$\begin{array}{lll}& & -\frac{4x^2+4x}{\left(2x^2+8x+6\right)\left(x+2\right)}\\ \text{•}& & \text{First, factor the numerator and denominator. If we can simplify by dividing terms, we do so.}\\ & & -\frac{2x}{\left(x+3\right)\left(x+2\right)}\\ \text{•}& & \text{Use the factors to write an equation for the partial fraction decomposition}\\ & & -\frac{2x}{\left(x+3\right)\left(x+2\right)}=\frac{A}{x+2}+\frac{B}{x+3}\\ \text{•}& & \text{Multiply by the LHS's denominator.}\\ & & -2x=\left(\frac{A}{x+2}+\frac{B}{x+3}\right)\left(x+3\right)\left(x+2\right)\\ \text{•}& & \text{Expand everything out.}\\ & & -2x=Ax+Bx+3A+2B\\ \text{•}& & \text{Equate the coefficients to create equations}\\ & & \left[0=3A+2B\,\mathrm{-2}=A+B\right]\\ \text{•}& & \text{Solve the equations.}\\ & & \left[A=4\,B=\mathrm{-6}\right]\\ \text{•}& & \text{Use these values in the partial fraction decomposition and simplify. The final partial fraction decomposition is}\\ & & -\frac{6}{x+3}+\frac{4}{x+2}\end{array}$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \ln \left(x^2\right)=\frac{{\ln \left(x^2\right)}^2}{2}\\ \text{•}& & \text{Simplify}\\ & & \ln \left(x^2\right)-\frac{{\ln \left(x^2\right)}^2}{2}=0\\ \text{•}& & \text{Factor}\\ & & \ln \left(x^2\right)-\frac{{\ln \left(x^2\right)}^2}{2}=0\\ \text{•}& & \text{Remove rationals and common factor}\\ & & -\frac{1}{2}\cdot \ln \left(x^2\right)\cdot \left(-2+\ln \left(x^2\right)\right)\\ \text{•}& & \text{The}\text{1st}\text{factor is}\ln \left(x^2\right)\text{which implies}\ln \left(x^2\right)\text{= 0 is a solution}\\ & & \ln \left(x^2\right)=0\\ \text{•}& & \text{Set}\text{2nd}\text{factor}-2+\ln \left(x^2\right)\text{to 0 to solve}\\ & & -2+\ln \left(x^2\right)=0\\ \text{•}& & \text{Solution of}-2+\ln \left(x^2\right)=0\\ & & \ln \left(x^2\right)=2\\ \text{•}& & \text{Solution}\\ & & \ln \left(x^2\right)=\left(0\,2\right)\\ \text{•}& & \text{Consider}\text{1st}\text{solution}\\ & & \ln \left(x^2\right)=0\\ \text{•}& & \text{Raise both sides to a power of e}\\ & & {\ⅇ}^{\ln \left(x^2\right)}={\ⅇ}^0\\ \text{•}& & \text{Log rule:}{\ⅇ}^{\ln \left(X\right)}=X\\ & & x^2={\ⅇ}^0\\ \text{•}& & \text{Simplify}\\ & & x^2=1\\ \text{•}& & \text{Take Square root of both sides}\\ & & x=\pm \sqrt{1}\\ \text{•}& & \text{Consider}\text{2nd}\text{solution}\\ & & \ln \left(x^2\right)=2\\ \text{•}& & \text{Raise both sides to a power of e}\\ & & {\ⅇ}^{\ln \left(x^2\right)}={\ⅇ}^2\\ \text{•}& & \text{Log rule:}{\ⅇ}^{\ln \left(X\right)}=X\\ & & x^2={\ⅇ}^2\\ \text{•}& & \text{Take Square root of both sides}\\ & & x=\pm \sqrt{{\ⅇ}^2}\\ \text{▫}& & \text{Check if}x=\mathrm{-1}\text{satisfies domain requirements}\\ & \text{◦}& \text{Domain requirement from}\ln \left(x^2\right)\text{, cannot take log of a negative number}\\ & & 0\le x^2\\ & \text{◦}& \text{Substitute}x=\mathrm{-1}\text{into}{x}^2\text{and evaluate}\\ & & {\left(\mathrm{-1}\right)}^2\\ & \text{◦}& \text{Evaluate exponents}\\ & & 1\\ & \text{◦}& \text{Domain requirement met}\\ & & 0\le 1\\ & & \text{Therefore the domain requirement is satisfied.}\\ & & \text{✓}\\ \text{•}& & \text{Substitute solution into equation and check if left-side = right-side}\\ & & 0=0\\ \text{•}& & \text{left-side = right-side}\\ & & 0=0\\ \text{•}& & x=\mathrm{-1}\text{satisfies all the domain requirements and left-side = right-side, it is a solution}\\ & & \text{✓}\\ \text{▫}& & \text{Check if}x=1\text{satisfies domain requirements}\\ & \text{◦}& \text{Domain requirement from}\ln \left(x^2\right)\text{, cannot take log of a negative number}\\ & & 0\le x^2\\ & \text{◦}& \text{Substitute}x=1\text{into}{x}^2\text{and evaluate}\\ & & 1^2\\ & \text{◦}& \text{Domain requirement met}\\ & & 0\le 1\\ & & \text{Therefore the domain requirement is satisfied.