Question

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Answer 1

Answer:

$\boldsymbol{ (a , b , c ) = (0,-4,4)}$

Step-by-step explanation:

We would like to solve the following system of equations ,

$\begin{cases} 2a + b + c = 0\\2b + c + a = -4 \\ 2c + a + b = 4 \end{cases}$

Firstly lets number the equations as ,

$\begin{cases} 2a + b + c = 0 \dots (i)\\2b + c + a = -4 \dots (ii)\\ 2c + a + b = 4 \dots (iii)\end{cases}$

Again , we can rewrite the first equation as , equations as ;

$\longrightarrow c = -2a - b \dots (iv)$

Now , substitute the value of equation (iv) in equation (ii) and (iii) .

$\longrightarrow 2b + c + a = -4$

$\longrightarrow 2b + (-2a - b) + a = -4$

$\longrightarrow 2b - 2a - b + a = -4$

$\longrightarrow (2b - b )+ (a - 2a) = -4$

$\longrightarrow b - a = -4 \dots (v)$

And ,

$\longrightarrow 2(-2a - b ) + a + b = 4$

$\longrightarrow -4a - 2b + a + b = 4$

$\longrightarrow (-4a + a )+(-2b + b) =4$

$\longrightarrow -3a - b = 4 \dots (vi)$

Now consider ,

$\begin{cases}b - a =-4\\ -3a - b =4\end{cases}$

On adding these two equations, we have ;

$\longrightarrow b - a -3a - b = -4+4$

$\longrightarrow -4a =0$

$\longrightarrow \underline{\underline{\boldsymbol{ a = 0}}}{}$

Now substitute this value in equation (v) ,

$\longrightarrow b - 0= -4$

$\longrightarrow \underline{\underline{\boldsymbol{ b = -4}}}{}$

Finally substitute the values of a and b in equation (iv) , we have ;

$\longrightarrow c = -2a - b$

$\longrightarrow c = -2(0)-(-4)$

$\longrightarrow c = 0+4$

$\longrightarrow \underline{\underline{\boldsymbol{ c = 4 }}}{}$

And we are done !

Answer 2

Answer:

• (a, b, c) = (0, -4, 4)

⠀

Step-by-step explanation:

⠀

${\longrightarrow{\qquad \sf \ 2a + b + c = 0\: \dashrightarrow \sf \: (i)}}$

${\longrightarrow{\qquad \sf \ 2b + c + a = - 4\dashrightarrow \sf \: (ii)}}$

${\longrightarrow{\qquad \sf \ 2c + a + b = 4\: \dashrightarrow \sf \: (iii)}}$

⠀

Adding equation (i), (ii) and (iii) :

${\longrightarrow{\qquad \sf \ 2a + b + c + (2b + c + a) + (2c + a + b)= 0 + ( - 4) + 4}}$

${\longrightarrow{\qquad \sf \ 3a + 3b + 3c= 0 }}$

${\longrightarrow{\qquad \sf \ 3(a + b + c)= 0 }}$

${\longrightarrow{\qquad \sf \ a + b + c= 0 \: \dashrightarrow \: (iv)}}$

⠀

Now, (i) – (ii) :

${\longrightarrow{\qquad \sf 2a + b + c - (2b + c + a) = 0 - ( - 4)\: }}$

${\longrightarrow{\qquad \sf 2a + b + c - 2b - c - a = 0 + 4\: }}$

${\longrightarrow{\qquad \sf \: a - b= 4\:\dashrightarrow \: (v) }}$

⠀

Again, (i) – (iii) :

${\longrightarrow{\qquad \sf 2a + b + c - (2c + a + b) = 0 - 4\: }}$

${\longrightarrow{\qquad \sf 2a + b + c - 2c - a - b = - 4\: }}$

${\longrightarrow{\qquad \sf \: a - c= - 4\:\dashrightarrow \: (vi) }}$

⠀

Now, (v) + (vi) :

${\longrightarrow{\qquad \sf \: a - b + (a + c)= 4 - 4\:}}$

${\longrightarrow{\qquad \sf \: 2a - b - c= 0\:\dashrightarrow \: (vii)}}$

⠀

Now, (iv) + (vii) :

${\longrightarrow{\qquad \sf \ a + b + c + (2a - b - c)= 0 + 0}} \:$

${\longrightarrow{\qquad \sf \ 3a= 0}} \:$

${\longrightarrow{\qquad \bf \ a= 0}} \:$

⠀

Now, substituting the value of a in equation (v) :

${\longrightarrow{\qquad \sf \: a - b= 4\:}}$

${\longrightarrow{\qquad \sf \: 0 - b= 4\:}}$

${\longrightarrow{\qquad \bf \: b= - 4\: }}$

⠀

Now, substituting the value of a in equation (vi) :

${\longrightarrow{\qquad \sf \: a - c= - 4\:}}$

${\longrightarrow{\qquad \sf \: 0 - c= - 4\:}}$

${\longrightarrow{\qquad \bf \: c= 4\:}}$