Step 1

The given system of equations are

x+2y=2.....(1)

and -3x-6y=-6

Step 2

To find: The solution for the given equations.

Step 3

Since, the given two equation are same just by a factor of multiple of -3 in second equation.

Hence, the given system of equations has infinitely many solutions.

ie.\(\displaystyle\frac{{1}}{{-{3}}}=\frac{{2}}{{-{6}}}=\frac{{2}}{{-{6}}}={k}\)

Step 4

Hence, the system of equations have infinite solutions.

Therefore, substituting x=k in (1) gives

k+2y=2

2y=2-k

\(\displaystyle{y}=\frac{{{2}-{k}}}{{2}}\)

(Hence for x = k, \(\displaystyle{y}=\frac{{{2}-{k}}}{{2}}\) is required solution for the system of equations.)

The given system of equations are

x+2y=2.....(1)

and -3x-6y=-6

Step 2

To find: The solution for the given equations.

Step 3

Since, the given two equation are same just by a factor of multiple of -3 in second equation.

Hence, the given system of equations has infinitely many solutions.

ie.\(\displaystyle\frac{{1}}{{-{3}}}=\frac{{2}}{{-{6}}}=\frac{{2}}{{-{6}}}={k}\)

Step 4

Hence, the system of equations have infinite solutions.

Therefore, substituting x=k in (1) gives

k+2y=2

2y=2-k

\(\displaystyle{y}=\frac{{{2}-{k}}}{{2}}\)

(Hence for x = k, \(\displaystyle{y}=\frac{{{2}-{k}}}{{2}}\) is required solution for the system of equations.)