Algebra/Systems of Equations

Systems of Simultaneous Equations
In a previous chapter, solving for a single unknown in one equation was already covered. However, there are situations when more than one unknown variable is present in more than one equation. When in a given problem, more than one algebraic equation is true at a time, it is said there is a system of simultaneous equations which are all true together at once. Such sets of multiple equations may help solve for more than one unknown variable in a problem, since having more than one unknown in one equation is typically not enough information to "solve" any of the unknowns.

An unknown quantity is something that needs algebraic information in order to solve it. An equation involving the unknown is typically a piece of information which may provide the information to "solve" the unknown, i. e. to determine a specific number value (or limited number of discrete values) that the unknown is (or can be) equal to. Some equations provide little or no information and so do little or nothing to narrow down the possibilities for solutions of the unknowns. Other equations make it impossible to satisfy an unknown with any real number, so the solution set for the unknown is an empty set. Many other useful equations make it possible to solve an unknown with one or just a few discrete solutions. Similar statements can be made for systems of simultaneous equations, especially regarding the relationships between them.

Linear Simultaneous Equations with Two Variables
In the previous module, linear equations with two variables were discussed. A single linear equation having two unknown variables is practically insufficient to solve or even narrow down the solutions for the two variables, although it does establish a relationship between them. The relationship is shown graphically as a line. Another linear equation with the same two variables may be enough to narrow down the solution to the two equations to one value for the first variable and one value for the second variable, i. e. to solve the system of two simultaneous linear equations. Let's see how two linear equations with the same two unknowns might be related to each other. Since we said it was given that both equations were linear, the graphs of both equations would be lines in the same two-dimensional coordinate plane (for a system with two variables). The lines could be related to each other in the following three ways:

1. The graphs of both equations could coincide giving the same line. This means that the two equations are providing the same information about how the variables are related to each other. The two equations are basically the same, perhaps just different versions or forms of each other. Either one could be mathematically manipulated to produce the other one. Both lines would have the same slope and the same y-intercept. Such equations are considered dependent on each other. Since no new information is provided, the addition of the second equation does not solve the problem by narrowing the solution set down to one solution.

Example: Dependent linear equations

$$ 6x - 3y = 12 \ $$

$$y = 2x - 4 \ $$

The above two equations provide the same information and result is the same graph, i. e. lines which coincide as shown in the following image.



Let's see how these equations can be mathematically manipulated to show they are basically the same.

Divide both sides of the first equation $$ 6x - 3y = 12 \ $$ by 3 to give $$ 2x -y = 4 \ $$


 * Now add y to both sides

$$ 2x = 4 + y \ $$


 * Now subtract 4 from both sides

$$ y = 2x - 4 \ $$

This is the same as the second equation in the example. This is the slope-intercept form of the equation, from which a slope and a y-intercept unique to the line can be compared with any other equations in the slope-intercept form.

2. The graphs of two lines could be parallel although not the same. The two lines do not intersect each other at any point. This means there is no solution which satisfies both equations simultaneously, i. e. at the same time. The solution set for this system of simultaneous linear equations is the empty set. Such equations are considered inconsistent with each other and actually give contradictory information if it is claimed they are both true at the same time in the same problem. The parallel lines have equal slopes but different y-intercepts.

Sets of equations which have at least one common point which might provide a solution set are consistent with each other. For example, the dependent equations mentioned previously are consistent with each other.