}\\ & & \text{✓}\\ \text{•}& & \text{Substitute solution into equation and check if left-side = right-side}\\ & & 0=0\\ \text{•}& & \text{left-side = right-side}\\ & & 0=0\\ \text{•}& & x=1\text{satisfies all the domain requirements and left-side = right-side, it is a solution}\\ & & \text{✓}\\ \text{▫}& & \text{Check if}x=\sqrt{{\ⅇ}^2}\text{satisfies domain requirements}\\ & \text{◦}& \text{Domain requirement from}\ln \left(x^2\right)\text{, cannot take log of a negative number}\\ & & 0\le x^2\\ & \text{◦}& \text{Substitute}x=\sqrt{{\ⅇ}^2}\text{into}{x}^2\text{and evaluate}\\ & & {\left(\sqrt{{\ⅇ}^2}\right)}^2\\ & \text{◦}& \text{Evaluate exponents}\\ & & {\ⅇ}^2\\ & \text{◦}& \text{Evaluate}\\ & & 7.389056099\\ & \text{◦}& \text{Domain requirement met}\\ & & 0\le 7.389056099\\ & & \text{Therefore the domain requirement is satisfied.}\\ & & \text{✓}\\ \text{•}& & \text{Substitute solution into equation and check if left-side = right-side}\\ & & 2=2\\ \text{•}& & \text{left-side = right-side}\\ & & 2=2\\ \text{•}& & x=\sqrt{{\ⅇ}^2}\text{satisfies all the domain requirements and left-side = right-side, it is a solution}\\ & & \text{✓}\\ \text{▫}& & \text{Check if}x=-\sqrt{{\ⅇ}^2}\text{satisfies domain requirements}\\ & \text{◦}& \text{Domain requirement from}\ln \left(x^2\right)\text{, cannot take log of a negative number}\\ & & 0\le x^2\\ & \text{◦}& \text{Substitute}x=-\sqrt{{\ⅇ}^2}\text{into}{x}^2\text{and evaluate}\\ & & {\left(-\sqrt{{\ⅇ}^2}\right)}^2\\ & \text{◦}& \text{Evaluate exponents}\\ & & {\ⅇ}^2\\ & \text{◦}& \text{Evaluate}\\ & & 7.389056099\\ & \text{◦}& \text{Domain requirement met}\\ & & 0\le 7.389056099\\ & & \text{Therefore the domain requirement is satisfied.}\\ & & \text{✓}\\ \text{•}& & \text{Substitute solution into equation and check if left-side = right-side}\\ & & 2=2\\ \text{•}& & \text{left-side = right-side}\\ & & 2=2\\ \text{•}& & x=-\sqrt{{\ⅇ}^2}\text{satisfies all the domain requirements and left-side = right-side, it is a solution}\\ & & \text{✓}\\ \text{•}& & \text{Exact solutions}\\ & & x=\left\{\mathrm{-1}\,1\,\sqrt{{\ⅇ}^2}\,-\sqrt{{\ⅇ}^2}\right\}\\ \text{•}& & \text{Approximate solutions}\\ & & x=\left\{\mathrm{-2.718281828}\,\mathrm{-1.}\,1.\,2.718281828\right\}\end{array}$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\log }_{11}\left(25\right)}{{\log }_{11}\left(5\right)}=6\cdot x\\ \text{•}& & \text{Rearrange expression}\\ & & 6\cdot x=\frac{{\log }_{11}\left(25\right)}{{\log }_{11}\left(5\right)}\\ \text{•}& & \text{Rewrite}{\log }_{11}\left(25\right)\text{as}{\log }_{11}\left(5^2\right)\text{, then apply the log rule}{\log }_{11}\left(5^2\right)=2{\log }_{11}\left(5\right)\\ & & 6x=2\\ \text{•}& & \text{Divide both sides by}6\\ & & \frac{6\cdot x}{6}=\frac{2}{6}\\ \text{•}& & \text{Simplify}\\ & & x=\frac{1}{3}\end{array}$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left[y=\left(\mathrm{-1}\right)\cdot x^2+6\cdot x-10\,y=2\cdot x^2+6\cdot x+5\right]\\ \text{•}& & \text{E1}\text{: Pick the}\text{2nd}\text{equation which is already solved for}y\\ & & y=2\cdot x^2+6\cdot x+5\\ \text{•}& & \text{Pick