\(\displaystyle 13+X-Y=2Y\)

We need to isolate \(\displaystyle X\). Move all other terms to the right hand side of the equation:

\(\displaystyle 13+X-Y-13+Y=2Y-13+Y\)

Combine like terms:

\(\displaystyle X=-13+3Y\)

\(\displaystyle |x-4|=-3x+5\)

To solve this absolute value equation, we need to set the right side of the equation equal to a positive and negative version in order to calculate the two options for absolute value.

\(\displaystyle x-4=-3x+5\)   OR      \(\displaystyle x-4=-(-3x+5)=3x-5\)

\(\displaystyle x+3x=5+4\)       OR      \(\displaystyle x-3x=-5+4\)

\(\displaystyle 4x=9\)                        OR      \(\displaystyle -2x=-1\)

\(\displaystyle x=\frac{9}{4}\)                          OR     \(\displaystyle x=\frac{1}{2}\)

We first solve each of the equations to find \(\displaystyle m\) and \(\displaystyle n\):

\(\displaystyle 4-n=5\)

\(\displaystyle -n=5-4\)

\(\displaystyle n=-1\)

\(\displaystyle 5m-8=12+3m\)

\(\displaystyle 5m-3m=12+8\)

\(\displaystyle 2m=20\)

\(\displaystyle m=10\)

\(\displaystyle m-n=10-(-1)=10+1=11\)

Therefore, \(\displaystyle m-n=11\).

First, we need to solve the provided equation for \(\displaystyle x\).

\(\displaystyle x=\frac{x-6}{3+x}\)

\(\displaystyle x(3+x)=x-6\)

\(\displaystyle 3x+x^{2}=x-6\)

\(\displaystyle 3x+x^{2}-x=-6\)

\(\displaystyle x^{2}+2x+6=0\)

We want to find the value of \(\displaystyle x^{2}+2x+3\). We can put the equation in this form by subtracting \(\displaystyle 3\) from each side:

\(\displaystyle x^{2}+2x+6-3=-3\)

\(\displaystyle x^{2}+2x+3=-3\)

Since we are solving an absolute value equation, \(\displaystyle \left | 2x+12\right |=20\), we must solve for both potential values of the equation:

1.) \(\displaystyle 2x+12=20\)

2.) \(\displaystyle 2x+12=-20\)

Solving Equation 1:

\(\displaystyle 2x+12=20\)

\(\displaystyle 2x=8\)

\(\displaystyle x=4\)

Solving Equation 2:

\(\displaystyle 2x+12=-20\)

\(\displaystyle 2x=-32\)

\(\displaystyle x=-16\)

Therefore, for \(\displaystyle \left | 2x+12\right |=20\), \(\displaystyle x=4\) or \(\displaystyle x=-16\). 

In order to solve \(\displaystyle 3y-4x=7x+33\) for \(\displaystyle y\), we need to isolate \(\displaystyle y\) on one side of the equation:

\(\displaystyle 3y-4x=7x+33\)

\(\displaystyle 3y=11x+33\)

\(\displaystyle y=\frac{11}{3}x+11\)

In order to find values of \(\displaystyle x\) and \(\displaystyle y\) for which \(\displaystyle 7x+3y=22\), we need to plug the values into the equation:

1.) \(\displaystyle x=1, y=5\): \(\displaystyle 7(1)+3(5)=7+15=22\) Correct

2.)\(\displaystyle x=5, y=1\): \(\displaystyle 7(5)+3(1)=35+3=38\) Incorrect

3.) \(\displaystyle x=2, y=3\): \(\displaystyle 7(2)+3(3)=14+9=23\) Incorrect

4.) \(\displaystyle x=4, y=2\): \(\displaystyle 7(4)+3(2)=28+6=34\) Incorrect

Therefore, the only correct answer is \(\displaystyle x=1, y=5\). 

In order to solve the equation for \(\displaystyle x\), we need to isolate \(\displaystyle x\) on one side of the equation:

\(\displaystyle \frac{1}{3}x+\frac{2}{3}y=\frac{7}{9}\)

\(\displaystyle \frac{1}{3}x=\frac{7}{9}-\frac{2}{3}y\)

\(\displaystyle x=\frac{21}{9}-2y\)

Reducing the fraction,

\(\displaystyle x=\frac{7}{3}-2y\)

\(\displaystyle 3d + ad + y = 10\)

\(\displaystyle 3d + ad + y - y= 10 - y\)

\(\displaystyle 3d + ad = 10 - y\)

\(\displaystyle \left (3 + a \right )d = 10 - y\)

\(\displaystyle \frac{\left (3 + a \right )d }{3+a}= \frac{10 - y }{3+a}\)

\(\displaystyle d= \frac{10 - y }{3+a}\)

\(\displaystyle v = 1+ \sqrt[3]{w-7}\)

\(\displaystyle v- 1 = 1+ \sqrt[3]{w-7} - 1\)

\(\displaystyle \sqrt[3]{w-7} = v- 1\)

\(\displaystyle \left (\sqrt[3]{w-7}\right )^{3} = \left (v- 1 \right )^{3}\)

\(\displaystyle w-7= v^{3} -3v^{2}+3v-1\)

\(\displaystyle w-7+ 7= v^{3} -3v^{2}+3v-1 + 7\)

\(\displaystyle w = v^{3} -3v^{2}+3v+6\)