Step 1

Given to solve the system of equations.

The system of equations can be solved using elimination.

To eliminate \(x_{3}\), the second equation is multiplied by 2 and subtracted from the first equation.

\(x_{1}+2x_{2}+6x_{3}=6\)

\(x_{1}+x_{2}+3x_{3}=3\)

\((x_{1}+2x_{2}+6x_{3})-2(x_{1}+x_{2}+3x_{3})=6-2(3)\)

\(x_{1}+2x_{2}+6x_{3}-2x_{1}-2x_{2}-6x_{3}=6-6-x_{1}=0\)

\(x_{1}=0\)

Step 2

Plugging the value of \(x_{1}\) in the first and second equations:

It is seen that the equations after plugging the value of \(x_{1}\) are same. Hence, there are infinitely many solutions that satisfy the equation

\(x_{2} + 3x_{3} = 3\)

Hence, let \(x_{3} = t.\)

So the value of \(x_{2}\) is given by:

Hence, the solution to the system of equations is given by

\((x_{1}, x_{2}, x_{3})= (0, 3 – 3t, t)\)

\((0)+2x_{2}+6x_{3}=6 \Rightarrow 2x_{2}+6x_{3}=6 \Rightarrow x_{2}+3x_{3}=3\)

\((0)+x_{2}+3x_{3}=3 \Rightarrow x_{2}+3x_{3}=3\)

\(x_{2}+3x_{3}=3\)

\(x_{2}+3(t)=3\)

\(x_{2}+3t=3\)

\(x_{2}=3-3t\)

Step 3

Result:

\((x_{1}, x_{2}, x_{3}) = (0, 3 – 3t, t)\)