Move everything to one side:

\[ x^2+x-12=0 \]

Factorise:

\[ (x+4)(x-3)=0 \]

So the roots are:

\[ x=-4 \quad \text{or} \quad x=3 \]

The larger root is:

\[ 3 \]

Expand the bracket:

\[ -x-6=2x+18 \]

Collect the \(x\)-terms on one side and the numbers on the other:

\[ -3x=24 \]

Hence:

\[ x=-8 \]

Add the two equations:

\[ (x+y)+(2x-y)=2+(-4) \]

This gives:

\[ 3x=-2 \]

So:

\[ x=-\frac{2}{3} \]

Substitute into the first equation:

\[ -\frac{2}{3}+y=2 \]

\[ y=2+\frac{2}{3}=\frac{8}{3} \]

Therefore the product is:

\[ x\cdot y=-\frac{2}{3}\cdot \frac{8}{3}=-\frac{16}{9} \]

Set each factor equal to zero:

\[ x-3=0 \Rightarrow x=3 \]

\[ 3x+15=0 \Rightarrow x=-5 \]

\[ 2x-4=0 \Rightarrow x=2 \]

The sum of the roots is:

\[ 3+(-5)+2=0 \]

Move everything to the left-hand side:

\[ x^2-2x-24=0 \]

Factorise:

\[ (x-6)(x+4)=0 \]

Hence:

\[ x=6 \quad \text{or} \quad x=-4 \]

The smaller root is:

\[ -4 \]