you the answer but it will be a lengthy process.There must be some quick succinct step to reach the answer. Please help me out.

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    $\begingroup$ Hint: add equations 2 and 3 first. Repeat with 1 and 4 next. $\endgroup$ – dxiv Jan 18 '18 at 5:16
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The first and the fourth give $$6(a+d)+10(b+c)=0$$ the second and the third give $$a+b+c+d=0$$

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