SolveSteps - Maple Help

Student[Basics]

 SolveSteps
 show steps in the solution of a specified problem

 Calling Sequence SolveSteps(ex, variable, opts)

Parameters

 ex - expression or equation variable - (optional) variable to solve for opts - options of the form keyword=value where keyword is one of displaystyle, output, trigpath, trigtimer, colorpack

Description

 • 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.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right):$
 > $\mathrm{SolveSteps}\left(5\mathrm{exp}\left(4x\right)=16\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {5}{\cdot }{{ⅇ}}^{{4}{\cdot }{x}}{=}{16}\\ \text{▫}& {}& \text{Convert from exponential equation}\\ {}& \text{◦}& \text{Divide both sides by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}5\\ {}& {}& \frac{{5}{\cdot }{{ⅇ}}^{{4}{\cdot }{x}}}{{5}}{=}\frac{{16}}{{5}}\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& {{ⅇ}}^{{4}{\cdot }{x}}{=}\frac{{16}}{{5}}\\ {}& \text{◦}& \text{Apply ln to each side}\\ {}& {}& {\mathrm{ln}}{}\left({{ⅇ}}^{{4}{}{x}}\right){=}{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)\\ {}& \text{◦}& \text{Apply ln rule: ln(e^b) = b}\\ {}& {}& {4}{}{x}{=}{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)\\ \text{•}& {}& \text{Divide both sides by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \frac{{4}{\cdot }{x}}{{4}}{=}\frac{{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)}{{4}}\\ \text{•}& {}& \text{Exact solution}\\ {}& {}& {x}{=}\frac{{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)}{{4}}\\ \text{•}& {}& \text{Approximate solution}\\ {}& {}& {x}{=}{0.2907877025}\end{array}$ (1)
 > $\mathrm{SolveSteps}\left({x}^{2}-b,x\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {{x}}^{{2}}{-}{1}{\cdot }{b}\\ \text{•}& {}& \text{Set expression equal to 0}\\ {}& {}& {{x}}^{{2}}{-}{1}{\cdot }{b}{=}{0}\\ \text{•}& {}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}b\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to both sides}\\ {}& {}& {{x}}^{{2}}{-}{1}{\cdot }{b}{+}{b}{=}{0}{+}{b}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {{x}}^{{2}}{=}{b}\\ \text{•}& {}& \text{Take Square root of both sides}\\ {}& {}& {x}{=}{±}\sqrt{{b}}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left(\sqrt{{b}}{,}{-}\sqrt{{b}}\right)\end{array}$ (2)
 > $\mathrm{SolveSteps}\left({x}^{3}+4{x}^{2}+4x,\mathrm{output}=\mathrm{typeset}\right)$
 $\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {x}{\cdot }\left({{x}}^{{2}}{+}{4}{}{x}{+}{4}\right)\\ \text{•}& {}& \text{Examine term:}\\ {}& {}& {{x}}^{{2}}{+}{4}{}{x}{+}{4}\\ \text{▫}& {}& \text{Factor using the AC Method}\\ {}& \text{◦}& \text{Examine quadratic}\\ {}& {}& \left({\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}\right)\\ {}& \text{◦}& \text{Look at the coefficients,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}A{}{x}^{2}+B{}x+C\\ {}& {}& \left[{"A"}{=}{1}{,}{"B"}{=}{4}{,}{"C"}{=}{4}\right]\\ {}& \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{,}{2}{,}{4}\right\}\\ {}& \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{\cdot }{4}{,}{2}{\cdot }{2}\right\}\\ {}& \text{◦}& \text{Which pairs of ± these factors have a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{sum}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of B =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\text{? Found:}\\ {}& {}& {2}{+}{2}{=}{4}\\ {}& \text{◦}& \text{Split the middle term to use above pair}\\ {}& {}& {{x}}^{{2}}{+}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){+}{4}\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the first group}\\ {}& {}& \left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\right){+}\left({2}{}{x}{+}{4}\right)\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the second group}\\ {}& {}& {x}{\cdot }\left({x}{+}{2}\right){+}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\right)\\ {}& \text{◦}& x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a common factor}\\ {}& {}& {x}{\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right){+}{2}{\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\\ {}& \text{◦}& \text{Group common factor}\\ {}& {}& \left({x}{+}{2}\right){\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\\ {}& {}& \text{This gives:}\\ {}& {}& {\left({x}{+}{2}\right)}^{{2}}\\ \text{•}& {}& \text{This gives:}\\ {}& {}& {x}{\cdot }{\left({x}{+}{2}\right)}^{{2}}\\ \text{•}& {}& \text{The}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{1st}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{factor is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{which implies}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{= 0 is a solution}\\ {}& {}& {x}{=}{0}\\ \text{•}& {}& \text{Set}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{2nd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to 0 to solve}\\ {}& {}& {x}{+}{2}{=}{0}\\ \text{▫}& {}& \text{Solution of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x+2=0\\ {}& \text{◦}& \text{Subtract}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from both sides}\\ {}& {}& {x}{+}{2}{-}{2}{=}{0}{-}{2}\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& {x}{=}{-2}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left({-2}{,}{0}\right)\end{array}$ (3)
 > $\mathrm{SolveSteps}\left({x}^{3}+4{x}^{2}+4x,\mathrm{mode}=\mathrm{Learn}\right)$
 $\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {x}{\cdot }\left({{x}}^{{2}}{+}{4}{}{x}{+}{4}\right)\\ \text{•}& {}& \text{Examine term:}\\ {}& {}& {{x}}^{{2}}{+}{4}{}{x}{+}{4}\\ \text{▫}& {}& \text{Factor using the AC Method}\\ {}& \text{◦}& \text{Examine quadratic}\\ {}& {}& \left({\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}\right)\\ {}& \text{◦}& \text{Look at the coefficients,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}A{}{x}^{2}+B{}x+C\\ {}& {}& \left[{"A"}{=}{1}{,}{"B"}{=}{4}{,}{"C"}{=}{4}\right]\\ {}& \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{,}{2}{,}{4}\right\}\\ {}& \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{\cdot }{4}{,}{2}{\cdot }{2}\right\}\\ {}& \text{◦}& \text{Which pairs of ± these factors have a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{sum}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of B =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\text{? Found:}\\ {}& {}& {2}{+}{2}{=}{4}\\ {}& \text{◦}& \text{Split the middle term to use above pair}\\ {}& {}& {{x}}^{{2}}{+}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){+}{4}\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the first group}\\ {}& {}& \left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\right){+}\left({2}{}{x}{+}{4}\right)\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the second group}\\ {}& {}& {x}{\cdot }\left({x}{+}{2}\right){+}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\right)\\ {}& \text{◦}& x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a common factor}\\ {}& {}& {x}{\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right){+}{2}{\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\\ {}& \text{◦}& \text{Group common factor}\\ {}& {}& \left({x}{+}{2}\right){\cdot }\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}\right)\\ {}& {}& \text{This gives:}\\ {}& {}& {\left({x}{+}{2}\right)}^{{2}}\\ \text{•}& {}& \text{This gives:}\\ {}& {}& {x}{\cdot }{\left({x}{+}{2}\right)}^{{2}}\\ \text{•}& {}& \text{The}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{1st}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{factor is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{which implies}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{= 0 is a solution}\\ {}& {}& {x}{=}{0}\\ \text{•}& {}& \text{Set}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{2nd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to 0 to solve}\\ {}& {}& {x}{+}{2}{=}{0}\\ \text{▫}& {}& \text{Solution of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x+2=0\\ {}& \text{◦}& \text{Subtract}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from both sides}\\ {}& {}& {x}{+}{2}{-}{2}{=}{0}{-}{2}\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& {x}{=}{-2}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left({-2}{,}{0}\right)\end{array}$ (4)