the}\text{1st}\text{equation to solve for}x\\ & & y=\left(\mathrm{-1}\right)\cdot x^2+6\cdot x-10\\ \text{•}& & \text{Substitute the value of}y=2x^2+6x+5\text{into the}\text{1st}\text{equation of the system}\\ & & 2x^2+6x+5=\left(\mathrm{-1}\right)\cdot x^2+6\cdot x-10\\ \text{▫}& & \text{Isolate for}x\\ & \text{◦}& \text{Add}{x}^2-6x+10\text{to both sides}\\ & & 2x^2+6x+5+\left(x^2-6x+10\right)=-x^2+6x-10+\left(x^2-6x+10\right)\\ & \text{◦}& \text{Simplify}\\ & & 3x^2+15=0\\ & \text{◦}& \text{Subtract}15\text{from both sides}\\ & & 3\cdot x^2+15-15=0-15\\ & \text{◦}& \text{Simplify}\\ & & 3\cdot x^2=\mathrm{-15}\\ & \text{◦}& \text{Divide both sides by}3\\ & & \frac{3\cdot x^2}{3}=\frac{\mathrm{-15}}{3}\\ & \text{◦}& \text{Simplify}\\ & & x^2=\mathrm{-5}\\ & \text{◦}& \text{Take Square root of both sides}\\ & & x=\pm \sqrt{\mathrm{-5}}\\ & \text{◦}& \text{No real solutions}\\ & & x=\mathrm{imaginary}\end{array}$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 2\le \left|2\cdot x+3\right|\\ \text{•}& & \text{Solve for}x\text{to find points to test for intervals}\\ & & 2=\left|2\cdot x+3\right|\\ \text{•}& & \text{To solve, we must drop the absolute values. To do this, we must determine the intervals where the expression within the absolute value becomes positive or negative}\\ & & \left[\right]\\ \text{•}& & \text{Thus our intervals are:}\\ & & \left[\left(-\mathrm{\infty }\&comma;-\frac{3}{2}\right)\&comma;\left[-\frac{3}{2}\&comma;\mathrm{\infty }\right)\right]\\ \text{▫}& & \text{Examine absolute values with}x\in \left(-\mathrm{\infty }\&comma;-\frac{3}{2}\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}2x+3<0\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & 2=-2x-3\\ & \text{◦}& \text{Solve the new equality}\\ & & x=-\frac{5}{2}\\ & \text{◦}& \text{Since}-\frac{5}{2}\in \left(-\mathrm{\infty }\&comma;-\frac{3}{2}\right)\text{we get that this is a solution}\\ & & x=-\frac{5}{2}\\ \text{▫}& & \text{Examine absolute values with}x\in \left[-\frac{3}{2}\&comma;\mathrm{\infty }\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}2x+3\ge 0\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & 2=2x+3\\ & \text{◦}& \text{Solve the new equality}\\ & & x=-\frac{1}{2}\\ & \text{◦}& \text{Since}-\frac{1}{2}\in \left[-\frac{3}{2}\&comma;\mathrm{\infty }\right)\text{we get that this is a solution}\\ & & x=-\frac{1}{2}\\ \text{•}& & \text{Solution}\\ & & x=\left(-\frac{5}{2}\&comma;-\frac{1}{2}\right)\\ \text{•}& & \text{Use the solutions to the equality as points to test for intervals}\\ & & \left[-\frac{5}{2}\&comma;-\frac{1}{2}\right]\\ \text{•}& & \text{Set up a table using the solutions as boundaries and find test points that are on either side}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{▫}& & \text{Sub each test point into the expression for}x\\ & \text{◦}& \text{Sub}x=\mathrm{-3}\text{into}2\le \left|2x+3\right|\\ & & 2\le 3\\ & & \mathrm{true}\\ & \text{◦}& \text{Sub}x=\mathrm{-1}\text{into}2\le \left|2x+3\right|\\ & & 2\le 1\\ & & \mathrm{false}\\ & \text{◦}& \text{Sub}x=0\text{into}2\le \left|2x+3\right|\\ & & 2\le 3\\ & & \mathrm{true}\\ \text{•}& & \text{Observe where the inequality holds true, these areas make up the intervals}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Plotted solution}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Solution}\\ & & \left[\left\{x\le -\frac{5}{2}\right\}\&comma;\left\{-\frac{1}{2}\le x\right\}\right]\end{array}$