# Solving Systems Of Linear Polynomial Equations Geometrically

In the last chapter, we were able to determine the tangent point ( 4 , 5 ), for one of the crease lines of a paper folded parabola.

As it turns out, the tangent point is an example of a solution to a system of polynomial equations (a group of two or more polynomial equations).

Actually, the two endpoints (0, 2) and (4, 0), the midpoint (2, 1), and the y-intercept (0, -3), are also solutions to systems of polynomial equations. Before we discover what it means to be a solution to a system, we have to make sure we remember what a solution to a polynomial equation is.

Below are snapshots of a program, which will allow you to explore solutions of linear polynomial equations.

Run  program 812  , until you can give and defend an answer to question number 1.

Use a copy of the pdf programnotes2g, to record and defend your answer to question number 1.

After recording and defending your answer to question number 1, along with an example, go on to the next page.

As we were just reminded, a solution of a linear polynomial equation in one variable, is a set of x- and y-values which makes the linear equation true. For example,  ( 2 , 5 )  is a solution of  y = 3x – 1 , because  ( 5 ) = 3( 2 ) – 1 .

To the right, are snapshots of a  program, which will allow you to  explore solutions of systems of  two linear polynomial equations which cross each other.

Run  program 813  , until you can give and defend an answer to question number 2.

Use a copy of the pdf  programnotes2g, to record and defend your answer to question number 2.

After recording and defending your answer to question number 2, along with an example, go on to the next page.

As we discovered, a solution of a system of two intersecting linear equations, is the one point where the two lines intersect (the intersection point is the only point which lies on both lines).

In other words, a solution of a system of polynomial equations, is a set of values which makes every polynomial equation in the system true, not just one or some of them.

As we discovered, the solution of a system of two intersecting linear polynomial equations, are the coordinates of the one and only one intersection point. This is only true however, when the two lines intersect each other at exactly one point.

There are actually two ways for two lines, to not intersect each other at one point. Your challenge is to use the next program to figure out how.

Below is a snapshot of question number 3, from the next program.

Run  program 814  , until you can give and defend an answer to question number 3.

Use a copy of the pdf  programnotes2g, to record examples and defend your answer to question number 3.

After recording and defending your answer to question number 3, along with one example for each of the three outcomes, go on to the next page.

Below are snapshots of a program, which will allow you to practice solving systems of two linear polynomials (written in slope-intercept form) by graphing, where the lines cross at one point.

Run  program 801  , until you can solve at least 3 out of the 4 given systems by graphing.

Use copies of the pdf grid1 to graph, solve, and record the systems of two linear polynomial equations.

After solving and recording at least 3 of the 4 systems of two linear polynomial equations, go on to the next page.

Below are snapshots of a program, which will allow you to practice solving systems of two linear polynomials (written in standard and slope-intercept form) by graphing, where the lines cross at one point.

Run  program 802  , until you can solve at least 3 out of the 4 given systems by graphing.

Use copies of the pdf  grid1 to graph, solve, and record the systems of two linear polynomial equations.

After solving and recording at least 3 of the 4 systems, read the following line before going on to the next page.

To discover how to solve systems of two linear inequalities, open the pdf  systemineqs.

To make it easier to understand solving systems by graphing, the intersection points (solutions) so far, have had integer value coordinates.  When this is not the case, the more accurately the lines are graphed, the better the estimates of the solutions will be.

Below are snapshots of a program, which will ask you to solve systems of linear equations to the nearest tenths place.  For each system, you will be given 3 chances to come within 2 tenths of the solution’s actual x- and y-values.

Run  program 803  , until you can solve at least 3 of the 5 systems of linear equations.

Use copies of the pdf grid3  to record, graph, and solve the systems to the nearest tenths place, +/- 2 tenths.

After recording and solving at least 3 of the 5 systems, to the nearest tenths place +/- .2, go on to the next page.

When rough estimates for the values of solutions are good enough, solving systems by graphing will do the trick.

When more precise or exact values of solutions are necessary, a different method will have to be found to solve systems of polynomial equations.

As it turns out, understanding the difference between variables and unknowns, is all we need to know to find another way.

A variable, is a letter which represents any and all possible values of a quantity.

For example, in the linear equation  y = 2x + 3 , the y and the x are variables, because they represent any and all of the infinite number of sets of possible values which will make the equation true.

