The SolveSteps command is used to show the steps of solving a basic student problem, such as an equation, system of equations, or inequality. It can also prove basic trigonometric identities.

If ex is an equation the variable in equation is solved for. If ex is given as an expression, the expression is solved for assuming ex=0.

If only one variable exists in ex, it is not necessary to specify a variable to solve for. If there are two or more variables in ex, a variable to solve for must be given for variable.

The displaystyle and output options can be used to change the output format.  See OutputStepsRecord for details.

The trigpath=n option, where n is a positive integer, can be used to view another way to prove a trigonometric identity.

The trigtimer option can be used to set the time limit for proving a trigonometric identity.  The value can be a positive integer or infinity.  The default is 60 (seconds).

The colorpack option can be used to specify an alternate color palette for inequality plots.  Valid options are the same as those accepted by the ColorTools:-GetPalette command.  If the colorpack option is not specified and Student:-SetColors has not been set then a custom palette is used.

This function is part of the Student:-Basics package.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{SolveSteps}\left(5\exp \left(4x\right)=16\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 5\cdot {\ⅇ}^{4\cdot x}=16\\ \text{▫}& & \text{Convert from exponential equation}\\ & \text{◦}& \text{Divide both sides by}5\\ & & \frac{5\cdot {\ⅇ}^{4\cdot x}}{5}=\frac{16}{5}\\ & \text{◦}& \text{Simplify}\\ & & {\ⅇ}^{4\cdot x}=\frac{16}{5}\\ & \text{◦}& \text{Apply ln to each side}\\ & & \ln \left({\ⅇ}^{4x}\right)=\ln \left(\frac{16}{5}\right)\\ & \text{◦}& \text{Apply ln rule: ln(e^b) = b}\\ & & 4x=\ln \left(\frac{16}{5}\right)\\ \text{•}& & \text{Divide both sides by}4\\ & & \frac{4\cdot x}{4}=\frac{\ln \left(\frac{16}{5}\right)}{4}\\ \text{•}& & \text{Exact solution}\\ & & x=\frac{\ln \left(\frac{16}{5}\right)}{4}\\ \text{•}& & \text{Approximate solution}\\ & & x=0.2907877025\end{array}$

$\mathrm{SolveSteps}\left(x^2-b\,x\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^2-1\cdot b\\ \text{•}& & \text{Set expression equal to 0}\\ & & x^2-1\cdot b=0\\ \text{•}& & \text{Add}b\text{to both sides}\\ & & x^2-1\cdot b+b=0+b\\ \text{•}& & \text{Simplify}\\ & & x^2=b\\ \text{•}& & \text{Take Square root of both sides}\\ & & x=\pm \sqrt{b}\\ \text{•}& & \text{Solution}\\ & & x=\left(\sqrt{b}\,-\sqrt{b}\right)\end{array}$

$\mathrm{SolveSteps}\left(x^3+4x^2+4x\,\mathrm{output}=\mathrm{typeset}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^3+4\cdot x^2+4\cdot x\\ \text{•}& & \text{Set expression equal to 0}\\ & & x^3+4\cdot x^2+4\cdot x=0\\ \text{•}& & \text{Common factor}x\\ & & x\cdot \left(x^2+4x+4\right)\\ \text{•}& & \text{Examine term:}\\ & & x^2+4x+4\\ \text{▫}& & \text{Factor using the AC Method}\\ & \text{◦}& \text{Examine quadratic}\\ & & \left(x^2+4x+4\right)\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\,''B''=4\,''C''=4\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}4\\ & & \left\{1\,2\,4\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}4\\ & & \left\{1\cdot 4\,2\cdot 2\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{sum}\text{of B =}4\text{? Found:}\\ & & 2+2=4\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(2x+2x\right)+4\\ & \text{◦}& \text{Factor}x\text{out of the first group}\\ & & \left(x\cdot \left(x+2\right)\right)+\left(2x+4\right)\\ & \text{◦}& \text{Factor}2\text{out of the second group}\\ & & x\cdot \left(x+2\right)+\left(2\cdot \left(x+2\right)\right)\\ & \text{◦}& x+2\text{is a common factor}\\ & & x\cdot \left(x+2\right)+2\cdot \left(x+2\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x+2\right)\cdot \left(x+2\right)\\ & & \text{This gives:}\\ & & {\left(x+2\right)}^2\\ \text{•}& & \text{This gives:}\\ & & x\cdot {\left(x+2\right)}^2\\ \text{•}& & \text{The}\text{1st}\text{factor is}x\text{which implies}x\text{= 0 is a solution}\\ & & x=0\\ \text{•}& & \text{Set}\text{2nd}\text{factor}x+2\text{to 0 to solve}\\ & & x+2=0\\ \text{▫}& & \text{Solution of}x+2=0\\ & \text{◦}& \text{Subtract}2\text{from both sides}\\ & & x+2-2=0-2\\ & \text{◦}& \text{Simplify}\\ & & x=\mathrm{-2}\\ \text{•}& & \text{Solution}\\ & & x=\left(\mathrm{-2}\,0\right)\end{array}$

