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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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# SOLVING SYSTEMS OF EQUATIONS

Many applied problems are modeled by two or more equations. When this happens, we talk about a
System of Equations. The equation in the system can be linear or they can be non-linear (quadratic,
rational, polynomial equations, etc.).

All systems of equations could be solved graphically, but this method tends to give us inaccurate
results. Therefore, it is NOT an acceptable solution method. Instead, we MUST solve systems of
equations using the Substitution or the Addition Method.

The Substitution Method works extremely well for finding solutions of Systems of Linear and Nonlinear
Equations. The Addition Method is often used for linear system, but cannot always be used for
non-linear systems.

## Strategy for Solving Systems of Equations by the Substitution Method

Step 1: Solve any one of the equations for one variable in terms of the other. If one of
the equations is already in this form, you can skip this step.

Step 2: Substitute the expression found in Step 1 into the other equation. You should
now have an equation in one variable. Find its value.

Please note that ONLY at the point of intersection two equations are equal to
each other. By setting the x- or y-value of one equation equal to the x- or yvalue
of the other equation, we are in effect finding the point of intersection.

Step 3: To find the value of the second variable, back-substitute the value of the variable
found in Step 2 into one of the original equations. If one of your equations is non-linear,
you must back-substitute into both equations and rule out "extraneous solutions."

Step 4: Form an ordered pair with the values found in Step 3 and Step 4. This is the
solution to your system of equations.

## Strategy for Solving Systems of Equations by the Addition Method

Step 1: If necessary, rewrite both equations in the same form so that the variables and
the constants match up when one equation is beneath the other.

Step 2: If necessary, multiply either equation or both equations by appropriate numbers
so that the coefficient of either x or y will be opposite in sign giving a sum of 0.

Step 3: Write the equations one below the other, draw a horizontal line, then add each of
their terms. The sum should be an equation in one variable. Find its value.

Step 4: To find the value of the second variable, back-substitute the value of the variable
found in Step 3 into one of the original equations. If one of your equations is non-linear,
you must back-substitute into both equations and rule out "extraneous solutions."

Step 5: Form an ordered pair with the values found in Step 4 and Step 5. This is the
solution to your system of equations.

Example 1:
Solve the following system of linear equations graphically, as well as using the Substitution and

-5x = y = 8

-4x + 5y = -1

Below is a pictorial representation of the system. The point of intersection is considered the solution
of the system.

Solving any System of Equations graphically often does not give use the correct solution.
For example, in the picture above we might believe that the solution is (-2, -2).

Actually, the solution isas you will see below.

## Substitution Method

Step 1:
Solve any one of the equations for one variable in terms of the other. By solving
-5x + y = 8 for y, we find

y = 5x + 8

Step 2:
Next, we back-substitute into the equation -4x + 5y = -1 as follows

-4x + 5(5x + 8) = -1
-4x + 25x + 40 = -1
21x = -41

x =-41/21

Step 3:
Since both equations are linear, we can use either one of the original equations to solve
for y. Let's calculate the y-coordinate using -5x + y = 8

Step 4:
Therefore, the solution to the linear system of equations or the point of intersection
of the two lines is

Step 1:
The equations are already rewritten in the same form so that the variables and the
constants match up when one equation is beneath the other.

Step 2:
We multiply both sides of the first equation by -5 so that the coefficient of y in both
equations will be opposite in sign giving a sum of 0.

-5(-5x + y) = -5(8)
25x - 5y = -40

Step 3:
Now, we will write the "new" first equation below the second equation, draw a horizontal
line, then add each of their terms.

The x-coordinate of the point of intersection

Step 4:
Since both equations are linear, we can use either original equation to solve for y. Let's
calculate the y-coordinate using -5x + y = 8

Step 5:
The solution to the system of linear equations or their point of intersection is

Example 2:
Solve the following system. It contains a quadratic equation and a circle.

We will use the Substitution Method, which means that we have to solve one of the equations for
either x or y.

Step 1:

Usingwe find

Step 2:
Next, we will back-substitute into the equationto get

Therefore, the x-coordinates for the solutions are

Step 3:
Since both equations are non-linear, we must find the values for y using both original
equations.

Thus, one solution is

andis another solution

and(3,-2)is yet another solution

thus,(-3,-2)is the last solution

Using the circle

Thus, one solution isbut there is another solution, namely,
This solution did not show up when we used the quadratic
equation, therefore, it must be a "ghost" or extraneous solution.

It can be concluded, that this will happen for every x-value, and further
investigation is optional.

Example 3:
Let's use a different method to solve the system in Example 2:

We will use the Substitution Method again, but this time, we will solve the second equations for x2.
That is,

Next, let's back-substitute into the equation

Thus, the y-coordinates for the solutions are 1 and -2. You can now find the corresponding xcoordinates.
See Example 2 for complete solutions!

Example 4:
Solve the following system. It contains a linear and a polynomial equation.

Let's use the Substitution Method and solve the first equation for x.

Using the Square Root Property we find

which means that y = 2 and y = -2.

Since NOT both equations are linear, we must find the values for y using both original equations.Using the linear equation x + y = 0:

If y = 0,then x + 0 = 0
Thus, one solution is (0,0)

If y = 2,then x + 2 = 0
and the second solution is(-2,2)

If y = -2,then x - 2 = 0
Thus, one solution is (2,-2)

Using the polynomial equation

Thus, one solution isbut there are two other solutions, namely
and These solutions did not show up when we used the linear equation,
therefore, they must be "ghost" or extraneous solutions.

t can be concluded, that this will happen for every x-value, and further investigation
is optional.

Example 5:
Solve the following system. It contains a linear equation and a circle.

We will use the Substitution Method, which means, that we have to solve one of the equations for
either x or y. In this example, y = -x + 3 is already solved for y.

Next, we will back-substitute into the equation

This is a quadratic equation that is not factorable. Therefore, we have to use the Quadratic Formula.

The x-coordinates for the points of intersection are imaginary numbers.

Therefore, there are no solutions to this system. The two graphs do not intersect.

Example 6:
Solve the following system. It contains a linear and a quadratic equation.

In this case, both equations are already solved for y. Therefore, we will definitely use the
Substitution Method.

Notice, that we are simply setting the two equations equal to each other

Since not both equations are linear, we must find the values for y using both original equations.
Using the linear equation y = x - 2

If x = 2,then y = 2 - 2 = 0
Thus, one solution is (2,0)

If x = 1,then y = 1 - 2 = -1
and the second solution is (1,-1)

If x = 2, then

Thus, just like in the case of the linear equation, the quadratic equation produces
exactly the same solution, namely (2, 0).

It can be concluded, that this will happen for every x-value, and further investigation
is optional.

Example 7:
Solve the following system. It contains a linear and a rational equation.

Let's use the Substitution Method since the first equation is already solved for y. Therefore, we can
back-substitute into the second equation as follows

Using the Square Root Property we find

which means that and

Since NOT both equations are linear, we must find the values for y using both original equations.

Using the linear equation y = 2x:

If,then
Thus, one solution is

If,then
and the second solution is

Using the rational equation xy = 4:

If,then
and

Thus, just like in the case of the linear equation, the rational equation produces
exactly the same solution, namely

It can be concluded, that this will happen for every x-value, and further investigation
is optional.