Question

Help! 100 pts

HelpHelpHelpHelpHelpHelpHelpHelpHelpHelpHelp Help Help Help

Answer 1

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

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

• (a, b, c) = (0, -4, 4)

⠀

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\:}}`