$\mathrm{SolveSteps}\left(x^3+4x^2+4x\,\mathrm{mode}=\mathrm{Learn}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^3+4\cdot x^2+4\cdot x\\ \text{•}& & \text{Set expression equal to 0}\\ & & x^3+4\cdot x^2+4\cdot x=0\\ \text{•}& & \text{Common factor}x\\ & & x\cdot \left(x^2+4x+4\right)\\ \text{•}& & \text{Examine term:}\\ & & x^2+4x+4\\ \text{▫}& & \text{Factor using the AC Method}\\ & \text{◦}& \text{Examine quadratic}\\ & & \left(x^2+4x+4\right)\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\,''B''=4\,''C''=4\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}4\\ & & \left\{1\,2\,4\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}4\\ & & \left\{1\cdot 4\,2\cdot 2\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{sum}\text{of B =}4\text{? Found:}\\ & & 2+2=4\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(2x+2x\right)+4\\ & \text{◦}& \text{Factor}x\text{out of the first group}\\ & & \left(x\cdot \left(x+2\right)\right)+\left(2x+4\right)\\ & \text{◦}& \text{Factor}2\text{out of the second group}\\ & & x\cdot \left(x+2\right)+\left(2\cdot \left(x+2\right)\right)\\ & \text{◦}& x+2\text{is a common factor}\\ & & x\cdot \left(x+2\right)+2\cdot \left(x+2\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x+2\right)\cdot \left(x+2\right)\\ & & \text{This gives:}\\ & & {\left(x+2\right)}^2\\ \text{•}& & \text{This gives:}\\ & & x\cdot {\left(x+2\right)}^2\\ \text{•}& & \text{The}\text{1st}\text{factor is}x\text{which implies}x\text{= 0 is a solution}\\ & & x=0\\ \text{•}& & \text{Set}\text{2nd}\text{factor}x+2\text{to 0 to solve}\\ & & x+2=0\\ \text{▫}& & \text{Solution of}x+2=0\\ & \text{◦}& \text{Subtract}2\text{from both sides}\\ & & x+2-2=0-2\\ & \text{◦}& \text{Simplify}\\ & & x=\mathrm{-2}\\ \text{•}& & \text{Solution}\\ & & x=\left(\mathrm{-2}\,0\right)\end{array}$

$\mathrm{SolveSteps}\left(\csc \left(x\right)\tan \left(x\right)\cos \left(x\right)=1\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Let's simplify the left side of the expression to match the right}\\ & & \left(\csc \left(x\right)\tan \left(x\right)\cos \left(x\right)\right)=1\\ \text{•}& & \text{Apply}\text{Quotient}\text{trig identity,}\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\\ & & \csc \left(x\right)\frac{\sin \left(x\right)}{\cos \left(x\right)}\cos \left(x\right)\\ \text{•}& & \text{Apply}\text{Reciprocal Function}\text{trig identity,}\csc \left(x\right)=\frac{1}{\sin \left(x\right)}\\ & & \frac{1}{\sin \left(x\right)}\sin \left(x\right)\\ \text{•}& & \text{Evaluate}\\ & & 1\\ \text{•}& & \text{Thus we have proved that the identity is true}\\ & & 1=1\end{array}$

