Mini Project

Robert L. Case, Jr.

Algebra I
Using Mathematica to teach Algebra I

Algebra I

Unity High School

Robert L. Case, Jr.

Unity High School does not offer Calculus as part of its high school curriculum.  Therefore, I intend to give some examples of possible uses in Algebra I.  In the teaching 
of Algebra I at Unity High School, Mathematica can be used to supplement the material 
in the text in the following manner.  I have chosen problems at random to demonstrate Mathematica's value.  The text used to teach Algebra I at Unity High School is the 
Merrill Algebra I book.  Chapter 1 involves the math topics concerning Expressions and Equations.  Section 5 first introduces the distributive property and simplifying expressions.  The following problems are from the text.

Simplify the expression.
3a + 7a

Clear[a]
Factor[3a + 7a]

Since this expression has a common variable, the coefficients can be added together to find a solution.

Simplify the expression.
5x + 3(x - y)

Clear[x,y]
Expand[3(x - y)]

First, you should distribute the 3 across the parentheses.  This should give the result of 3x - 3y.

Factor[5x + 3(x - y)]

Simplify the expression.
3/4y + x/4 + 3x

Clear[x]
Factor[x/4 + 3x]

The first part of this simplification is to combine the terms with the variable x.  To accomplish this task, the terms must have a common denominator.  This is done by multiplying 3x by 4/4.

Factor[3/4y + x/4 + 3x]

This final step shows the adding of 3/4y to the previosly simplified expression.  Since the denominators are already identical, the numerators can be combined if possible. This problem could have been solved in one step instead of two, but the first step was added to illustrate the two parts of this solution that would needed to be solved if this was done by hand.

If the hardware is available, the value of doing these problems on Mathematica is that the students would be able to check their work instantly.  The math concepts behind simplifying expressions could be covered in the traditional lecture session.  Then each student would be required to work practice problems in class.  The students would then have to explain their solutions verbally or by showing the work on the chalkboard.  The previous problems are examples of how the student can use Mathematica. What follows are examples of how Mathematica can be used to present math topics.

The equation for finding the surface area of a rectangular solid can be computed by using the formula 
2(lh + wh + lw).
Open this cell to get a 3-D drawing of a cube with dimensions of 30. After the drawing, the program computes the surface area using the formula above.

Clear[l,w,h]
l = 30;
w = 30;
h = 30;
Show[Graphics[Line[{{0,0},{0,h},{w,h},{w,0},{0,0},{0,h},
{w/2,h+h/2},{w+w/2,h+h/2},{w+w/2,h/2},{w,0},{w,h},
{w+w/2,h+h/2}}]],PlotRange->{{0,90},{0,45}},
AspectRatio->Automatic];

N[2(l h + w h + l w)]

Now go back and change the width of the rectangular solid to 15 and have the computer draw the figure.  Notice that the figure is half the size of the cube.  Does this mean that the surface area should also be half, or equal to 2700?  If you have the program figure out the surface area of the new rectangular solid, you will see that the answer is no.  Why not?

The same visual representation as above can be used in finding the area of a trapezoid.  The equation for the area of a trapezoid is 1/2h(a + b) where a and b are the lengths of the parallel sides and h is the height.  A common trapezoid is a rectangle.  Open this files to find the area of the trapezoid below with parallel sides of lenght 30 and a height of 40.

Clear[h,a,b]
h = 40;
a = 30;
b = 30;

If[a >= b,perp := b,perp := a];
If[a >= b,x := 2a,x := 2b];
center := x/2;
pos := h/20
Show[Graphics[{Line[{{center - b/2,0},{center - a/2,h},
                    {center + a/2,h},{center + b/2,0},
                    {center - b/2,0}}],
              Text[b,{center,pos}],
              Text[a,{center,h-pos}],
              Text[h,{center + pos - perp/2,h/2}],
              RGBColor[1,0,0],
              Line[{{center - perp/2,0},{center - perp/2,h}}],
              Line[{{center - perp/2,pos},{center + pos - perp/2,pos}
                  ,{center + pos - perp/2,0}}],
              Line[{{center - perp/2,h-pos},
                   {center + pos - perp/2,h-pos},
                   {center + pos - perp/2,h}}]}],
PlotRange->{{0,x},{0,h}},
AspectRatio->Automatic];

N[h/2(a + b)]

If you multiply the length and width of this rectangle you also get 1200 for your area.  Change the value of a to 15 instead of 30 and have Mathematica draw your figure and compute the area.  Do you get half the area since you divided the length of one of the sides in half?  Now make the height (h) 20 instead of 40.  Do you get half the area now?