Example: Inconsistent linear equations

$$ 3x - 2y = -2 \ $$

$$ 3x -2y = 2 \ $$

To compare slopes and y-intercepts for these two linear equations, we place them in the slope-intercept forms. Subtract 3x from both sides of both equations. $$ 3x - 2y = -2 \qquad \qquad 3x -2y = 2 $$ $$-2y = -3x - 2 \qquad \qquad -2y = -3x + 2$$ Divide both sides of both equations by -2 and simplify to get slope-intercept forms for comparison. $$ (-2y)/(-2) = (-3x - 2)/(-2) \qquad (-2y)/(-2) = (-3x + 2)/(-2) $$
 * $$y = \frac{3}{2} x + 1 \qquad \qquad \qquad \qquad \qquad

y = \frac{3}{2} x - 1$$ Now, both slopes are equal at 3/2, but the y-intercepts at 1 and -1 are different. The lines are parallel. The graphs are shown here:



3. If the two lines are not the same and are not parallel, then they would intersect at one point because they are graphed in the same two-dimensional coordinate plane. The one point of intersection is the ordered pair of numbers which is the solution to the system of two linear equations and two unknowns. The two equations provide enough information to solve the problem and further equations are not needed. Such equations intersecting at a point providing a solution to the problem are considered independent of each other. The lines have different slopes but may or may not have the same y-intercept. Because such equations provide at least one solution point, they are consistent with each other.

Example: Consistent independent linear equations

$$ y = 3x - 5 \ $$

$$ y = -x - 1 \ $$

Both of these equations are given in the slope-intercept, so it is easy to compare slopes and y-intercepts. For these two linear functions, both slopes are different and both y-intercepts are different. This means the lines are neither dependent nor inconsistent, so on a two-dimensional graph they must intersect at some point. In fact, the graph shows the lines intersecting at (1,-2), which is the ordered pair solution to this system of independent simultaneous equations. Visual inspection of a graph cannot be relied on to give perfectly accurate coordinates every time, so either the point is tested with both equations or one of the following two methods is used to determine accurate coordinates for the intersection point.



Solving Linear Simultaneous Equations
Two ways to solve a system of linear equations are presented here, the addition method and the substitution method. Examples will show how two independent linear simultaneous equations with two unknown variables could be solved for both unknown variables using these methods.

Elimination by Addition Method
The elimination by addition method is often simply called the addition method. Using the addition method, one of the equations is added (or subtracted) to the other equation(s), usually after multiplying the entire equation by a constant, in order to eliminate one of the unknowns. If the equations are independent, then the resulting equation(s) should be one(s) which will have one less unknown. For an original system of two equations and two unknowns, the resulting equation with one less unknown would have one unknown left which could easily be solved for. For systems with more than two equations and two unknowns, the process of elimination by addition continues until an equation with one unknown results. This unknown could then be solved for and the solved value then substituted into the other equations resulting in a system with one less unknown. The elimination by addition process is repeated until all of the unknowns are solved.

If a system has two equations which are dependent, then the addition of the equations could or would eliminate both unknowns at once. If the equations are parallel lines which are inconsistent, then a contradictory equation could result. The addition method is useful for solving systems of simultaneous linear equations, particularly if the equations are given in the form Ax + By = C, where x and y are the two unknown variables and A, B, and C are constants.

Example: Solve the following system of two equations for unknowns x and y using the addition method:

$$ x + 2y = 4 \ $$

$$ 3x - y = 5 \ $$

Solution: We can either multiply the first equation by -3 and add the result to the second equation to eliminate x, or we can multiply the second equation by 2 and add the result to the first equation to eliminate y. Let's multiply [both sides of ] the second equation by 2.


 * $$ 2 \cdot (3x - y) = 2 \cdot 5 $$

$$ 2 \cdot 3x - 2 \cdot y = 10 $$


 * $$ 6x \ - 2y = 10 \ $$

Now we add this resulting equation to the first equation; i. e. each of the two sides of the equations are added together to give a combined equation as shown here:

$$ x \ + \ 2y = 4 \ $$ $$ + \ ( 6x - 2y = 10 ) \ $$ _____________________
 * $$ 7x + 0\cdot y = 14 \ $$

This means that we add x + 2y and 6x – 2y to get 7x + 0·y and we add 4 and 10 to get 14.