$\mathrm{SolveSteps}\left(\frac{{\cos \left(x\right)}^2}{1+\sin \left(x\right)}=1-\sin \left(x\right)\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Let's simplify the left side of the expression to match the right}\\ & & \frac{{\cos \left(x\right)}^2}{1+\sin \left(x\right)}=1-\sin \left(x\right)\\ \text{•}& & \text{Apply}\text{Pythagoras}\text{trig identity,}{\cos \left(x\right)}^2=1-{\sin \left(x\right)}^2\\ & & \frac{\left(1-{\sin \left(x\right)}^2\right)}{1+\sin \left(x\right)}\\ \text{•}& & \text{Factor the numerator}\\ & & \frac{\left(-\left(\sin \left(x\right)-1\right)\left(1+\sin \left(x\right)\right)\right)}{1+\sin \left(x\right)}\\ \text{•}& & \text{Cancel out a factor of}1+\sin \left(x\right)\\ & & \left(1-\sin \left(x\right)\right)\\ \text{•}& & \text{Evaluate}\\ & & 1-\sin \left(x\right)\\ \text{•}& & \text{Thus we have proved that the identity is true}\\ & & 1-\sin \left(x\right)=1-\sin \left(x\right)\end{array}$

$\mathrm{SolveSteps}\left(\sin \left(5x\right)=16{\sin \left(x\right)}^5-20{\sin \left(x\right)}^3+5\sin \left(x\right)\,\mathrm{trigtimer}=\mathrm{\infty }\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Let's simplify the right-side of the expression to match the left}\\ & & \sin \left(5x\right)=\left(16{\sin \left(x\right)}^5-20{\sin \left(x\right)}^3+5\sin \left(x\right)\right)\\ \text{•}& & \text{Apply}\text{Full Power Reduction}\text{trig identity,}{\sin \left(x\right)}^3=-\frac{\sin \left(3x\right)}{4}+\frac{3\sin \left(x\right)}{4}\\ & & 16{\sin \left(x\right)}^5-20\left(-\frac{\sin \left(3x\right)}{4}+\frac{3\sin \left(x\right)}{4}\right)+5\sin \left(x\right)\\ \text{•}& & \text{Apply}\text{Full Power Reduction}\text{trig identity,}{\sin \left(x\right)}^5=\frac{\sin \left(5x\right)}{16}-\frac{5\sin \left(3x\right)}{16}+\frac{5\sin \left(x\right)}{8}\\ & & 16\left(\frac{\sin \left(5x\right)}{16}-\frac{5\sin \left(3x\right)}{16}+\frac{5\sin \left(x\right)}{8}\right)+5\sin \left(3x\right)-10\sin \left(x\right)\\ \text{•}& & \text{Evaluate}\\ & & \sin \left(5x\right)\\ \text{•}& & \text{Thus we have proved that the identity is true}\\ & & \sin \left(5x\right)=\sin \left(5x\right)\end{array}$

