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[SOLVED] Simultaneous equations

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anemone

MHB POTW Director
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Feb 14, 2012
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The real numbers $a,\,b,\,c$ and $d$ satisfy the following simultaneous equations:

$abc-d=1\\bcd-a=2\\cda-b=3\\dab-c=-6$

Prove that $a+b+c+d \ne 0$.
 

solakis

Active member
Dec 9, 2012
352
The real numbers $a,\,b,\,c$ and $d$ satisfy the following simultaneous equations:

$abc-d=1\\bcd-a=2\\cda-b=3\\dab-c=-6$

Prove that $a+b+c+d \ne 0$.
abc-d=1.......................(1)

bcd-a=2..........................(2)

cda-b=3.............................(3)

dab-c=-6.............................(4)

Supose a+b+c+d=0................(5)
then by adding all the above equations we get:
abc+bcd+cda+dab=0=> bc(a+d)+da(c+b)=0......................(6)
But from(5) we have :a+d=-(c+b)......................................................(7)
Substituting (7) into (6) we have : -bc(c+b)+da(c+b)=0=>bc(c+b)-da(c+b)=0=>(c+b)(bc-ad)=0.........................(8)
This equality now must give us a contradiction
But(c+b)(bc-ad)=0.=> c=-b or bc=ad
Hence for c=-b the equality (3) becomes : -bda+c=3 and adding to that eqaulity (4) we have:-3=0 a contradiction

Now we have find out if bc=ad give us a contradiction
Now again from(5) we have : $(b+c)^2=(-(a+d))^2\Leftrightarrow b^2+c^2+2bc=a^2+d^2+2ad$................(9)
But since bc=ad (9) becomes :$(b^2-a^2)+(c^2-d^2)=0\Rightarrow (b+a)(b-a)+(c+d)(c-d)=0$.................................(10)
But from (5) again we have: (b+a)=-(c+d)...........................................................................(11)
And substituting (11) into (10)we have :-(c+d)(b-a)+(c+d)(c-d)=0=>(c+d)(c-d)-(c+d)(b-a)=0=>(c+d)(c-d-(b-a))=0=>(c+d)(c-d-b+a)=(c+d)(c+a-(d+b))=(c+a)(-2(d+b))=0 since from (5) we have (c+a)=-(d+b)

Hence(c+a)(d+b)=0 => c=-a or d=-b
For c=-a (2) becomes -dab+c=2 and adding to it equality (6) wehave: -4=0 a contradiction
for d=-b (3) becomes -cba+d=2 and adding to it equality(1) we have : 4=0 a contradiction again

Hence a+b+c+d is different from zero
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,812
Thanks for participating, solakis !

Suppose $a+b+c+d=0$.

Adding the original equations yields $abc+bcd+cda+dab=0$.

Replacing $d=-a-b-c$, we get

$-b^2c-bc^2-a^2c-ac^2-abc-a^2b-ab^2-abc=0\\-(a+b)(b+c)(c+a)=0$

So we must have either $a=-b,\,a=-c$ or $a=-d$ (since $b+c=0\implies a+d=0)$

$a=-b$ gives $bcd+b=2$ and $-bcd-b=3$, leads to a contradiction.
$a=-c$ gives $bcd+c=2$ and $-bcd-c=-6$, leads to a contradiction.
$a=-d$ gives $bcd+d=2$ and $-bcd-d=1$, leads to a contradiction.

Therefore, $a+b+c+d\ne 0$.