An Unknown, is a letter which represents a specific value, or the specific values, of a quantity.

For example, in the linear equation  2x + 3 = 11 , the x is an unknown, because it represents a specific value which will make the equation true, namely 4.

Up until now, knowing this difference wasn’t too important, because the definition of a solution is the same for polynomial equations with either variables or unknowns.

Now however, knowing the difference between variables and unknowns, will lead us to another way of solving systems of two linear equations.

Given two linear polynomial equations, such as y = 2x + 5 and y = x + 3 , the x’s and y’s are variables.

This means, if we subtract them (like we did in chapter 1), we will end up with another linear polynomial equation with variables x and y.

y = 2x + 5                                                                                                                                   -(y =  x + 3)                                                                                                                                       y = x + 2

The infinite output values of 2x + 5, minus each of the infinite corresponding output values of x + 3, gives you the infinite output values of x + 2.

If the two linear polynomial equations y = 2x + 5  and  y = x + 3  are a system however, the x’s and y’s are no longer variables, but rather unknowns.

This is because the x’s and the y’s represent the specific values  of the solution of the system (the coordinates where the lines cross).

This time when we subtract, we’re focussing on a specific x-value and a specific y-value.

y = 2x + 5                                                                                                                                   -(y =  x + 3)                                                                                                                                       0 = x + 2

The specific y-value of the solution, minus the specific y-value of the solution, gives you a y-value of zero.

The value y is still unknown at this point, but no matter what y is, yy=0 .

Solving for the unknown x (the specific x-value of the solution), we get the following.

The specific x-value of the solution, is -2.

Since -2 is the x-value of the solution for both linear equations, you can use either  y = 2x + 5  or  y = x + 3  to determine the y-value of the solution.

So the solution to the system  y = 2x + 5  and  y = x + 3  is  ( -2 , 1 ) .

In other words, you can solve a system of two linear equations (written in slope-intercept form), by subtracting them to find the value of the unknown x, and then substituting this value into one of the original equations to find the value of the unknown y.

Because this can be done algebraically, without the use of a graph, we can get exact values for the solution.

Example:  Given the system  y = 5x – 5  and  y = 3x – 1 , we get the solution  ( 2 , 5 ) .

If we subtract the two linear equations in a system, but don’t actually do the subtraction, we get the following.

By adding ( x + 3 ) to both sides, which is just a vertical translation, we get the following, which can be solved for the unknown x.

In other words, instead of subtracting the two linear equations,  we can just substitute the specific y-value in the first linear equation, with the linear polynomial in the second equation.

Geometrically, at the solution, both polynomials have equivalent output values, so they’re equivalent to each other.

Algebraically, if two expressions are both equivalent to a third quantity, they’re equivalent to each other.

Substituting equivalent expressions for quantities in a system, is known as, solving the system by substitution.

Solving systems of two linear equations by substitution, still works when the slopes of the polynomials are not integers.

Below are snapshots of a program, which will give you 4 systems of two linear equations, to be solved by substitution.

Run  program 804  , until you can solve at least 3 of the 4 systems, by substitution.

Use a copy of the pdf  programnotes, to record the systems, and to show the steps needed to solve them.

After solving at least 3 of the systems, and neatly showing all of your work, go on to the next page.

Below are snapshots of a program, which will randomly generate 4 systems of two linear equations (one in standard form, and one in slope-intercept form), to be solved by substitution.  The form of the line doesn’t matter, because at the solution of the system, the unknowns x and y represent specific values, which can be substituted for equivalent expressions.

Run  program 805  , until you can solve at least 3 of the 4 systems, by substitution.

Use a copy of the pdf programnotes, to record the systems, and to show the steps needed to solve them.

After solving at least 3 of the systems, and neatly showing all of your work, go on to the next page.

Below are snapshots of two programs, which will randomly generate 4 systems of two linear equations, to be solved by substitution.  This time the equations which are written in slope-intercept form, will have more realistic rational slopes.  After solving at least 3 of the 4 systems in the first program, you’ll have the option to go to the second program, where the y-intercepts are rational as well.

Run  program 806  , until you can solve at least 3 of the 4 systems, by substitution.

Use a copy of the pdf  programnotes, to record the systems, and to neatly show all of your steps.

