Algebra

Explore Linear Equations And And The Pythagorean Theorem See Quadratic Areas Before Parabolic Equations
	Click Application If Distorted! Click Here, For Video and Mobile Display To Explore Linear Algebra on the Pythabacus The Steps Below Generate a Linear Equation and (X), its Solution. STEP1:Click Application, Then Press Enter/Return or Click Check Button. STEP2:Counting Left From the Rightmost Gold Bead, Left Click a Gold Bead at a position Greater Than two and Less Than Eight. Click Right Arrow (CRA). This Step Generates (X - C), Which Equals the Gold Beads Against the Right Post. STEP3:Counting Left From the Rightmost Gold Bead of the Next Group, Left Click a Gold Beads at any position. CRA. This Step Generates (M), the Gold Beads Displaced to the Middle of the Frame and (Y), Which Equals the Quadrilateral of Brown Beads to the Right and Above the Triangle of Brown Beads Above the Gold Beads. STEP4:Counting Right From the Leftmost Gold Bead of the Group Against the Right Post, Left Click a Gold Beads at any position. CLA, or Counting Left From the Rightmost Gold Bead of the Middle or Last Group, Left Click a Gold Bead at any position CRA. This Step in the First Case Generates a Positive Value for (C) or in the Second Case a Negative Value for (C). Left Click right most bead CLA

Hover mouse over gold beads until one bead flashes and beeps. If no bead flashes click rightmost bead again. StepA: Counting left from the rightmost gold bead left click a bead at a position equal a number that times the number by which (X) is multiplied (M) equals (Y) and CRA until this number of gold beads rest against the right post. In other words left click the bead you now hover over and CRA.	To convert equation to its inverse, click minus button in top right corner of application. StepA: If the number in the second line of the equation is substracted count from the leftmost bead against the right post and left click a gold bead at a position equal the number and CLA to push beads into middle group. Also, push any associated brown beads to the left. If number in second line of equation is added count from the righttmost bead in the middle group and left click a gold bead at a position equal the number and CRA to push beads into rightpost group.
Hover mouse over gold beads until one bead flashes and beeps. StepB: If (M) is positive, count left from the rightmost bead remaining against the left post left click a gold bead at a position equal (M) and CRA. If (M) is negative, count right from the lefttmost bead against the right post left click a gold bead at a position equal (M) and CLA. A number of beads equal (M) are pused midway t0 the right or left . In other words left click the bead you now hover over and CRA or CLA.	StepB: If the number multiplying in the third line of the equation is positive count from the rightmost bead in the middle to beads against the left post and left click a gold bead at a position equal the number and CRA. If number multiplying in the third line of equation is negative click the rightmost bead and CLA to push group into the middle group, count from the lefttmost bead of this group and left click a gold bead to the right of the position equal the number and CRA to push remainder of group back agaist right post.
Hover mouse over gold beads until one bead flashes and beeps. StepC: If the sign on the constant on the left side of the equation is negative, then counting left from the rightmost gold bead of the middle group equal (M), count to a position equal a number that times (M) equals the constant. Left click and CRA until this number of beads rest against the right post. If the sign on the constant on the left side of the equation is positive, then counting right from the leftmost bead of the gold beads against the right post, count to a position equal a number that times (M) equals the constant. Left click and CLA until all beads to the left rest against the left post. In other words left click the bead you now hover over and proceed according to directions above. The number of beads against the right post will equal the solution (X). Press down arrow to see algebraic solution.	StepC: If X is positive, X will equal the number of beads in the rectantgles above the middle group and group against the right post. If X is negative, X will equal the number of beads in the rectantgles above the middle group and group against the right post plus the square of gold beads in the middle group..
Click Here To See Program Below in New Window.
Algebra Areas Video Displayed Below. To See Program Use Scroll Bar Below Frame.

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  Click Application If Distorted!

To Explore Quadratics
      on the Pythabacus

STEP1:Left Click Application. Press Enter/Return.
STEP2:Counting From the Rightmost Gold Bead, Left Click a Gold Bead at a position equal any Number. Press Right Arrow (CRA). In the equation
X"- BX + C = Y, this step generates (B), which equals the displaced gold beads.
STEP3:Left click another gold beads in or out of the displaced group (B). CRA. Of the equation
(x +/- a)*(x +/- b) = y, this step generates (a), the bead clicked in step 3 plus the total number of gold beads to its right, while (b) equals (B) - (a). If (B)>(a), the quadrilateral of brown beads above and to the left of of the a-group equals the product of (a) times (b), equals (C) in the equation
X" - BX + C = Y and C is positive. If (B)<(a) the quadrilateral of brown beads above and to the right of the displaced group plus the square of the number of gold beads in the displaced group equals (C) and (C) is negative.
STEP4:Left click another gold bead in any group. CRA. In the equation
(x +/- e)*(x +/- f)=0, this step generates (e), the bead clicked in step 4 plus the total number of gold beads to its right, while (f) equals (B)-(e).
STEP5:Press Space Bar to review steps below.
Left Click right most bead CLA

