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:

\(\displaystyle x - 5 = 7 \Rightarrow x = 7 + 5 \Rightarrow x= 12\)

\(\displaystyle x - 5 = -7 \Rightarrow x = -7 + 5\Rightarrow x= -2\)

\(\displaystyle 5 (N - 6)- 2(N + 4) = 7 (N + 5)\)

\(\displaystyle \left (5 N - 30 \right ) - \left (2 N + 8 \right )= 7N + 35\)

\(\displaystyle 5 N - 30 - 2 N - 8= 7N + 35\)

\(\displaystyle 3 N - 3 8= 7N + 35\)

\(\displaystyle 3 N - 3 8+ 38 = 7N + 35 + 38\)

\(\displaystyle 3 N = 7N + 73\)

\(\displaystyle 3 N-7N = 7N + 73 -7N\)

\(\displaystyle -4N = 73\)

\(\displaystyle N = \frac{73}{-4}= - 18 \frac{1}{4}\)

\(\displaystyle \left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)\)

\(\displaystyle \left (4x + 7 \right ) + (2x+30) = 3x-51\)

\(\displaystyle 6x+37 = 3x-51\)

\(\displaystyle 6x+37 - 3x = 3x-51 - 3x\)

\(\displaystyle 3x+37 = -51\)

\(\displaystyle 3x = -51 - 37\)

\(\displaystyle 3x = -88\)

\(\displaystyle x = \frac{-88}{3}= -29 \frac{1}{3}\)