This eliminates y from the combined equation to give an equation in x only:

$$ 7x = 14 \ $$


 * Now we solve for x:

$$ x = 14/7 = 2 \ $$

Now that we have x, we can substitute the value for x into either of the original two equations and then solve for y. Let's pick the first equation for the substitution into x.

$$ 2 + 2y = 4 \ $$


 * Solving for y:

$$ 2y = 4 - 2 = 2\ $$

$$ y = 2/2 = 1 \ $$

So the solution set consists of the ordered pair ( 2,1) which is the point of intersection for the two linear functions as shown here:



Elimination by Substitution Method
The elimination by substitution method is often simply called the substitution method. With the substitution method, one of the equations is solved for one of the unknowns in terms of the other unknown(s). Then that expression for the first unknown is substituted into the other equation(s) to eliminate it such that the equation(s) then have only the other unknown(s) left. If the equations are independent, then the resulting equation(s) should be one(s) which will have one less unknown. For an original system of two equations and two unknowns, the resulting equation with one less unknown would have one unknown left which could easily be solved for. For systems with more than two equations and two unknowns, the process of elimination by substitution is repeated until an equation with one unknown results. This unknown could then be solved for and the solved value then substituted into the other equation(s), resulting in a system with one less unknown. The process of elimination by substitution continues until all of the unknowns are solved.

If a system has two equations which are dependent, then applying the substitution method would either eliminate two unknowns at once or result in an equation which do not yield single values for the remaining unknown(s). If the equations are parallel lines which are inconsistent, then a contradictory equation could result.

Example: Solve the following system of two equations for unknowns x and y using the substitution method:

$$ x - y = -1 \ $$

$$ x + 2y = -4 \ $$

Solution: We can start by solving for either x or y in terms of the other unknown in either one of the equations. Let's start by solving for x in terms of y in the first equation.

$$ x = y - 1 \ $$

Next, we substitute this expression for x into the other equation in order to eliminate x from the equation.

$$ (y - 1) + 2y = -4 \ $$

$$ 3y - 1 = -4 \ $$

We have eliminated x and now we have an equation in terms of y only. We now solve for y in this equation.

$$ 3y = -4 + 1 = -3 \ $$

$$ y = -3/3 = -1 \ $$

We have found the solution for y to be -1. We substitute this value for y into the expression for x in terms of y we determined from the first equation earlier.

$$ x = y - 1 = -1 - 1\ $$

Finally, we calculate the value of x.

$$ x = -2 \ $$

So the solution set consists of the ordered pair (-2,-1) which is the point of intersection for the two linear functions as shown here:



Slopes of Parallel and Perpendicular Lines
''This paragraph restates itself. It should be reworded.''
 * In a two-dimensional Cartesian coordinate plane, linear functions which are dependent or whose graphs are parallel lines will have the same slope . CONVERSELY, linear functions having equal slopes are either dependent or have graphs that are parallel lines in a two-dimensional Cartesian coordinate plane. Of course, vertical parallel lines of the general form x=c are not functions and have no defined slopes.


 * In a two-dimensional Cartesian coordinate plane, two lines that are perpendicular to each other will form right angles (90° angles) with each other at the point where they intersect. When the slopes of two linear functions whose graphs are lines that are perpendicular are multiplied together, the product of the two slopes equals -1 . Conversely, if multiplying the slopes of two linear functions gives a product equal to -1, then their graphs are perpendicular lines on a two-dimensional Cartesian coordinate plane.


 * In other words, if two perpendicular lines have slopes m1 and m2, then

$$m_1 m_2 = -1 \ $$.


 * If a pair of perpendicular lines consists of a horizontal line (of the form y = c) and a vertical line (of the form x = c), then the preceding rule does not apply. A vertical line has no slope and the slope of a horizontal line = 0.

Example: Find the slope-intercept form of a [new] line which intersects y = (1/2)x – 3 at (4,-1) and is perpendicular to it.