Advancing through the text, the student will come upon this problem in the homework set for Chapter 3.

A trapezoid has an area of 117 sq. ft. and a height of 9 ft.  One base is 4 ft. shorter than the other base.  How long is each base?

If the student remembers the previous problem from chapter one, they can use the work already done to help solve this problem.

Assign the height to 9, the side a to be 4 less than the side b, and the area A to 117.  Then plug it into the equation A = 1/2h(a + b) and solve.  Mathematica gave the solution of one of the sides equal to 15.  Then the student must realize that the answer to the question requires the measures of a and b.

Clear[h,a,b,A]
h = 9;
a = b - 4;
A = 117
Solve[A==1/2 h(a + b),b]

b = Solve[A==1/2 h(a + b),b][[1,1,2]]
N[a]

Now Mathematica gave the answer to both sides.  An advantage of using Mathematica is that now a student can see what the figure looks just for fun.

If[a >= b,perp := b,perp := a];
If[a >= b,x := 2a,x := 2b];
center := x/2;
pos := h/20
Show[Graphics[{Line[{{center - b/2,0},{center - a/2,h},
                    {center + a/2,h},{center + b/2,0},
                    {center - b/2,0}}],
              Text[b,{center,pos}],
              Text[a,{center,h-pos}],
              Text[h,{center + pos - perp/2,h/2}],
              RGBColor[1,0,0],
              Line[{{center - perp/2,0},{center - perp/2,h}}],
              Line[{{center - perp/2,pos},{center + pos - perp/2,pos}
                  ,{center + pos - perp/2,0}}],
              Line[{{center - perp/2,h-pos},
                   {center + pos - perp/2,h-pos},
                   {center + pos - perp/2,h}}]}],
PlotRange->{{0,x},{0,h}},
AspectRatio->Automatic];

Show[Graphics[{Line[{{8,0},{10,9},{21,9},{23,0},{8,0}}]}],
              PlotRange->{{0,30},{0,9}},
              AspectRatio->Automatic];

Notice that there are two graphings of the same trapezoid.  The second trapezoid is exactly the same size as the first trapezoid, but it is devoid of the labels and the height.  The coding for the second trapezoid is obviously much easier.  A student would be able to enhance their output according to how well they understand Mathematica. The sky's the limit!

By chapter 5, the students should know how to use the distributive property to solve equations.  By using Mathematica, the students can work at a faster pace, because they do not have to do the repetitive work.  As an example, when the students are doing age and mixture problems, the area that gives the student the most difficulty is interpreting the word problem into an equation.  Solving the equation is the easy part.  By letting Mathematica solve the equations, the assignments can be longer and will let the students concentrate on setting up equations which is where the emphasis should be.

Matt is 5 years older than his sister, and 23 years younger than his mother.  If the sum of Matt's age and his mother's age is 35, how old is Matt?

Clear[sister,Matt,mother]
sister := Matt - 5
Mother := Matt + 23
Solve[Matt + Mother == 35,Matt]

At the end of Chapter 5, the students are required to solve several science problems involving percents.  The processes involved in solving the problems are very similar to the age problems.  Again, Mathematica would assist the students in solving the equations after the students interpreted them from the word problem.

An aluminum alloy containing 48% aluminum is to be made by combining 30% and 60% alloys.  How many pounds of the 60% alloy must be added to 24 pounds of the 30% alloy to produce the desired alloy?

Clear[alloy1,alloy2,pounds]
alloy1 := 24 .30
alloy2 := .60 pounds
alumalloy := .48(pounds + 24)
Solve[alumalloy == alloy1 + alloy2,pounds]

The generic solution is: The amount of aluminum in the mixture is equal to the sum of the amounts of aluminum in each solution.  Once the student solves this problem, the coding can be used to solve the very next problem, with some minor alterations.

A car radiator has a capacity of 16 quarts and is filled with 25% antifreeze solution.  How much must be drained off and replaced with pure antifreeze to obtain a 40% antifreeze solution?

Clear[alloy1, quarts]
alloy1 := .25
Solve[.40 16 == alloy1(16 - quarts) + quarts,quarts]

For the students to change the aluminum solution program to solve the antifreeze problem, they must have a firm understanding of the procedures involved in solving these problems.