$\mathrm{SolveSteps}\left(\frac{\sin \left(2x\right)}{\sin \left(x\right)}-\frac{\cos \left(2x\right)}{\cos \left(x\right)}=\sec \left(x\right)\,\mathrm{trigpath}=2\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Let's simplify the left side of the expression to match the right}\\ & & \left(\frac{\sin \left(2x\right)}{\sin \left(x\right)}-\frac{\cos \left(2x\right)}{\cos \left(x\right)}\right)=\sec \left(x\right)\\ \text{•}& & \text{Find fractions to get lowest common denominator of}\sin \left(x\right)\cos \left(x\right)\\ & & \frac{\cos \left(x\right)}{\cos \left(x\right)}\cdot \left(\frac{\sin \left(2x\right)}{\sin \left(x\right)}\right)+\frac{\sin \left(x\right)}{\sin \left(x\right)}\cdot \left(-\frac{\cos \left(2x\right)}{\cos \left(x\right)}\right)\\ \text{•}& & \text{Multiply}\\ & & \frac{\cos \left(x\right)\cdot \sin \left(2x\right)}{\sin \left(x\right)\cos \left(x\right)}+\frac{\sin \left(x\right)\cdot \left(-\cos \left(2x\right)\right)}{\sin \left(x\right)\cos \left(x\right)}\\ \text{•}& & \text{Add fractions}\\ & & \frac{\cos \left(x\right)\sin \left(2x\right)-\sin \left(x\right)\cos \left(2x\right)}{\sin \left(x\right)\cos \left(x\right)}\\ \text{•}& & \text{Apply}\text{Reciprocal Function}\text{trig identity,}\frac{1}{\cos \left(x\right)}=\sec \left(x\right)\\ & & \frac{\left(\cos \left(x\right)\sin \left(2x\right)-\sin \left(x\right)\cos \left(2x\right)\right)\left(\sec \left(x\right)\right)}{\sin \left(x\right)}\\ \text{•}& & \text{Apply}\text{Double Angle}\text{trig identity,}\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)\\ & & \frac{\left(\cos \left(x\right)\left(2\sin \left(x\right)\cos \left(x\right)\right)-\sin \left(x\right)\cos \left(2x\right)\right)\sec \left(x\right)}{\sin \left(x\right)}\\ \text{•}& & \text{Factor the numerator}\\ & & \frac{\left(\sin \left(x\right)\left(2{\cos \left(x\right)}^2-\cos \left(2x\right)\right)\sec \left(x\right)\right)}{\sin \left(x\right)}\\ \text{•}& & \text{Cancel out a factor of}\sin \left(x\right)\\ & & \left(\left(2{\cos \left(x\right)}^2-\cos \left(2x\right)\right)\sec \left(x\right)\right)\\ \text{•}& & \text{Apply}\text{Half Angle}\text{trig identity,}{\cos \left(x\right)}^2=\frac{\cos \left(2x\right)}{2}+\frac{1}{2}\\ & & \left(2\left(\frac{\cos \left(2x\right)}{2}+\frac{1}{2}\right)-\cos \left(2x\right)\right)\sec \left(x\right)\\ \text{•}& & \text{Evaluate}\\ & & \sec \left(x\right)\\ \text{•}& & \text{Thus we have proved that the identity is true}\\ & & \sec \left(x\right)=\sec \left(x\right)\end{array}$

$\mathrm{SolveSteps}\left(\left[\frac{1}{2}<x\&comma;2\le x\right]\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Examine the}\text{1st}\text{inequality and solve for}x\\ & & \frac{1}{2}<x\\ \text{•}& & \text{Examine the}\text{2nd}\text{inequality and solve for}x\\ & & 2\le x\\ \text{•}& & \text{The solved system is:}\\ & & \left[\begin{array}{cc}x>\frac{1}{2}& \mathrm{L1}\\ x\ge 2& \mathrm{L2}\end{array}\right]\\ \text{•}& & \text{Graph the boundary lines of the inequalites}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Show inequalities}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Solution is where the inequalities overlap}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\end{array}$

$\mathrm{SolveSteps}\left(\left[y\le 2x+\frac{7}{2}\&comma;y\le -2x+\frac{7}{2}\&comma;-\frac{2x}{3}-1\le y\&comma;\frac{2x}{3}-1\le y\&comma;y\le \frac{3}{2}\right]\&comma;\mathrm{colorpack}=''MapleV''\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ \text{•}& & \text{Examine the}\text{1st}\text{inequality and solve for}y\\ & & y\le 2\cdot x+\frac{7}{2}\\ \text{•}& & \text{Examine the}\text{2nd}\text{inequality and solve for}y\\ & & y\le \left(\mathrm{-2}\right)\cdot x+\frac{7}{2}\\ \text{•}& & \text{Examine the}\text{3rd}\text{inequality and solve for}y\\ & & \frac{\mathrm{-2}}{3}\cdot x-1\le y\\ \text{•}& & \text{Examine the}\text{4th}\text{inequality and solve for}y\\ & & \frac{2}{3}\cdot x-1\le y\\ \text{•}& & \text{Examine the}\text{5th}\text{inequality and solve for}y\\ & & y\le \frac{3}{2}\\ \text{•}& & \text{The solved system is:}\\ & & \left[\begin{array}{cc}y\le 2x+\frac{7}{2}& \mathrm{L1}\\ y\le -2x+\frac{7}{2}& \mathrm{L2}\\ y\ge -\frac{2x}{3}-1& \mathrm{L3}\\ y\ge \frac{2x}{3}-1& \mathrm{L4}\\ y\le \frac{3}{2}& \mathrm{L5}\end{array}\right]\\ \text{•}& & \text{Graph the boundary lines of the inequalites}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Show inequalities}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\\ \text{•}& & \text{Solution is where the inequalities overlap}\\ & & \mathrm{PLOT}\left(\mathrm{...}\right)\end{array}$

