SolveSteps - Maple Help
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Student[Basics]

  

SolveSteps

  

show steps in the solution of a specified problem

 

Calling Sequence

Parameters

Description

Examples

Compatibility

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

withStudent:-Basics:

SolveSteps5exp4x=16

Let's solve5ⅇ4x=16Convert from exponential equationDivide both sides by55ⅇ4x5=165Simplifyⅇ4x=165Apply ln to each sidelnⅇ4x=ln165Apply ln rule: ln(e^b) = b4x=ln165Divide both sides by44x4=ln1654Exact solutionx=ln1654Approximate solutionx=0.2907877025

(1)

SolveStepsx2b,x

Let's solvex21bSet expression equal to 0x21b=0Addbto both sidesx21b+b=0+bSimplifyx2=bTake Square root of both sidesx=±bSolutionx=b,b

(2)

SolveStepsx3+4x2+4x,output=typeset

Let's solvex3+4x2+4xSet expression equal to 0x3+4x2+4x=0Common factorxxx2+4x+4Examine term:x2+4x+4Factor using the AC MethodExamine quadraticx2+4x+4Look at the coefficients,Ax2+Bx+CA=1,B=4,C=4Find factors of |AC| = |14| =41,2,4Find pairs of the above factors, which, when multiplied equal414,22Which pairs of ± these factors have asumof B =4? Found:2+2=4Split the middle term to use above pairx2+2x+2x+4Factorxout of the first groupxx+2+2x+4Factor2out of the second groupxx+2+2x+2x+2is a common factorxx+2+2x+2Group common factorx+2x+2This gives:x+22This gives:xx+22The1stfactor isxwhich impliesx= 0 is a solutionx=0Set2ndfactorx+2to 0 to solvex+2=0Solution ofx+2=0Subtract2from both sidesx+22=02Simplifyx=−2Solutionx=−2,0

(3)

SolveStepsx3+4x2+4x,mode=Learn

Let's solvex3+4x2+4xSet expression equal to 0x3+4x2+4x=0Common factorxxx2+4x+4Examine term:x2+4x+4Factor using the AC MethodExamine quadraticx2+4x+4Look at the coefficients,Ax2+Bx+CA=1,B=4,C=4Find factors of |AC| = |14| =41,2,4Find pairs of the above factors, which, when multiplied equal414,22Which pairs of ± these factors have asumof B =4? Found:2+2=4Split the middle term to use above pairx2+2x+2x+4Factorxout of the first groupxx+2+2x+4Factor2out of the second groupxx+2+2x+2x+2is a common factorxx+2+2x+2Group common factorx+2x+2This gives:x+22This gives:xx+22The1stfactor isxwhich impliesx= 0 is a solutionx=0Set2ndfactorx+2to 0 to solvex+2=0Solution ofx+2=0Subtract2from both sidesx+22=02Simplifyx=−2Solutionx=−2,0

(4)

SolveSteps is also capable of proving trigonometric identities

SolveStepscscxtanxcosx=1

Let's solveLet's simplify the left side of the expression to match the rightcscxtanxcosx=1ApplyQuotienttrig identity,tanx=sinxcosxcscxsinxcosxcosxApplyReciprocal Functiontrig identity,cscx=1sinx1sinxsinxEvaluate1Thus we have proved that the identity is true1=1

(5)

SolveStepscosx21+sinx=1sinx

Let's solveLet's simplify the left side of the expression to match the rightcosx21+sinx=1sinxApplyPythagorastrig identity,cosx2=1sinx21sinx21+sinxFactor the numeratorsinx11+sinx1+sinxCancel out a factor of1+sinx1sinxEvaluate1sinxThus we have proved that the identity is true1sinx=1sinx

(6)

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

SolveStepssin5x=16sinx520sinx3+5sinx,trigtimer=

Let's solveLet's simplify the right-side of the expression to match the leftsin5x=16sinx520sinx3+5sinxApplyFull Power Reductiontrig identity,sinx3=sin3x4+3sinx416sinx520sin3x4+3sinx4+5sinxApplyFull Power Reductiontrig identity,sinx5=sin5x165sin3x16+5sinx816sin5x165sin3x16+5sinx8+5sin3x10sinxEvaluatesin5xThus we have proved that the identity is truesin5x=sin5x

(7)

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

SolveStepssin2xsinxcos2xcosx=secx,trigpath=2

Let's solveLet's simplify the left side of the expression to match the rightsin2xsinxcos2xcosx=secxFind fractions to get lowest common denominator ofsinxcosxcosxcosxsin2xsinx+sinxsinxcos2xcosxMultiplycosxsin2xsinxcosx+sinxcos2xsinxcosxAdd fractionscosxsin2xsinxcos2xsinxcosxApplyReciprocal Functiontrig identity,1cosx=secxcosxsin2xsinxcos2xsecxsinxApplyDouble Angletrig identity,sin2x=2sinxcosxcosx2sinxcosxsinxcos2xsecxsinxFactor the numeratorsinx2cosx2cos2xsecxsinxCancel out a factor ofsinx2cosx2cos2xsecxApplyHalf Angletrig identity,cosx2=cos2x2+122cos2x2+12cos2xsecxEvaluatesecxThus we have proved that the identity is truesecx=secx

(8)

SolveSteps is also capable of solving systems of linear inequalities

SolveSteps12<x&comma;2x

Let's solveExamine the1stinequality and solve forx12<xExamine the2ndinequality and solve forx2xThe solved system is:Graph the boundary lines of the inequalitesPLOT...Show inequalitiesPLOT...Solution is where the inequalities overlapPLOT...

