Prove Simultaneous Equations: $a+b+c+d \ne 0$

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• anemone
In summary, the real numbers $a,\,b,\,c$ and $d$ satisfy the given simultaneous equations and it is required to prove that $a+b+c+d \ne 0$. To prove this, we assume that $a+b+c+d=0$ and derive a contradiction by adding all the given equations and obtaining $(c+b)(bc-ad)=0$. We then consider two cases where $c=-b$ and $bc=ad$ and show that both lead to a contradiction. Therefore, $a+b+c+d$ cannot be equal to zero.
anemone
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MHB
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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 $a+b+c+d \ne 0$.

anemone said:
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 $a+b+c+d \ne 0$.

[sp]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
Hence for c=-b the equality (3) becomes : -bda+c=3 and adding to that eqaulity (4) we have:-3=0 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[/sp]

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

1. What are simultaneous equations?

Simultaneous equations are a set of equations that are solved together to find the values of multiple variables. In other words, they are equations that have multiple unknown variables and can be solved at the same time.

2. Why is it important to prove simultaneous equations?

Proving simultaneous equations is important because it allows us to confirm that the solutions we have found for the variables are correct. It also helps us to check for any errors in our calculations.

3. How do you prove that $a+b+c+d \ne 0$ in simultaneous equations?

To prove that $a+b+c+d \ne 0$ in simultaneous equations, we can use the substitution method. We substitute the values of the variables into the equations and solve for the left side. If the left side is equal to 0, then the equations are not simultaneous.

4. Can simultaneous equations have more than four variables?

Yes, simultaneous equations can have any number of variables. However, the more variables there are, the more complex the equations become and the more difficult they are to solve.

5. What are some real-life applications of simultaneous equations?

Simultaneous equations have many real-life applications, such as in engineering, economics, and physics. They can be used to model and solve problems involving multiple variables, such as calculating the optimal production levels for a company or predicting the trajectory of a projectile.

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