After solving at least 3 of the systems, read the following line before going on to the next page.

Run  program 807  , until you can solve at least 3 of the 4 systems, by substitution.

It’s trickier working with fractions, but being able to algebraically manipulate expressions is a powerful skill to have.  Fractions or not, the beauty of solving systems by substitution is, we can always get the exact values of the solutions (something solving systems by graphing can’t always claim).  Below are snapshots of a program, which will randomly generate 4 systems of two linear equations, where the solutions have rational values.

Run  program 808  , until you can solve at least 3 of the 4 systems, by substitution.

Use a copy of the pdf  programnotes, to record the systems, and to show the steps needed to solve them.

After solving at least 3 of the systems, and neatly showing all of your work, go on to the next page.

Solving systems of polynomial equations by substitution, has another advantage over solving systems by graphing as well.  Graphing gets more challenging as the values of the equations get bigger.  For instance, below is a system with y-intercepts which are a little over 49 units away from each other.  49 isn’t a huge number, but it’s big enough to make it hard to use a scale of 1.  When we use a larger scale though, we lose accuracy when plotting points.  When solving systems of equations by substitution though, big numbers can easily be handled with calculators.

As it turns out, the first equation is a Celcius-to-Fahrenheit formula, and the second equation is a Fahrenheit-to-Celcius formula.  This means the value -40º, is the one temperature reading which is the same in both Fahrenheit and in Celcius.

Back in chapter 4, we explored the roots of quadratic equations.  We discovered that roots are solutions which have output values of zero.

y = x2 + 2x – 3

For example, the roots of the  polynomial x2 + 2x – 3  , are  -3  and  1 , because that’s where the parabola intersects the x-axis y = 0 (points on the x-axis have output values of zero).

In other words, finding roots is just like  solving a system of two polynomial equations, by substitution.

In the last chapter, we were able to determine the linear equation for a tangent line to our paper folded parabola.  If we determine the equation for a second tangent line, the intersection point of the two tangent lines and the two tangent points will form a triangle (shown in blue).

Archimedes discovered that the area of the parabola within the triangle, is 2/3 of the area of the entire triangle.

As it turns out, we now know enough math to be able to determine the area of the triangle.

The first thing we need to do, is solve the following system of two linear equations, to determine the coordinates of the third vertex of the triangle.

Use a copy of the pdf  booknotes, and solve the above system of two linear equations, by substitution.

After solving the above system, read the following line before going on to the next page.

To determine the area of the section of the parabola, open the pdf  “Archimedes”  .

Being able to solve systems of linear polynomial equations by substitution, opens the door to a lot of interesting mathematics.  That door opens even wider next chapter, when we discover how to solve systems of linear equations, written in standard form.

As an example, along with using the distance formula from chapter 6 and finding perpendicular lines using the point-slope form of a line from chapter 7, being able to solve a system of two linear polynomial equations is enough for us to discover a formula which gives the acute angle between any two lines.

All we have to do, is determine a formula for the slope of the second line, relative to the first line.  This describes the tangent of the angle between the two lines, which can then be used to determine the actual angle.

In other words, given any two lines, we need to be able to determine the lengths of the opposite and adjacent sides of a particular right triangle formed by the two lines.

Read the following line before going on to the next page.

To explore a formula for the angle between any two lines, open the pdf tanformula

If you thought the image at the beginning of the chapter was interesting (shown again below), run  program 815 to explore Lissajous curves, named after Jules Antoine Lissajous (LEE-suh-zhoo).

Below is a snapshot of the chapter 8 programs menu program.

Run the program  “8menu.bas”   , or its alias to access every program used in chapter 8.

“This, therefore, is mathematics: . . . she gives life to her own discoveries; she awakens the mind and purifies the intellect; she brings light to our intrinsic ideas; she abolishes oblivion and ignorance which are ours by birth” — Proclus —

“…there is no study in the world which brings into more harmonious action  all the faculties of the mind than [mathematics]” — J.J. Sylvester —

“Mathematical knowledge adds vigour to the mind, frees it from prejudice, credulity, and superstition.” — John Arbuthnot —

“The mathematician has reached the highest rung on the ladder of human thought” — Havelock Ellis —

“It is clear that the chief end of mathematical study must be to make the students think.”  — John Wesley Young —