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, so it can be used in the short form SolveSteps(..) only after executing the command with(Student[Basics]). However, it can always be accessed through the long form of the command by using Student[Basics][SolveSteps](..).

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

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

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

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

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

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

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

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

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

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

For more information on Maple 2021 changes, see Updates in Maple 2021.