Solution: First, find slope of the new line from slope of the given line. Let m = slope of the new line.

$$\left ( \frac{1}{2} \right ) m = -1 $$

$$2\cdot \left ( \frac{1}{2} \right ) m = 2\cdot (-1) $$

$$m = -2 \ $$

The slope-intercept form of the new line will be:

$$ y = -2x + b \ $$

where b is the y-intercept of the new line. Next, solve for y-intercept of new line using the intersecting point (4,-1) and the new slope of -2. Substitute x = 4 and y = -1 into the preceding equation and solve for b.

$$-1 = -2 \cdot 4 + b \ $$

$$ -1 = -8 + b \ $$

$$ b = -1 + 8 = 7 \ $$

Finally, the slope-intercept form of the new perpendicular line is :

$$ y = -2x + 7 \ $$.

Graph showing perpendicular lines in above example.

Solving Systems of Simultaneous Equations Involving Equations Of Degree 2
The substitution method should be used for efficiency when solving nonlinear simultaneous equations, unless other methods such as the graphing method provide clear and simple solutions quickly (when they would be faster than substitution).

Example: Solve the system of simultaneous equations. $$y^2 + (2x+3)^2 = 10 \ $$

$$2x + y = 1 \ $$

With the second equation, make a given term (here, 2x should be used) the subject.

$$2x = 1 - y \ $$

Substitute the third equation into the first, and through factorization of the resulting, simplified quadratic with one variable the solutions can be found.

$$y^2 + ((1-y)+3)^2=10 \ $$ $$y^2 + (4-y)^2=10 \ $$ $$2y^2 - 8y + 16 =10 \ $$ $$y^2 - 4y + 3 = 0 \ $$ $$(y-1)(y-3) = 0 \ $$ Hence we know $$y = 1 \ $$ or $$y=3 \ $$

Then, we calculate that the two possibilities are: $$y = 1 \ $$, $$x = 0 \ $$ or; $$y = 3 \ $$, $$x = -1 \ $$

Solving Systems of Simultaneous Equations Using a Graphing Calculator
TI-83 (Plus) and TI-84 Plus:

1. Press "Y="

2. Enter both equations, solved for Y

3. Press "GRAPH"

4. If all intersection points are not visible, press "ZOOM" then 0 or select "0: ZoomFit"

5. Press "2nd" then "TRACE"

6. Press 5 or select "5: intersect"

7. Move the cursor to one of the intersection points. (There may be only one) Each of these points represents one solution to the system.

8. Press "ENTER" three times

9. The coordinates of the intersection are shown at the bottom of the screen. Repeat steps 5-8 for other solutions.

TI-89 (Titanium):

via Graphing:

1. Press the green "diamond key", located directly beneath the "2nd" (blue) button.

2. Follow steps 2-5 as listed above. To access "Y=" and "GRAPH", press the green "diamond key", then press F1 (it activates the tertiary function, "Y=") and F3 ( "GRAPH"). To access "ZOOM" and "TRACE", press F2 and F3 (diamond function activated), respectively. For "ZoomFit", press F2, then "ALPHA" (white), then "=" (for A).

3. To locate the point of intersection, manually use the directional keypad (arrow keys), or press F5 for "Math", then 5 for "Intersection". (The second option is more difficult to use, however; manual searching and zooming is recommended.)

4. The coordinates are displayed on the bottom of the screen. Repeat steps 2 and 3 until all desired solutions have been found. For new or additional equations, return to the "Y=" as described above.

via Simultaneous Equation Solver:

Note:This is a default App on the TI-89 Titanium. If you are using the TI-89 or no longer have the Solver, visit the Texas Instruments site for a free download.

1. On the APPS screen, select "Simultaneous Equation Solver" and press enter. Press "3" when the next screen appears.

2. Enter the number of equations you wish to solve and the corresponding number of solutions.

3. The two equations are represented simultaneously in a 2 x 3 matrix (assuming that you are solving two equations and searching for two solutions. The size of the matrix depends on the number of equations you wanted to solve). In the corresponding boxes, enter the coefficients/constants of your equations, pressing "ENTER" every time you submit a value. (Remember that all equations must be converted into standard form – Ax + By = C – first!)

4. Once all values have been entered, press F5 to solve.