(9)

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

SolveStepsy2x+72&comma;y2x+72&comma;2x31y&comma;2x31y&comma;y32&comma;colorpack=MapleV

Let's solveExamine the1stinequality and solve foryy2x+72Examine the2ndinequality and solve foryy−2x+72Examine the3rdinequality and solve fory−23x1yExamine the4thinequality and solve fory23x1yExamine the5thinequality and solve foryy32The solved system is:Graph the boundary lines of the inequalitesPLOT...Show inequalitiesPLOT...Solution is where the inequalities overlapPLOT...

(10)

SolveSteps is also capable of solving nonlinear inequalities

SolveStepsx^2 - 4*x + 4 > 7

Let's solve7<x24x+4Solve forxto find points to test for intervals7=x24x+4Rearrange expressionx24x+4=7Subtract7from both sidesx24x+47=77Simplifyx24x3=0Since we can't factor we'll use the quadratic formulax=b±b24ac2aUse quadratic formula to solve forxSubstitute a=1, b=−4, c=−3x=4±−4241−321Evaluate under discriminantx=4±16−1221Perform remaining operationsx=2±7Solutionx=27&comma;2+7Use the solutions to the equality as points to test for intervals27&comma;2+7Set up a table using the solutions as boundaries and find test points that are on either sidePLOT...Sub each test point into the expression forxSubx=−1into0<x24x30<2trueSubx=2into0<x24x30<−7falseSubx=5into0<x24x30<2trueObserve where the inequality holds true, these areas make up the intervalsPLOT...Plotted solutionPLOT...Solutionx<27&comma;2+7<x

(11)

SolveStepsx + 4/x > 4

Let's solve4<x+4xNote the values forxwhich causes the expression to be undefined. These values will be used later to identify the solution intervalsx=0Solve forxto find points to test for intervals4=x+4xRearrange expressionx+4x=4Multiply both sides byxxx+x4x=x4Evaluatex2+4=4xSubtract4xfrom both sidesx2+44x=4x4xSimplifyx24x+4=0Factor using the AC MethodLook at the coefficients,Ax2+Bx+CA=1&comma;B=−4&comma;C=4Find factors of |AC| = |14| =41&comma;2&comma;4Find pairs of the above factors, which, when multiplied equal414&comma;22Which pairs of ± these factors have asumof B =−4? Found:−22=−4Split the middle term to use above pairx2+2x2x+4Factorxout of the first groupxx2+2x+4Factor−2out of the second groupxx22x2x2is a common factorxx22x2Group common factorx2x2This gives:x22=0Examine factor1x2Solution ofx2=0Add2to both sidesx2+2=0+2Simplifyx=2Use the solutions and undefined values as points to test for intervals0&comma;2Set up a table using the solutions as boundaries and find test points that are on either sidePLOT...Sub each test point into the expression forxSubx=−1into4<x+4x4<−5falseSubx=1into4<x+4x4<5trueSubx=3into4<x+4x4<133trueObserve where the inequality holds true, these areas make up the intervalsPLOT...Plotted solutionPLOT...Solution0<x<2&comma;2<x

(12)

SolveSteps is also capable of solving expressions with absolute values

SolveStepsabsx+1=4x

Let's solvex+1=4xTo 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 negativeThus our intervals are:&comma;−1&comma;−1&comma;Examine absolute values withx&comma;−1Determine whether the inside of the absolute value will be positive or negativex+1<0Drop the absolute values and multiply the expressions that would be negative by -1Sub the new expressions in where the absolute values used to bex1=4xSolve the new equalityx=15Since15&comma;−1we get that this is not a solutionx15Examine absolute values withx−1&comma;Determine whether the inside of the absolute value will be positive or negativex+10Drop the absolute values and multiply the expressions that would be negative by -1Sub the new expressions in where the absolute values used to bex+1=4xSolve the new equalityx=13Since13−1&comma;we get that this is a solutionx=13Solutionx=13

(13)

SolveStepsabs2x+6=absx+7

Let's solve2x+3=x+7To 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 negativeThus our intervals are:&comma;−7&comma;−7&comma;−3&comma;−3&comma;Examine absolute values withx&comma;−7Determine whether the inside of the absolute value will be positive or negativex+3<0x+7<0Drop the absolute values and multiply the expressions that would be negative by -1Sub the new expressions in where the absolute values used to be2x6=x7Solve the new equalityx=1Since1&comma;−7we get that this is not a solutionx1Examine absolute values withx−7&comma;−3Determine whether the inside of the absolute value will be positive or negativex+3<0x+70Drop the absolute values and multiply the expressions that would be negative by -1Sub the new expressions in where the absolute values used to be2x6=x+7Solve the new equalityx=133Since133−7&comma;−3we get that this is a solutionx=133Examine absolute values withx−3&comma;Determine whether the inside of the absolute value will be positive or negativex+30x+70Drop the absolute values and multiply the expressions that would be negative by -1Sub the new expressions in where the absolute values used to be2x+6=x+7Solve the new equalityx=1Since1−3&comma;we get that this is a solutionx=1Solutionx=133&comma;1

(14)

Compatibility

• 

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

• 

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

• 

The Student[Basics][SolveSteps] command was updated in Maple 2024.

• 

The trigpath, trigtimer and colorpack options were introduced in Maple 2024.

• 

For more information on Maple 2024 changes, see Updates in Maple 2024.

See Also

Student:-Basics

Student:-Basics:-FactorSteps

Student:-Basics:-LinearSolveSteps

Student:-Basics:-OutputStepsRecord

Student:-Calculus1:-ShowSolution

Student:-Calculus1:-ShowSteps