SolveSteps is also capable of proving trigonometric identities

 > $\mathrm{SolveSteps}\left(\mathrm{csc}\left(x\right)\mathrm{tan}\left(x\right)\mathrm{cos}\left(x\right)=1\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Let's simplify the left side of the expression to match the right}\\ {}& {}& \left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{csc}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{tan}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right){=}{1}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Quotient}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{tan}{}\left(x\right)=\frac{\mathrm{sin}{}\left(x\right)}{\mathrm{cos}{}\left(x\right)}\\ {}& {}& {\mathrm{csc}}{}\left({x}\right){}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{}{\mathrm{cos}}{}\left({x}\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Reciprocal Function}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{csc}{}\left(x\right)=\frac{1}{\mathrm{sin}{}\left(x\right)}\\ {}& {}& \frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{}{\mathrm{sin}}{}\left({x}\right)\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {1}\\ \text{•}& {}& \text{Thus we have proved that the identity is true}\\ {}& {}& {1}{=}{1}\end{array}$ (5)
 > $\mathrm{SolveSteps}\left(\frac{{\mathrm{cos}\left(x\right)}^{2}}{1+\mathrm{sin}\left(x\right)}=1-\mathrm{sin}\left(x\right)\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Let's simplify the left side of the expression to match the right}\\ {}& {}& \frac{{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{=}{1}{-}{\mathrm{sin}}{}\left({x}\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Pythagoras}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cos}{}\left(x\right)}^{2}=1-{\mathrm{sin}{}\left(x\right)}^{2}\\ {}& {}& \frac{\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\right)}{{1}{+}{\mathrm{sin}}{}\left({x}\right)}\\ \text{•}& {}& \text{Factor the numerator}\\ {}& {}& \frac{\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\right){}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)\right)}{{1}{+}{\mathrm{sin}}{}\left({x}\right)}\\ \text{•}& {}& \text{Cancel out a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1+\mathrm{sin}{}\left(x\right)\\ {}& {}& \left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {1}{-}{\mathrm{sin}}{}\left({x}\right)\\ \text{•}& {}& \text{Thus we have proved that the identity is true}\\ {}& {}& {1}{-}{\mathrm{sin}}{}\left({x}\right){=}{1}{-}{\mathrm{sin}}{}\left({x}\right)\end{array}$ (6)

Use the optional parameter trigtimer, which takes a positive integer, to set the allowed time for solving. By default it is 60 seconds.

 > $\mathrm{SolveSteps}\left(\mathrm{sin}\left(5x\right)=16{\mathrm{sin}\left(x\right)}^{5}-20{\mathrm{sin}\left(x\right)}^{3}+5\mathrm{sin}\left(x\right),\mathrm{trigtimer}=\mathrm{\infty }\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Let's simplify the right-side of the expression to match the left}\\ {}& {}& {\mathrm{sin}}{}\left({5}{}{x}\right){=}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{16}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{20}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{3}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Full Power Reduction}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{sin}{}\left(x\right)}^{3}=-\frac{\mathrm{sin}{}\left(3{}x\right)}{4}+\frac{3{}\mathrm{sin}{}\left(x\right)}{4}\\ {}& {}& {16}{}{{\mathrm{sin}}{}\left({x}\right)}^{{5}}{-}{20}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{3}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{3}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{4}}}\right){+}{5}{}{\mathrm{sin}}{}\left({x}\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Full Power Reduction}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{sin}{}\left(x\right)}^{5}=\frac{\mathrm{sin}{}\left(5{}x\right)}{16}-\frac{5{}\mathrm{sin}{}\left(3{}x\right)}{16}+\frac{5{}\mathrm{sin}{}\left(x\right)}{8}\\ {}& {}& {16}{}\left(\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{16}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{3}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{16}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{8}}}\right){+}{5}{}{\mathrm{sin}}{}\left({3}{}{x}\right){-}{10}{}{\mathrm{sin}}{}\left({x}\right)\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {\mathrm{sin}}{}\left({5}{}{x}\right)\\ \text{•}& {}& \text{Thus we have proved that the identity is true}\\ {}& {}& {\mathrm{sin}}{}\left({5}{}{x}\right){=}{\mathrm{sin}}{}\left({5}{}{x}\right)\end{array}$ (7)

Use the optional parameter trigpath, which takes a positive integer, to view different ways to prove the identity

