The LinearSolveSteps command accepts a linear equation expr in the given variable, var, and displays the steps required to solve for that variable.

Note that this command also accepts some nonlinear equations that can be reduced down to linear equations (in other words, you can isolate $x$ on one side of the equation, and there is only one solution).

If expr is a string, then it is parsed into an expression using InertForm:-Parse so that no automatic simplifications are applied, and thus no steps are missed.  

The implicitmultiply option is only relevant when expr is a string.  This option is passed directly on to the InertForm:-Parse command and will cause things like "2x" to be interpreted as 2*x, but also, "xyz" to be interpreted as x*y*z.

A step may show up where the expression is not obviously different from the previous step.  This can happen when the underlying data structure is transformed during the step, and it is not obvious that the resulting structure is the same as the original, but just expressed differently.  This becomes more apparent when looking at the inert form of the raw data.

The return value is a module that display annotated steps by default.  This module also has callable methods and data members: data, numsteps, step, and toMathML.  

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

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

$\mathrm{LinearSolveSteps}\left(\frac{x+1}{2yz}=\frac{4y^2}{z}+\frac{3x}{y}\,x\right)$

$\begin{array}{c}\frac{x+1}{2\cdot y\cdot z}=\frac{4\cdot y^2}{z}+\frac{3\cdot x}{y}\\ \frac{x+1}{2\cdot y\cdot z}+\frac{\left(\mathrm{-3}\right)\cdot x}{y}=\frac{4\cdot y^2}{z}+\frac{3\cdot x}{y}+\frac{\left(\mathrm{-3}\right)\cdot x}{y}& \left(\text{Subtract}\frac{3\cdot x}{y}\text{from both sides}\right)\\ \frac{x+1}{2\cdot y\cdot z}+\frac{\left(\mathrm{-3}\right)\cdot x}{y}=\frac{4\cdot y^2}{z}& \left(\text{Simplify}\right)\\ \frac{y\cdot \left(x+1\right)}{2\cdot y\cdot z\cdot y}+\frac{2\cdot y\cdot z\cdot \left(\left(\mathrm{-3}\right)\cdot x\right)}{2\cdot y\cdot z\cdot y}=\frac{4\cdot y^2}{z}& \left(\text{Find common denominator}\right)\\ \frac{y\cdot \left(x+1\right)+2\cdot y\cdot z\cdot \left(\left(\mathrm{-3}\right)\cdot x\right)}{2\cdot y\cdot z\cdot y}=\frac{4\cdot y^2}{z}& \left(\text{Sum over common denominator}\right)\\ \frac{xy+y+2\cdot y\cdot z\cdot \left(\left(\mathrm{-3}\right)\cdot x\right)}{2\cdot y\cdot z\cdot y}=\frac{4\cdot y^2}{z}& \left(\text{Multiply through:}y\cdot \left(x+1\right)\text{=}xy+y\right)\\ \frac{-6xyz+xy+y}{2\cdot y\cdot z\cdot y}=\frac{4\cdot y^2}{z}& \left(\text{Reorder terms}\right)\\ \frac{y\cdot \left(-6xz+x+1\right)}{y\cdot \left(2yz\right)}=\frac{4\cdot y^2}{z}& \left(\text{Factor}\right)\\ \frac{-6xz+x+1}{2yz}=\frac{4\cdot y^2}{z}& \left(\text{Divide}\right)\\ \frac{-6xz+x+1}{2yz}\cdot \left(2yz\right)=2yz\cdot \left(\frac{4\cdot y^2}{z}\right)& \left(\text{Multiply rhs by denominator of lhs}\right)\\ -6xz+x+1=2yz\cdot \left(\frac{4\cdot y^2}{z}\right)& \left(\text{Simplify}\right)\\ -6xz+x+1-1=2yz\cdot \left(\frac{4\cdot y^2}{z}\right)-1& \left(\text{Subtract}1\text{from both sides}\right)\\ -6xz+x=\mathrm{-1}+2yz\cdot \left(\frac{4\cdot y^2}{z}\right)& \left(\text{Simplify}\right)\\ -6xz+x=\mathrm{-1}+\frac{8y^{3}z}{z}& \left(\text{Multiply fraction}\right)\\ -6xz+x=\mathrm{-1}+8y^3& \left(\text{divide}\right)\\ -6xz+x=8y^3-1& \left(\text{Reorder terms}\right)\\ x\cdot \left(1-6z\right)=8y^3-1& \left(\text{Factor}\right)\\ \frac{x\cdot \left(1-6z\right)}{1-6z}=\frac{8y^3-1}{1-6z}& \left(\text{Divide both sides by}1-6z\right)\\ x=\frac{8\cdot y^3-1}{1-6\cdot z}& \left(\text{Simplify}\right)\end{array}$

