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 Depdendent Variable

 Number of equations to solve: 23456789
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 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

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 Solve for:

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# Solving Systems of Linear Equations

The following pages contain some instructions on the usage of the TI-83/83 Plus
graphing calculator.
The example used below is taken out of the text titled “College Algebra - Enhanced with
Graphing Utilities, 3rd Edition” by Sullivan and Sullivan, III.

Example#3 Solving a System of Linear Equations p. 515: Solve the following system of
equations.

2x + y = 5 (1)
− 4x + 6y =12 (2)

First, each equation in the above system must be solved for y (i.e., written in slope-intercept
form) before it can be entered into function Editor . Hence, the system of
equations to be entered into the graphing calculator becomes: Press to go into the function Editor menu. If you have any previously entered
functions showing up on the function editor menu, press to move the cursor down by
that function, and press delete it.

While the blinking cursor is by \Y1 =, start entering the expression − 2x + 5 , which is the
right side of the first equation. Press . Type in 2. Press for the generic input
variable x. Press . Type in 5. At this point, your screen should look like the screen on
the left given below. Press to move the cursor down by \Y2 =. Start entering the expression which
is the right side of the second equation. Press Type in 2. Press Type in 3. Press . Press for the generic input variable x. Press . Type in 2. At this point,
your screen should look like the screen on the right given above.

The solution of this system of linear equations is the same as the points of intersection of
the graphs of the equations, which have been entered into the \Y1 = and \Y2 = locations of
the function editor, respectively. Press to type in the following WINDOW
settings, which are given on the left below. Press see the two graphs. At this
point, your screen should look like the screen on the right given below. From the graph on the right above, there appears to be a point of intersection of these two
graphs, which means the system is consistent and the equations are independent.
Press and to access the CALC menu. At this point, your screen should
look like the screen on the left given below. Press four times to move the cursor down to 5:intersect and press to select
this option from the CALCULATE menu. At this point, your screen should look like the
screen on the right given above with the calculator prompting you for the First Curve?.
Be sure that the blinking trace cursor is on the first graph, which is indicated by a small 1
showing up in the upper-right corner. Press to confirm. At this point, your screen
should look like the screen on the left given below with the calculator prompting you for
the Second Curve?. Be sure that the blinking trace cursor is on the second graph, which is indicated by a
small 2 showing up in the upper-right corner. Press to confirm. At this point,
your screen should look like the screen on the right given above with the calculator
prompting you for a Guess? value for the point of intersection of the two graphs. Type in
1 for a guess value of the point of intersection. At this point, your screen should look like
the screen on the left given below. Press to see that the location of the
intersection point of these two graphs. At this point, your screen should look like the
screen on the right given below. The intersection point of these two graphs or the solution of this linear system of
equations is obtained as (1.125, 2.75).