$\mathrm{SolveSteps}\left(''x^2\; -\; 4*x\; +\; 4\; >\; 7''\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 7<x^2-4\cdot x+4\\ \text{•}& & \text{Solve for}x\text{to find points to test for intervals}\\ & & 7=x^2-4\cdot x+4\\ \text{•}& & \text{Rearrange expression}\\ & & x^2-4\cdot x+4=7\\ \text{•}& & \text{Subtract}7\text{from both sides}\\ & & x^2-4x+4-7=7-7\\ \text{•}& & \text{Simplify}\\ & & x^2-4x-3=0\\ \text{•}& & \text{Since we can't factor we'll use the quadratic formula}\\ & & x=\frac{\left(-b\right)\pm \left(\sqrt{b^2-4\cdot a\cdot c}\right)}{2\cdot a}\\ \text{▫}& & \text{Use quadratic formula to solve for}x\\ & \text{◦}& \text{Substitute a=}1\text{, b=}\mathrm{-4}\text{, c=}\mathrm{-3}\\ & & x=\frac{4\pm \left(\sqrt{{\left(\mathrm{-4}\right)}^2-4\cdot 1\cdot \left(\mathrm{-3}\right)}\right)}{2\cdot 1}\\ & \text{◦}& \text{Evaluate under discriminant}\\ & & x=\frac{4\pm \left(\sqrt{16-\left(\mathrm{-12}\right)}\right)}{2\cdot 1}\\ & \text{◦}& \text{Perform remaining operations}\\ & & x=2\pm \left(\sqrt{7}\right)\\ \text{•}& & \text{Solution}\\ & & x=\left(2-\sqrt{7}\&comma;2+\sqrt{7}\right)\\ \text{•}& & \text{Use the solutions to the equality as points to test for intervals}\\ & & \left[2-\sqrt{7}\&comma;2+\sqrt{7}\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{-1}\text{into}0<x^2-4x-3\\ & & 0<2\\ & & \mathrm{true}\\ & \text{◦}& \text{Sub}x=2\text{into}0<x^2-4x-3\\ & & 0<\mathrm{-7}\\ & & \mathrm{false}\\ & \text{◦}& \text{Sub}x=5\text{into}0<x^2-4x-3\\ & & 0<2\\ & & \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<2-\sqrt{7}\right\}\&comma;\left\{2+\sqrt{7}<x\right\}\right]\end{array}$