 > $\mathrm{SolveSteps}\left(\frac{\mathrm{sin}\left(2x\right)}{\mathrm{sin}\left(x\right)}-\frac{\mathrm{cos}\left(2x\right)}{\mathrm{cos}\left(x\right)}=\mathrm{sec}\left(x\right),\mathrm{trigpath}=2\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Let's simplify the left side of the expression to match the right}\\ {}& {}& \left(\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}\right){=}{\mathrm{sec}}{}\left({x}\right)\\ \text{•}& {}& \text{Find fractions to get lowest common denominator of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sin}{}\left(x\right){}\mathrm{cos}{}\left(x\right)\\ {}& {}& \frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\cdot }\left(\frac{{\mathrm{sin}}{}\left({2}{}{x}\right)}{{\mathrm{sin}}{}\left({x}\right)}\right){+}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\cdot }\left({-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{\mathrm{cos}}{}\left({x}\right)}\right)\\ \text{•}& {}& \text{Multiply}\\ {}& {}& \frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{+}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\cdot }}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}\\ \text{•}& {}& \text{Add fractions}\\ {}& {}& \frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Reciprocal Function}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{1}{\mathrm{cos}{}\left(x\right)}=\mathrm{sec}{}\left(x\right)\\ {}& {}& \frac{\left({\mathrm{cos}}{}\left({x}\right){}{\mathrm{sin}}{}\left({2}{}{x}\right){-}{\mathrm{sin}}{}\left({x}\right){}{\mathrm{cos}}{}\left({2}{}{x}\right)\right){}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sec}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)}{{\mathrm{sin}}{}\left({x}\right)}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Double Angle}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sin}{}\left(2{}x\right)=2{}\mathrm{sin}{}\left(x\right){}\mathrm{cos}{}\left(x\right)\\ {}& {}& \frac{\left({\mathrm{cos}}{}\left({x}\right){}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right){-}{\mathrm{sin}}{}\left({x}\right){}{\mathrm{cos}}{}\left({2}{}{x}\right)\right){}{\mathrm{sec}}{}\left({x}\right)}{{\mathrm{sin}}{}\left({x}\right)}\\ \text{•}& {}& \text{Factor the numerator}\\ {}& {}& \frac{\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right){}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sec}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)}{{\mathrm{sin}}{}\left({x}\right)}\\ \text{•}& {}& \text{Cancel out a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sin}{}\left(x\right)\\ {}& {}& \left(\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right){}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sec}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Half Angle}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cos}{}\left(x\right)}^{2}=\frac{\mathrm{cos}{}\left(2{}x\right)}{2}+\frac{1}{2}\\ {}& {}& \left({2}{}\left(\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}{}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\right){-}{\mathrm{cos}}{}\left({2}{}{x}\right)\right){}{\mathrm{sec}}{}\left({x}\right)\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {\mathrm{sec}}{}\left({x}\right)\\ \text{•}& {}& \text{Thus we have proved that the identity is true}\\ {}& {}& {\mathrm{sec}}{}\left({x}\right){=}{\mathrm{sec}}{}\left({x}\right)\end{array}$ (8)

SolveSteps is also capable of solving systems of linear inequalities

 > $\mathrm{SolveSteps}\left(\left[\frac{1}{2}
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{1st}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& \frac{{1}}{{2}}{<}{x}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{2nd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {2}{\le }{x}\\ \text{•}& {}& \text{The solved system is:}\\ {}& {}& \left[\right]\\ \text{•}& {}& \text{Graph the boundary lines of the inequalites}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Show inequalities}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Solution is where the inequalities overlap}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (9)

Use the optional parameter colorpack to specify an alternate color palette for inequality plots.

 > $\mathrm{SolveSteps}\left(\left[y\le 2x+\frac{7}{2},y\le -2x+\frac{7}{2},-\frac{2x}{3}-1\le y,\frac{2x}{3}-1\le y,y\le \frac{3}{2}\right],\mathrm{colorpack}="MapleV"\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{1st}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {y}{\le }{2}{\cdot }{x}{+}\frac{{7}}{{2}}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{2nd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {y}{\le }\left({-2}\right){\cdot }{x}{+}\frac{{7}}{{2}}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{3rd}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& \frac{{-2}}{{3}}{\cdot }{x}{-}{1}{\le }{y}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{4th}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& \frac{{2}}{{3}}{\cdot }{x}{-}{1}{\le }{y}\\ \text{•}& {}& \text{Examine the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{5th}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{inequality and solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {y}{\le }\frac{{3}}{{2}}\\ \text{•}& {}& \text{The solved system is:}\\ {}& {}& \left[\right]\\ \text{•}& {}& \text{Graph the boundary lines of the inequalites}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Show inequalities}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Solution is where the inequalities overlap}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (10)

