By the Binomial Theorem, if you expand $\displaystyle (A+B)^{n}$, writing the result in standard form, the $\displaystyle r \mathrm{th}$ term (with the terms being numbered from 0 to $\displaystyle n$ ) is

$\displaystyle C(n,r)A^{n-r}B^{r}$

Set $\displaystyle A=x$, $\displaystyle B=-3$, $\displaystyle n=10$, and $\displaystyle r=2$ (again, the terms are numbered 0 through $\displaystyle n$, so the third term is numbered 2) to get

$\displaystyle C(10,2)\cdot x^{10-2}\cdot (-3)^{2}$

$\displaystyle =\frac{10!}{8! 2!}\cdot x^{8}\cdot 9$

$\displaystyle =45\cdot x^{8}\cdot 9$

$\displaystyle =405x^{8}$

$\displaystyle \frac{(x+y)^{2} +(x - y)^{2}}{x^{2}y^{2}}$

$\displaystyle =\frac{(x^{2}+2xy+y^{2}) + (x^{2}-2xy+y^{2}) }{x^{2}y^{2}}$

$\displaystyle =\frac{x^{2}+x^{2}+2xy-2xy+y^{2}+y^{2} }{x^{2}y^{2}}$

$\displaystyle =\frac{2x^{2}+2y^{2} }{x^{2}y^{2}}$

$\displaystyle =\frac{2x^{2} }{x^{2}y^{2}} + \frac{2y^{2} }{x^{2}y^{2}}$

$\displaystyle =\frac{2 }{y^{2}} + \frac{2 }{x^{2}} = \frac{2 }{x^{2}} + \frac{2 }{y^{2}}$

Simplify both sides of the equation as much as possible, and solve for $\displaystyle \small x$ in the equation in terms of $\displaystyle \small N$:

$\displaystyle \small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\displaystyle \small 3 \cdot 4x - 3 \cdot 7 + 12 = 2 \cdot 5x - 2 \cdot 3 + N \cdot x - N \cdot 3$

$\displaystyle \small 12x - 21 + 12 = 10x - 6 + Nx - 3N$

$\displaystyle \small 12x - 9 = (10+N)x + (- 6 - 3N)$

$\displaystyle \small \small 12x - (10+N)x = (- 6 - 3N) + 9$

$\displaystyle \small (2-N)x = 3 - 3N$

$\displaystyle \small x = \frac{3 - 3N}{2 -N}$

$\displaystyle \small x$ has exactly one solution unless the denominator is 0 - that is, $\displaystyle \small N = 2$. We make sure that this value renders no solution by substituting:

$\displaystyle \small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\displaystyle \small \small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + 2 (x - 3)$

$\displaystyle \small 12x - 21 + 12 = 10x - 6 + 2x - 6$

$\displaystyle \small 12x - 9 = 12x - 12$

$\displaystyle \small - 9 = - 12$

The equation has no solution, and $\displaystyle \small N = 2$ is the correct answer.

$\displaystyle n+2=-14-n$

$\displaystyle n+n=-14-2$

$\displaystyle 2n=-16$

$\displaystyle n=-8$

$\displaystyle -6x-20=-2x+4(1-3x)$

$\displaystyle -6x-20=-2x+4-12x$

$\displaystyle -6x-20=-14x+4$

$\displaystyle -6x+14x=4+20$

$\displaystyle 8x=24$

$\displaystyle x=3$

$\displaystyle -14+6b+7-2b=1+5b$

$\displaystyle -7+4b=1+5b$

$\displaystyle 4b-5b=1+7$

$\displaystyle -b=8$

$\displaystyle b=-8$

Midpoint formula:

$\displaystyle \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$\displaystyle (1,4)\ and\ (7,10)$

$\displaystyle \left(\frac{1+7}{2},\frac{4+10}{2}\right)$

$\displaystyle \left(\frac{8}{2},\frac{14}{2}\right)$

$\displaystyle (4,7)$

Midpoint formula:

$\displaystyle \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$\displaystyle (1,2)\ and\ (5,2)$

$\displaystyle \left(\frac{1+5}{2},\frac{2+2}{2}\right)$

$\displaystyle \left(\frac{6}{2},\frac{4}{2}\right)$

$\displaystyle (3,2)$

Midpoint formula:

$\displaystyle \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$\displaystyle (-2,-1)\ and\ (-8,7)$

$\displaystyle \left(\frac{-2+-8}{2},\frac{-1+7}{2}\right)$

$\displaystyle \left(\frac{-10}{2},\frac{6}{2}\right)$

$\displaystyle (-5,3)$

We start by isolating the absolute value expression:

$\displaystyle 2\left | x-5\right |+16=30 \Leftrightarrow 2\left | x-5\right |=30-16=14\Leftrightarrow \left | x-5\right |=7$

This gives us two cases when we remove the absolute value:

$\displaystyle x - 5 = 7$ and $\displaystyle x - 5 = -7$

Then we solve for each case:

$\displaystyle x - 5 = 7 \Rightarrow x = 7 + 5 \Rightarrow x= 12$

$\displaystyle x - 5 = -7 \Rightarrow x = -7 + 5\Rightarrow x= -2$