$\mathrm{ex}≔\mathrm{LinearSolveSteps}\left(\frac{1}{x}-\frac{1}{2}=\frac{3}{4}-\frac{2}{x}\,x\right)$

$\mathrm{ex}≔\begin{array}{c}\frac{1}{x}-\frac{1}{2}=\frac{3}{4}-\frac{2}{x}\\ \frac{1}{x}-\frac{1}{2}+\left(\frac{1}{2}+\frac{2}{x}\right)=\frac{3}{4}-\frac{2}{x}+\left(\frac{1}{2}+\frac{2}{x}\right)& \left(\text{Add}\frac{1}{2}+\frac{2}{x}\text{to both sides}\right)\\ \frac{1}{x}+\frac{2}{x}=\frac{3}{4}+\frac{1}{2}& \left(\text{Simplify}\right)\\ \frac{3}{x}=\frac{3}{4}+\frac{1}{2}& \left(\text{Add terms}\right)\\ \frac{3}{x}=\frac{5}{4}& \left(\text{Add terms}\right)\\ \frac{x}{3}=\frac{4}{5}& \left(\text{Reciprocal of both sides}\right)\\ \frac{x}{3}\cdot 3=3\cdot \left(\frac{4}{5}\right)& \left(\text{Multiply rhs by denominator of lhs}\right)\\ x=\frac{12}{5}& \left(\text{Simplify}\right)\end{array}$

$\mathrm{ex}:-\mathrm{numsteps}$

$8$

$\mathrm{ex}:-\mathrm{step}\left(2\right)$

$\begin{array}{cc}\frac{1}{x}-\frac{1}{2}+\left(\frac{1}{2}+\frac{2}{x}\right)=\frac{3}{4}-\frac{2}{x}+\left(\frac{1}{2}+\frac{2}{x}\right)& \left(\text{Add}\frac{1}{2}+\frac{2}{x}\text{to both sides}\right)\end{array}$

$\mathrm{ex}:-\mathrm{toMathML}\left(\right)$

$\mathrm{LinearSolveSteps}\left(''x\; +\; 3^2\; =\; 12''\,x\right)$

$\begin{array}{c}x+3^2=12\\ x+3^2-3^2=12-3^2& \left(\text{Subtract}{3}^2\text{from both sides}\right)\\ x=12-3^2& \left(\text{Simplify}\right)\\ x=12-9& \left(\text{Evaluate power}\right)\\ x=3& \left(\text{Add terms}\right)\end{array}$

$\mathrm{LinearSolveSteps}\left(''3(x-2)\; =\; 0''\,x\,'\mathrm{implicitmultiply}'\right)$

$\begin{array}{c}3\cdot \left(x-2\right)=0\\ 3x-6=0& \left(\text{Multiply through:}3\cdot \left(x-2\right)\text{=}3x-6\right)\\ 3x-6+6=0+6& \left(\text{Add}6\text{to both sides}\right)\\ 3x=6& \left(\text{Simplify}\right)\\ \frac{3\cdot x}{3}=\frac{6}{3}& \left(\text{Divide both sides by}3\right)\\ x=2& \left(\text{Simplify}\right)\end{array}$

The Student[Basics][LinearSolveSteps] command was introduced in Maple 18.

For more information on Maple 18 changes, see Updates in Maple 18.