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

Description

 • The SolveSteps command is used to show the steps of solving a basic student problem.
 • 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.
 • This function is part of the Student:-Basics package.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right):$
 > $\mathrm{SolveSteps}\left(5{ⅇ}^{4x}=16\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left[{}\right]{=}{16}\\ \text{▫}& {}& \text{Convert from exponential equation}\\ {}& \text{◦}& \text{Divide both sides by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}5\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& \left[{}\right]{=}\frac{{16}}{{5}}\\ {}& \text{◦}& \text{Apply ln to each side}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ {}& \text{◦}& \text{Apply ln rule: ln(e^b) = b}\\ {}& {}& {4}{}{x}{=}{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)\\ \text{•}& {}& \text{Divide both sides by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Exact solution}\\ {}& {}& {x}{=}\frac{{\mathrm{ln}}{}\left(\frac{{16}}{{5}}\right)}{{4}}\\ \text{•}& {}& \text{Approximate solution}\\ {}& {}& {x}{=}{0.2907877025}\end{array}$ (1)
 > $\mathrm{SolveSteps}\left({x}^{2}-b,x\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Set expression equal to 0}\\ {}& {}& \left[{}\right]{=}{0}\\ \text{•}& {}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}b\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to both sides}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Simplify}\\ {}& {}& \left[{}\right]{=}{b}\\ \text{•}& {}& \text{Take Square root of both sides}\\ {}& {}& {x}{=}{±}\left[{}\right]\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left(\sqrt{{b}}{,}{-}\sqrt{{b}}\right)\end{array}$ (2)
 > $\mathrm{SolveSteps}\left({x}^{3}+4{x}^{2}+4x,\mathrm{output}=\mathrm{typeset}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Set expression equal to 0}\\ {}& {}& \left[{}\right]{=}{0}\\ \text{•}& {}& \text{Factor}\\ {}& {}& {{x}}^{{3}}{+}{4}{}{{x}}^{{2}}{+}{4}{}{x}{=}{0}\\ \text{•}& {}& \text{Common factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Examine term:}\\ {}& {}& {{x}}^{{2}}{+}{4}{}{x}{+}{4}{=}{0}\\ \text{▫}& {}& \text{Apply the AC Method}\\ {}& \text{◦}& \text{Examine quadratic}\\ {}& {}& \left[{}\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\{\left[{}\right]{,}\left[{}\right]\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:}\\ {}& {}& \left[{}\right]{=}{4}\\ {}& \text{◦}& \text{Split the middle term to use above pair}\\ {}& {}& \left[{}\right]\\ {}& \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 pair}\\ {}& {}& \left[{}\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 pair}\\ {}& {}& \left[{}\right]\\ {}& \text{◦}& x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a common factor}\\ {}& {}& \left[{}\right]\\ {}& \text{◦}& \text{Group common factor}\\ {}& {}& \left[{}\right]\\ {}& {}& \text{This gives:}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{This gives:}\\ {}& {}& \left[{}\right]\\ \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}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ {}& \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}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Set expression equal to 0}\\ {}& {}& \left[{}\right]{=}{0}\\ \text{•}& {}& \text{Factor}\\ {}& {}& {{x}}^{{3}}{+}{4}{}{{x}}^{{2}}{+}{4}{}{x}{=}{0}\\ \text{•}& {}& \text{Common factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Examine term:}\\ {}& {}& {{x}}^{{2}}{+}{4}{}{x}{+}{4}{=}{0}\\ \text{▫}& {}& \text{Apply the AC Method}\\ {}& \text{◦}& \text{Examine quadratic}\\ {}& {}& \left[{}\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\{\left[{}\right]{,}\left[{}\right]\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:}\\ {}& {}& \left[{}\right]{=}{4}\\ {}& \text{◦}& \text{Split the middle term to use above pair}\\ {}& {}& \left[{}\right]\\ {}& \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 pair}\\ {}& {}& \left[{}\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 pair}\\ {}& {}& \left[{}\right]\\ {}& \text{◦}& x+2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is a common factor}\\ {}& {}& \left[{}\right]\\ {}& \text{◦}& \text{Group common factor}\\ {}& {}& \left[{}\right]\\ {}& {}& \text{This gives:}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{This gives:}\\ {}& {}& \left[{}\right]\\ \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}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ {}& \text{◦}& \text{Simplify}\\ {}& {}& {x}{=}{-2}\\ \text{•}& {}& \text{Solution}\\ {}& {}& {x}{=}\left({-2}{,}{0}\right)\end{array}$ (4)

Compatibility

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