To find the point where these two lines intersect, set the equations equal to each other, such that \(\displaystyle y\) is substituted with the \(\displaystyle x\) side of the second equation. Solving this new equation for \(\displaystyle x\) will give the \(\displaystyle x\)-coordinate of the point of intersection.

\(\displaystyle 3x = x - 2\)

Subtract \(\displaystyle x\) from both sides.

\(\displaystyle (3x) - x = (x - 2) - x\)

\(\displaystyle 2x = - 2\)

Divide both sides by 2.

\(\displaystyle \frac{2x}{2} = \frac{-2}{2}\)

\(\displaystyle x = - 1\)

Now substitute \(\displaystyle -1\) into either equation to find the \(\displaystyle y\)-coordinate of the point of intersection.

\(\displaystyle y = 3x\)

\(\displaystyle y= 3(-1)\)

\(\displaystyle y = -3\)

With both coordinates, we know the point of intersection is \(\displaystyle (-1,-3)\). One can plug in \(\displaystyle -1\) for \(\displaystyle x\) and \(\displaystyle -3\) for \(\displaystyle y\) in both equations to verify that this is correct.

\(\displaystyle 3x - 5y = 5\)

\(\displaystyle -2x + 5y = 0\)

Add the equations together.

\(\displaystyle 3x+(-2x)=x\)

\(\displaystyle -5y+5y=0\)

\(\displaystyle 5+0=5\)

Put the terms together to see that \(\displaystyle x+0=5\ \textup{or}\ x=5\).

Substitute this value into one of the original equaitons and solve for \(\displaystyle y\).

\(\displaystyle 3x - 5y = 5\)

\(\displaystyle 3(5) - 5y = 5\)

\(\displaystyle -5y=-10\)

\(\displaystyle y=2\)

Now we know that \(\displaystyle x=5\ \textup{and}\ y=2\), thus we can find the sum of \(\displaystyle x\) and \(\displaystyle y\).

\(\displaystyle x+y=5+2=7\)

We add the two systems of equations:

For the Left Hand Side:

\(\displaystyle (8x+3y)+(2x-3y)=10x\)

For the Right Hand Side:

\(\displaystyle 2+18=20\)

So our resulting equation is:

\(\displaystyle 10x=20\)

Divide both sides by 10:

For the Left Hand Side:

\(\displaystyle \frac{10x}{10}=x\)

For the Right Hand Side:

\(\displaystyle \frac{20}{10}=2\)

Our result is:

\(\displaystyle x=2\)

Reduce the second system by dividing by 3.

Second Equation:

\(\displaystyle 12x+6y=24\)     We this by 3.

\(\displaystyle \frac{12x}{3}+\frac{6y}{3}=\frac{24}{3}\)

\(\displaystyle 4x+2y=8\)

Then we subtract the first equation from our new equation.

First Equation:

\(\displaystyle 5x+2y=9\)

First Equation - Second Equation:

Left Hand Side:

\(\displaystyle (5x+2y)-(4x+2y)=x\)

Right Hand Side:

\(\displaystyle 9-8=1\)

Our result is:

\(\displaystyle x=1\)

We first add both systems of equations.

Left Hand Side:

\(\displaystyle (2x-y)+(x+y)=3x\)

Right Hand Side:

\(\displaystyle 2+4=6\)

Our resulting equation is:

\(\displaystyle 3x=6\)

We divide both sides by 3.

Left Hand Side:

\(\displaystyle \frac{3x}{3}=x\)

Right Hand Side:

\(\displaystyle \frac{6}{3}=2\)

Our resulting equation is:

\(\displaystyle x=2\)

We first multiply the second equation by 4.

So our resulting equation is:

\(\displaystyle x\cdot4+4y\cdot4=17\cdot4\)

\(\displaystyle 4x+16y=68\)

Then we subtract the first equation from the second new equation.

Left Hand Side:

\(\displaystyle (4x+y)-(4x+16y)=-15y\)

Right Hand Side:

\(\displaystyle 6-68=-60\)

Resulting Equation:

\(\displaystyle -15y=-60\)

We divide both sides by -15

Left Hand Side:

\(\displaystyle \frac{-15y}{-15}=y\)

Right Hand Side:

\(\displaystyle \frac{-60}{-15}=4\)

Our result is:

\(\displaystyle y=4\)

If we multiply both sides of our bottom equation by \(\displaystyle 2\), we get \(\displaystyle -10x-2y=-2\). We can now add our two equations, and eliminate \(\displaystyle y\), leaving only one variable. When we add the equations, we get \(\displaystyle 3x=9\). Therefore, \(\displaystyle x=3\). Finally, we go back to either of our equations, and plug in \(\displaystyle x=3\) so we can solve for \(\displaystyle y\).

\(\displaystyle 13(3)+2y=11\)

\(\displaystyle 39+2y=11\)

\(\displaystyle 2y=-28\)

\(\displaystyle y=-14\) 

\(\displaystyle 4x-3y=11\)

\(\displaystyle 2x+y=13\)

Solve the second equation for \(\displaystyle y\), allowing us to solve using the substitution method.

\(\displaystyle 2x+y=13\)

\(\displaystyle y=13-2x\)

Substitute for  \(\displaystyle y\) in the first equation, and solve for \(\displaystyle x\).

\(\displaystyle 4x-3(13-2x)=11\)

\(\displaystyle 4x-39+6x=11\)

\(\displaystyle 10x-39=11\)

\(\displaystyle 10x=50\)

\(\displaystyle x=5\)

Now, substitute for \(\displaystyle x\) in either equation; we will choose the second. This allows us to solve for \(\displaystyle y\).

\(\displaystyle 2x+y=13\)

\(\displaystyle 2\cdot 5+y=13\)

\(\displaystyle 10+y=13\)

\(\displaystyle y=3\)

Now we can write the solution in the notation \(\displaystyle (x,y)\), or \(\displaystyle (5,3)\).

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

This inequality could be rewritten as:

4x + 14 > 30  OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.