SolveSteps is also capable of solving nonlinear inequalities

 > $\mathrm{SolveSteps}\left("x^2 - 4*x + 4 > 7"\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {7}{<}{{x}}^{{2}}{-}{4}{\cdot }{x}{+}{4}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}7\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from both sides}\\ {}& {}& {{x}}^{{2}}{-}{4}{}{x}{+}{4}{-}{7}{=}{7}{-}{7}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {{x}}^{{2}}{-}{4}{}{x}{-}{3}{=}{0}\\ \text{•}& {}& \text{Since we can't factor we'll use the quadratic formula}\\ {}& {}& {x}{=}\frac{\left({-}{b}\right){±}\left(\sqrt{{{b}}^{{2}}{-}{4}{\cdot }{a}{\cdot }{c}}\right)}{{2}{\cdot }{a}}\\ \text{▫}& {}& \text{Use quadratic formula to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& \text{◦}& \text{Substitute a=}1\text{, b=}-4\text{, c=}-3\\ {}& {}& {x}{=}\frac{{4}{±}\left(\sqrt{{\left({-4}\right)}^{{2}}{-}{4}{\cdot }{1}{\cdot }\left({-3}\right)}\right)}{{2}{\cdot }{1}}\\ {}& \text{◦}& \text{Evaluate under discriminant}\\ {}& {}& {x}{=}\frac{{4}{±}\left(\sqrt{{16}{-}\left({-12}\right)}\right)}{{2}{\cdot }{1}}\\ {}& \text{◦}& \text{Perform remaining operations}\\ {}& {}& {x}{=}{2}{±}\left(\sqrt{{7}}\right)\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left({2}{-}\sqrt{{7}}{,}{2}{+}\sqrt{{7}}\right)\\ \text{•}& {}& \text{Use the solutions to the equality as points to test for intervals}\\ {}& {}& \left[{2}{-}\sqrt{{7}}{,}{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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& \text{◦}& \text{Sub}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}0<{x}^{2}-4{}x-3\\ {}& {}& {0}{<}{2}\\ {}& {}& {\mathrm{true}}\\ {}& \text{◦}& \text{Sub}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}0<{x}^{2}-4{}x-3\\ {}& {}& {0}{<}{-7}\\ {}& {}& {\mathrm{false}}\\ {}& \text{◦}& \text{Sub}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=5\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}0<{x}^{2}-4{}x-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\}{,}\left\{{2}{+}\sqrt{{7}}{<}{x}\right\}\right]\end{array}$ (11)
 > $\mathrm{SolveSteps}\left("x + 4/x > 4"\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {4}{<}{x}{+}\frac{{4}}{{x}}\\ \text{•}& {}& \text{Note the values for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4{}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from both sides}\\ {}& {}& {{x}}^{{2}}{+}{4}{-}{4}{}{x}{=}{4}{}{x}{-}{4}{}{x}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {{x}}^{{2}}{-}{4}{}{x}{+}{4}{=}{0}\\ \text{▫}& {}& \text{Factor using the AC Method}\\ {}& \text{◦}& \text{Look at the coefficients,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}A{}{x}^{2}+B{}x+C\\ {}& {}& \left[{"A"}{=}{1}{,}{"B"}{=}{-4}{,}{"C"}{=}{4}\right]\\ {}& \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{,}{2}{,}{4}\right\}\\ {}& \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left\{{1}{\cdot }{4}{,}{2}{\cdot }{2}\right\}\\ {}& \text{◦}& \text{Which pairs of ± these factors have a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{sum}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of B =}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-4\text{? Found:}\\ {}& {}& {-2}{-}{2}{=}{-4}\\ {}& \text{◦}& \text{Split the middle term to use above pair}\\ {}& {}& {{x}}^{{2}}{+}\left({-}{2}{}{x}{-}{2}{}{x}\right){+}{4}\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the first group}\\ {}& {}& {x}{\cdot }\left({x}{-}{2}\right){+}\left({-}{2}{}{x}{+}{4}\right)\\ {}& \text{◦}& \text{Factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{out of the second group}\\ {}& {}& {x}{\cdot }\left({x}{-}{2}\right){-}{2}{\cdot }\left({x}{-}{2}\right)\\ {}& \text{◦}& x-2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& {x}{-}{2}\\ \text{▫}& {}& \text{Solution of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x-2=0\\ {}& \text{◦}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to both sides}\\ {}& {}& {x}{-}{2}{+}{2}{=}{0}{+}{2}\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& {x}{=}{2}\\ \text{•}& {}& \text{Use the solutions and undefined values as points to test for intervals}\\ {}& {}& \left[{0}{,}{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}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& \text{◦}& \text{Sub}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4 (12)