The last section of Chapter 5 involves the students solving problems involving levers.  The graphic capabilities can be used to illustrate what happens to fulcrum as the weights on the system are changed.

Mary Jo weighs 120 pounds and Doug weighs 160 pounds.  They are seated at opposite ends of a seesaw.  Doug and Mary Jo are 14 feet apart, and the seesaw is balanced.  How far is Mary Jo from the fulcrum?

Run the following animation by hitting the apple key and the y key after executing the following two cells.  Observe the distance of the fulcrum when the weight on the left is 120.

Clear[length,dist1,ansdist1,weight1,dist2,weight2]
length = 14
dist2 = length - dist1
weight2 = 160;

Do[
  ansdist1:=Solve[weight1 dist1 == weight2 dist2, dist1][[1,1,2]]//N;
  Show[Graphics[{Line[{{0,2},{ansdist1,2},{ansdist1 + 1,1.6},
                    {ansdist1 - 1,1.6},{ansdist1,2},{length,2}}],
               Text[ansdist1,{ansdist1,1.5}],
               Text[weight1,{0,2.1},{-1,0}],
               Text[weight2,{length,2.1},{1,0}]},
               PlotRange -> {{0,length + 2},{0,3}}]]  ,{weight1,5,320,5}]

By this time of the year, several students will be able to design the coding to find the solution on their own.  Part of the class time would still be used for instruction and Mathematica would be used to get the solution and a visual representation of what is actually happening.

Clear[length,dist1,weight1,dist2,weight2]
length = 14
dist2 = length - dist1
weight1 = 120
weight2 = 160;

Solve[dist1 weight1 == dist2 weight2,dist1]//N

Length is the distance apart of the objects, weight1 is the weight of the object on the left, weight2 is the weight of the object on the right, dist1 is the distance of the left object from the fulcrum and dist2 is the distance of the object on the right from the fulcrum.

Once the students determine the relationship between the different variables, Mathematica would solve the equation and get a solution.  The really exciting part will be the picture of the lever.  The animation shows the students what must happen to the position of the fulcrum as the weights change to keep it balanced.

Chapter 6 concerns solving equations involving the product of polynomials.  The area of a rectangle is used in the following homework assignment.

The length of a rectangle is 20 yards greater than the width.  If the length was decreased by 5 yards, and the width increased by 4 yards, the area would remain unchanged.  Find the original dimensions of the rectangle.

Again, the students must be able to recognize the relationships between the various sides before they can get Mathematica to find the solution.  Mathematica was used to make the figure above, but the coding has been deleted and only the figure remains.

Clear[length,newlength,width,newwidth]
length := width + 20
newlength := length - 5
newwidth := width + 4
Solve[newlength newwidth == length width,width]

width = Solve[newlength newwidth == length width,width][[1,1,2]]

N[length]

Mathematica was used to create the following figures for the sole purpose of generating either a handout or overhead.  What follows is a geometric representation of the difference of squares.

Several sections in chapter 13 involves solving quadratics by interpreting the graphs of the functions.  Few, if any, of our students have graphing calculators.  The problem in determining the solutions of a quadratic by graphing is that the graphs must be perfect to get a reasonable approximation.  Mathematica can be easily programmed to produce the graph of a quadratic.

4x¤ + 4x - 35 = 0

Clear[f,x]
f[x_] := 4x^2 + 4x - 35
Plot[f[x],{x,-20,20},AxesLabel->{"x","f[x]"},PlotRange->{-40,25}];

Notice that it is difficult to determine approximately where the function crosses the x-axis.  Change the ranges of x from (-20 to 20) to (-10 to 10) to (-5 to 5) and then to (-4 to 3).  Each time the graph is changed, it becomes easier to determine an aproximate solution.  Doing this procedure by hand would take more time than could be realistically devoted to one problem in a typical class period.  Now change the ranges of x to (-4 to -3).  This will just concentrate on one solution, but will give you a much more acurate picture of a possible solution.  Mathematica could then be programmed to find the exact solution to see if the students approximate solutions are reasonable.

Solve[f[x] == 0,{x}]

I chose to concentrate on Algebra I as the area of study to demonstrate the possible uses of Mathematica in classes that students take prior to Calculus in the average high school curriculum.  Since starting this project, I have thought of several areas of study in Pre-Algebra and Refresher Math that could benefit from using Mathematica.