$\mathrm{SolveSteps}\left(''x\; +\; 4/x\; >\; 4''\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 4<x+\frac{4}{x}\\ \text{•}& & \text{Note the values for}x\text{which causes the expression to be undefined. These values will be used later to identify the solution intervals}\\ & & x=\left[0\right]\\ \text{•}& & \text{Solve for}x\text{to find points to test for intervals}\\ & & 4=x+\frac{4}{x}\\ \text{•}& & \text{Rearrange expression}\\ & & x+\frac{4}{x}=4\\ \text{•}& & \text{Multiply both sides by}x\\ & & x\cdot x+x\cdot \left(\frac{4}{x}\right)=x\cdot 4\\ \text{•}& & \text{Evaluate}\\ & & x^2+4=4\cdot x\\ \text{•}& & \text{Subtract}4x\text{from both sides}\\ & & x^2+4-4x=4x-4x\\ \text{•}& & \text{Simplify}\\ & & x^2-4x+4=0\\ \text{▫}& & \text{Factor using the AC Method}\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\&comma;''B''=\mathrm{-4}\&comma;''C''=4\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}4\\ & & \left\{1\&comma;2\&comma;4\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}4\\ & & \left\{1\cdot 4\&comma;2\cdot 2\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{sum}\text{of B =}\mathrm{-4}\text{? Found:}\\ & & \mathrm{-2}-2=\mathrm{-4}\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(-2x-2x\right)+4\\ & \text{◦}& \text{Factor}x\text{out of the first group}\\ & & x\cdot \left(x-2\right)+\left(-2x+4\right)\\ & \text{◦}& \text{Factor}\mathrm{-2}\text{out of the second group}\\ & & x\cdot \left(x-2\right)-2\cdot \left(x-2\right)\\ & \text{◦}& x-2\text{is a common factor}\\ & & x\cdot \left(x-2\right)-2\cdot \left(x-2\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x-2\right)\cdot \left(x-2\right)\\ & & \text{This gives:}\\ & & {\left(x-2\right)}^2=0\\ \text{•}& & \text{Examine factor}1\\ & & x-2\\ \text{▫}& & \text{Solution of}x-2=0\\ & \text{◦}& \text{Add}2\text{to both sides}\\ & & x-2+2=0+2\\ & \text{◦}& \text{Simplify}\\ & & x=2\\ \text{▫}& & \text{Check if}x=2\text{satisfies domain requirements}\\ & \text{◦}& \text{Domain requirement from}\frac{4}{x}\text{: cannot divide by 0}\\ & & x\ne 0\\ & \text{◦}& \text{Substitute}x=2\text{into}4\text{and evaluate}\\ & & 2\\ & \text{◦}& \text{Domain requirement met}\\ & & 2\ne 0\\ & & \text{Therefore the domain requirement is satisfied.}\\ & & \text{✓}\\ \text{•}& & \text{Substitute solution into equation and check if left-side = right-side}\\ & & 4<4\\ \text{•}& & \text{left-side ≠ right-side}\\ & & 4\ne 4\\ \text{•}& & x=2\text{is not a solution}\\ & & \text{⨉}\\ \text{•}& & \text{Use the solutions and undefined values as points to test for intervals}\\ & & 0\\ \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{-1}\text{into}4<x+\frac{4}{x}\\ & & 4<\mathrm{-5}\\ & & \mathrm{false}\\ & \text{◦}& \text{Sub}x=1\text{into}4<x+\frac{4}{x}\\ & & 4<5\\ & & \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\{0<x\right\}\end{array}$

$\mathrm{SolveSteps}\left(\mathrm{abs}\left(x+1\right)=4x\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left|x+1\right|=4\cdot x\\ \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[\begin{array}{ccc}x+1<0& \mathrm{when}& x\in \left(-\mathrm{\infty }\&comma;\mathrm{-1}\right)\\ x+1\ge 0& \mathrm{when}& x\in \left[\mathrm{-1}\&comma;\mathrm{\infty }\right)\end{array}\right]\\ \text{•}& & \text{Thus our intervals are:}\\ & & \left[\left(-\mathrm{\infty }\&comma;\mathrm{-1}\right)\&comma;\left[\mathrm{-1}\&comma;\mathrm{\infty }\right)\right]\\ \text{▫}& & \text{Examine absolute values with}x\in \left(-\mathrm{\infty }\&comma;\mathrm{-1}\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}\mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%<}\left(x\&plus;1\&comma;0\right)\right]\right)\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\begin{array}{c}\left|x+1\right|=-x-1\end{array}\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & -x-1=4x\\ & \text{◦}& \text{Solve the new equality}\\ & & x=-\frac{1}{5}\\ & \text{◦}& \text{Since}-\frac{1}{5}\text{∉}\left(-\mathrm{\infty }\&comma;\mathrm{-1}\right)\text{we get that this is not a solution}\\ & & x\ne -\frac{1}{5}\\ \text{▫}& & \text{Examine absolute values with}x\in \left[\mathrm{-1}\&comma;\mathrm{\infty }\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}\mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%>=}\left(x\&plus;1\&comma;0\right)\right]\right)\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\begin{array}{c}\left|x+1\right|=x+1\end{array}\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & x+1=4x\\ & \text{◦}& \text{Solve the new equality}\\ & & x=\frac{1}{3}\\ & \text{◦}& \text{Since}\frac{1}{3}\in \left[\mathrm{-1}\&comma;\mathrm{\infty }\right)\text{we get that this is a solution}\\ & & x=\frac{1}{3}\\ \text{•}& & \text{Solution}\\ & & x=\frac{1}{3}\end{array}$