SolveSteps is also capable of solving expressions with absolute values

 > $\mathrm{SolveSteps}\left(\mathrm{abs}\left(x+1\right)=4x\right)$
 $\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[\right]\\ \text{•}& {}& \text{Thus our intervals are:}\\ {}& {}& \left[\left({-}{\mathrm{\infty }}{,}{-1}\right){,}\left[{-1}{,}{\mathrm{\infty }}\right)\right]\\ \text{▫}& {}& \text{Examine absolute values with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\in \left(-\mathrm{\infty },-1\right)\\ {}& \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ {}& {}& \left[\begin{array}{c}x+1{<}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}\\ {}& {}& {-}{x}{-}{1}{=}{4}{}{x}\\ {}& \text{◦}& \text{Solve the new equality}\\ {}& {}& {x}{=}{-}\frac{{1}}{{5}}\\ {}& \text{◦}& \text{Since}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-\frac{1}{5}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{∉}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\left(-\mathrm{\infty },-1\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{we get that this is not a solution}\\ {}& {}& {x}{\ne }{-}\frac{{1}}{{5}}\\ \text{▫}& {}& \text{Examine absolute values with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\in \left[-1,\mathrm{\infty }\right)\\ {}& \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ {}& {}& \left[\begin{array}{c}x+1\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}\\ {}& {}& {x}{+}{1}{=}{4}{}{x}\\ {}& \text{◦}& \text{Solve the new equality}\\ {}& {}& {x}{=}\frac{{1}}{{3}}\\ {}& \text{◦}& \text{Since}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{1}{3}\in \left[-1,\mathrm{\infty }\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{we get that this is a solution}\\ {}& {}& {x}{=}\frac{{1}}{{3}}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\frac{{1}}{{3}}\end{array}$ (13)
 > $\mathrm{SolveSteps}\left(\mathrm{abs}\left(2x+6\right)=\mathrm{abs}\left(x+7\right)\right)$
 $\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[\right]\\ \text{•}& {}& \text{Thus our intervals are:}\\ {}& {}& \left[\left({-}{\mathrm{\infty }}{,}{-7}\right){,}\left[{-7}{,}{-3}\right){,}\left[{-3}{,}{\mathrm{\infty }}\right)\right]\\ \text{▫}& {}& \text{Examine absolute values with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\in \left(-\mathrm{\infty },-7\right)\\ {}& \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ {}& {}& \left[\begin{array}{c}x+3{<}0\\ x+7{<}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}{}{x}{-}{6}{=}{-}{x}{-}{7}\\ {}& \text{◦}& \text{Solve the new equality}\\ {}& {}& {x}{=}{1}\\ {}& \text{◦}& \text{Since}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{∉}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\left(-\mathrm{\infty },-7\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{we get that this is not a solution}\\ {}& {}& {x}{\ne }{1}\\ \text{▫}& {}& \text{Examine absolute values with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\in \left[-7,-3\right)\\ {}& \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ {}& {}& \left[\begin{array}{c}x+3{<}0\\ x+7\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}{}{x}{-}{6}{=}{x}{+}{7}\\ {}& \text{◦}& \text{Solve the new equality}\\ {}& {}& {x}{=}{-}\frac{{13}}{{3}}\\ {}& \text{◦}& \text{Since}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-\frac{13}{3}\in \left[-7,-3\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{we get that this is a solution}\\ {}& {}& {x}{=}{-}\frac{{13}}{{3}}\\ \text{▫}& {}& \text{Examine absolute values with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\in \left[-3,\mathrm{\infty }\right)\\ {}& \text{◦}& \text{Determine whether the inside of the absolute value will be positive or negative}\\ {}& {}& \left[\begin{array}{c}x+3\ge 0\\ x+7\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}{}{x}{+}{6}{=}{x}{+}{7}\\ {}& \text{◦}& \text{Solve the new equality}\\ {}& {}& {x}{=}{1}\\ {}& \text{◦}& \text{Since}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\in \left[-3,\mathrm{\infty }\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{we get that this is a solution}\\ {}& {}& {x}{=}{1}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left({-}\frac{{13}}{{3}}{,}{1}\right)\end{array}$ (14)

Compatibility

 • The Student[Basics][SolveSteps] command was introduced in Maple 2021.