Hover mouse over gold beads until bead flashes and beeps. Step A: To solve a given factorable equation, such as the equation above generated by our manipulations of the Pythabacus, first push to the right a group of beads equal the factor of the equation's second term. In other words, left click the bead your mouse now hovers over and CRA six times. Press Space Bar to find next bead.	Hover mouse over gold beads until bead flashes and beeps. Step B: Second, if the third term of the equation is positive, then partition the first group of beads to reveal a quadrilateral of brown beads equal the third term, the product of the parted beads. If the third term of the equation is negative, then to the left of the first group bead push to the right a group of beads that includes a triangle with a number of base beads that when multiplied by the total number of bead not still resting on the left post equals the third term. In other words, left click the bead your mouse now hovers over and CRA Press Space Bar to find next bead.	Hover mouse over gold beads until bead flashes and beeps. Step C: Repeat step B, but for the third term substitute the third term minus the value of (Y) or (C1-Y). In other words, left click the bead your mouse now hovers over and CRA. (X1) and (X2), the roots to the equation will equal, for (X1), the total number of gold beads to the right of the last bead pushed, and for (X2), the first group of beads minus the last group of beads. Press down arrow to See the Algebraic Solution.
	Refresh webpage to generate another example.
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After step 4, three subgroups of the displaced gold beads can be identified. These subgroups can be termed by the order in which the first bead of each group was displaced: First=F, Second=S and Third=T. *Click one of the subgroups in the list and click next 4 times. The areas for a given equation will be defined. *Click the pythabacus and press enter to position the beards against the left post. *Counting from right to left: First click a bead at the count equal the length of a side of the blue square,PRA.* Second click a bead at the count equal the length of a side of the red square, PRA. *Third click a bead at the count equal the length of a side of the green square, PRA. *Slowly move mouse over displaced gold beads and subgroups will be identified by text appearing below the Pythabacus.

The order of the subgroups from left to right will determine the mathematical character of X" - BX + C and therefore the value of (Y). In the chart above, see the formula X"-BX+C and Y written in terms of F, S and T. Where there are no math signs between variables these variables or summed and where parenthesis separate groups of variables the sums of the groups are multiplied. Press the NEXT AREA button to cycle through examples of areas corresponding to the terms of a Parabola X"+BX + C =Y. Click on the list left of the graph area to see varying examples.
Click Here to display webpage that displays areas as you generate terms of Parabolas using Pythabacus.

Proving The Pythagorean Theorem On The Pythabacus

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(Press space bar to switch arrow key control between farmer and beads on Pythabacus. Press escape to restart and review.)
(Left click mouse to correct or reverse the direction arrow keys move farmer)

The figure above represents a farm with four tenant farmers. Each tenant farmer has a rectangle plot of land composed of a triangle of farm land and a triangle of grazing land. In the middle of these farm plots is a shared square of grazing land with a watering pond in its center. Lets designate the shorter sides of the rectangles (A), the longer sides (B) and the diagonals dividing the rectangle into two triangles (C). Then, the area including the pond and grazing land bordered by the diagonals will equal C-squared. The tenant farmer of the upper right hand rectangle is the farm owners land master. He monitors the growth of the crops of the other tenants and collects from them part of what they grow in trade for the use of the land and watering pond. Pressing the enter key will allow you to use the arrow keys to take the land master to each plot of land. After leaving the land masters plot, walk counter clockwise and visit each of the plots in turn. The corner stones will turn to point your way.
Notice on the simulation of the Pythabacus the 4 times 4 quadrilateral; it represents the square of farm land illustrated above. As the land master walks from farm plot to farm plot, columns of beads will separate from the quadrilateral to represent each plot. The (A) length of each plot will be one bead and the (B) length of each plot will be three beads. The Pythagorean Theorem states that lengths A-squared plus B-squared will equal C-squared.

When the land master reached the last tenant, he found that the last tenant did not grow enough crops to trade for the use of the land and the watering pond. The land master decided to take part of the last tenants land to grow crops for himself. He left the last tenant only a small plot equal A-squared parts of the whole farm. If you include the square of watering land the land master now controls B-squared parts of the whole farm. Each rectangle equals two triangles and two rectangles equal four triangles If you look closely at the C-squared part of the farm you will see it is bordered by four triangles with the watering square in the middle. So C-squared is equal two rectangles plus the watering square. If you now look closely at the A-squared part plus the B-squared part of the whole farm, you will see that these parts also equal two rectangles and the watering square. Therefore A-squared plus B-squared equal C-squared. On the Pythabacus simulation the four beads in the middle equal the watering square and the columns of beads to its right equal two rectangles. Now walk the land master through the gate and into the A-squared part of the farm.
The watering square beads plus the two rectangle columns became two groups of beads equal A-squared and B-squared. You have proven the Pythagorean Theorem on the Pythabacus! So, the cows now drink happily at the pond.
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