$\mathrm{SolveSteps}\left(\mathrm{abs}\left(2x+6\right)=\mathrm{abs}\left(x+7\right)\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 2\cdot \left|x+3\right|=\left|x+7\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[\begin{array}{ccc}x+3<0& \mathrm{when}& x\in \left(-\mathrm{\infty }\&comma;\mathrm{-3}\right)\\ x+3\ge 0& \mathrm{when}& x\in \left[\mathrm{-3}\&comma;\mathrm{\infty }\right)\\ x+7<0& \mathrm{when}& x\in \left(-\mathrm{\infty }\&comma;\mathrm{-7}\right)\\ x+7\ge 0& \mathrm{when}& x\in \left[\mathrm{-7}\&comma;\mathrm{\infty }\right)\end{array}\right]\\ \text{•}& & \text{Thus our intervals are:}\\ & & \left[\left(-\mathrm{\infty }\&comma;\mathrm{-7}\right)\&comma;\left[\mathrm{-7}\&comma;\mathrm{-3}\right)\&comma;\left[\mathrm{-3}\&comma;\mathrm{\infty }\right)\right]\\ \text{▫}& & \text{Examine absolute values with}x\in \left(-\mathrm{\infty }\&comma;\mathrm{-7}\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}\mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%<}\left(x\&plus;3\&comma;0\right)\right]\right)\\ \mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%<}\left(x\&plus;7\&comma;0\right)\right]\right)\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\begin{array}{c}\left|x+3\right|=-x-3\\ \left|x+7\right|=-x-7\end{array}\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & -2x-6=-x-7\\ & \text{◦}& \text{Solve the new equality}\\ & & x=1\\ & \text{◦}& \text{Since}1\text{∉}\left(-\mathrm{\infty }\&comma;\mathrm{-7}\right)\text{we get that this is not a solution}\\ & & x\ne 1\\ \text{▫}& & \text{Examine absolute values with}x\in \left[\mathrm{-7}\&comma;\mathrm{-3}\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}\mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%<}\left(x\&plus;3\&comma;0\right)\right]\right)\\ \mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%>=}\left(x\&plus;7\&comma;0\right)\right]\right)\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\begin{array}{c}\left|x+3\right|=-x-3\\ \left|x+7\right|=x+7\end{array}\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & -2x-6=x+7\\ & \text{◦}& \text{Solve the new equality}\\ & & x=-\frac{13}{3}\\ & \text{◦}& \text{Since}-\frac{13}{3}\in \left[\mathrm{-7}\&comma;\mathrm{-3}\right)\text{we get that this is a solution}\\ & & x=-\frac{13}{3}\\ \text{▫}& & \text{Examine absolute values with}x\in \left[\mathrm{-3}\&comma;\mathrm{\infty }\right)\\ & \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ & & \left[\begin{array}{c}\mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%>=}\left(x\&plus;3\&comma;0\right)\right]\right)\\ \mathrm{Typesetting}:-\mathrm{\_Hold}\left(\left[\mathrm{\%>=}\left(x\&plus;7\&comma;0\right)\right]\right)\end{array}\right]\\ & \text{◦}& \text{Drop the absolute values and multiply the expressions that would be negative by -1}\\ & & \left[\begin{array}{c}\left|x+3\right|=x+3\\ \left|x+7\right|=x+7\end{array}\right]\\ & \text{◦}& \text{Sub the new expressions in where the absolute values used to be}\\ & & 2x+6=x+7\\ & \text{◦}& \text{Solve the new equality}\\ & & x=1\\ & \text{◦}& \text{Since}1\in \left[\mathrm{-3}\&comma;\mathrm{\infty }\right)\text{we get that this is a solution}\\ & & x=1\\ \text{•}& & \text{Solution}\\ & & x=\left(-\frac{13}{3}\&comma;1\right)\end{array}$