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:

\[ -4 \]

Expand the bracket:

\[ 4x+8=2x-22 \]

Rearrange:

\[ 2x=-30 \]

Therefore:

\[ x=-15 \]

From the first equation, express \(y\):

\[ y=x-3 \]

Substitute into the second equation:

\[ 2x+(x-3)=12 \]

\[ 3x-3=12 \]

\[ 3x=15 \]

\[ x=5 \]

Then:

\[ y=5-3=2 \]

The sum is:

\[ x+y=5+2=7 \]

Set each factor equal to zero:

\[ x+1=0 \Rightarrow x=-1 \]

\[ 2x+16=0 \Rightarrow x=-8 \]

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

The sum of the roots is:

\[ -1-8+5=-4 \]

Move everything to one side:

\[ x^2-5x-6=0 \]

Factorise:

\[ (x-6)(x+1)=0 \]

Therefore:

\[ x=6 \quad \text{or} \quad x=-1 \]

The larger root is:

\[ 6 \]

Expand the bracket:

\[ -2x+3=x+27 \]

Rearrange:

\[ -3x=24 \]

So:

\[ x=-8 \]

From the second equation, express \(y\):

\[ 2x-y=-1 \Rightarrow y=2x+1 \]

Substitute into the first equation:

\[ x+2(2x+1)=7 \]

\[ x+4x+2=7 \]

\[ 5x=5 \]

\[ x=1 \]

Then:

\[ y=2\cdot 1+1=3 \]

The product is:

\[ x\cdot y=1\cdot 3=3 \]

Set each factor equal to zero:

\[ x-7=0 \Rightarrow x=7 \]

\[ 3x+12=0 \Rightarrow x=-4 \]

\[ 2x-10=0 \Rightarrow x=5 \]

The sum of the roots is:

\[ 7-4+5=8 \]

Move everything to one side:

\[ x^2+3x-10=0 \]

Factorise:

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

Therefore:

\[ x=-5 \quad \text{or} \quad x=2 \]

The smaller root is:

\[ -5 \]

Expand the bracket:

\[ 4x+8=2x-14 \]

Rearrange:

\[ 2x=-22 \]

Therefore:

\[ x=-11 \]

From the first equation, express \(x\):

\[ x=y-1 \]

Substitute into the second equation:

\[ (y-1)+2y=5 \]

\[ 3y-1=5 \]

\[ 3y=6 \]

\[ y=2 \]

Then:

\[ x=2-1=1 \]

The sum is:

\[ x+y=1+2=3 \]

Set each factor equal to zero:

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

\[ 4x+32=0 \Rightarrow x=-8 \]

\[ 2x-12=0 \Rightarrow x=6 \]

The sum of the roots is:

\[ -2-8+6=-4 \]