Methods for Solving Systems of Equations
Substitution:
Problem:
Solve the following system of equations using substitution.
x2  y2  9
x2  y2  3
Solution:
The first step to solving a problem like this is to choose one of the
equations to solve for a particular variable. Let’s choose the top
equation and solve for the variable y.
x2  y2  9
y2  9  x2
y   9  x2
This expression CANNOT be simplified to y  (3  x) . It must be
used as the written expression.
The second step is to plug in the expression for y into the other (in
our case, second) equation and solve for the other variable (in our
case, the variable x). We have two expressions y  9  x 2 and
y   9  x 2 that we will need to plug in independently and solve each
for x. Once we get x values, we plug those into the expression for y
to get the associated y-values.
Part 1 of the second step:
Plug in y  9  x 2 into x 2  y 2  3 for y and solve for x.
x2  ( 9  x2 )2  3
x 2  (9  x 2 )  3
x2  9  x2  3
2x 2  9  3
2x 2  6
x2  3
x 3
We now have two values for the variable x: x   3 and x   3 . We
now need to plug these values into the expression for y ( y  9  x 2 )
to get the associated y-values.
We will first plug in x  
3 into y  9  x 2 and solve for y.

y  9  3

2
y  93
y 6

This gives us the solutions point
3, 6

We will first plug in x   3 into y  9  x 2 and solve for y.

y  9  3

2
y  93
y 6

This gives us the solutions point  3 , 6

Part 2 of the second step:
Plug in y   9  x 2 into x 2  y 2  3 for y and solve for x.
x 2  ( 9  x 2 ) 2  3
x 2  (9  x 2 )  3
x2  9  x2  3
2x 2  9  3
2x 2  6
x2  3
x 3
We now have two values for the variable x: x   3 and x   3 . We
now need to plug these values into the expression for y ( y   9  x 2 )
to get the associated y-values.
We will first plug in x  
3 into y   9  x 2 and solve for y.

y   9  3

2
y   93
y 6

This gives us the solutions point
3,  6

We will first plug in x   3 into y   9  x 2 and solve for y.

y   9  3

2
y   93
y 6

This gives us the solutions point  3 ,  6
This gives us 4 solutions to the system of equations:
3, 6 ,  3, 6 , 3,  6 ,  3,  6

 
 
 


When to use substitution:
Substitution is an ALGEBRAIC method of solution and can be used
with both linear and non-linear systems of equations. This is a method
that will always work.
Graphing:
Problem:
Solve the following system of equations with your calculator.
x2  y2  9
x2  y2  3
Solution:
The first step when using the calculator is to put each equation into a
form that it can be put into the calculator as an equation to graph.
This means that we must solve each equation in the system for y.
Let’s start with the top equation and solve for the variable y.
x2  y2  9
y2  9  x2
y   9  x2
This expression CANNOT be simplified to y  (3  x) . It must be
used as the written expression. Since the calculator does not have a
way to graph the y expression as written, we must graph it as two
separate equations. Thus we would put y  9  x 2 as one of the
equations to graph and y   9  x 2 as the another equation to graph.
Now we will solve the bottom equation for the variable y.
x2  y2  3
 y2  3  x2
y 2  3  x 2
y    3  x2
y   x2  3
Since the calculator does not have a way to graph the y expression as
written, we must graph it as two separate equations. Thus we would
put y  x 2  3 as a third equation to graph and y   x 2  3 as a
fourth equation to graph.
This means that we must put 4 equations into our calculator and graph
them.
y  9  x2
y   9  x2
y  x2  3
y   x2  3
If you have a TI-83, you will need to put these expressions into the
Y= window. If you have a Casio, you must choose GRAPH (5) from the
main menu.
Now you want to choose GRAPH or DRAW. We might as well use the
standard window [(—10, 10, 1) by (—10, 10, 1)]. We can see the graphs
of the two equations as follows:
The part that we are interested in are the points where the graphs
intersect. These points are circled.
We will need to use the intersection function of the calculator (2nd
CALC 5:intersect on the TI)
We are then asked for the First Curve (the number in the top right
corner tells us which graph we are currently on).
This indicates that we are
currently on Y1 ( 9  x 2 ).
This is asking if we want to use this curve
as one of the two that we are going to find
the intersection of.
This indicates where the cursor is now.
We need to decide what which of the equations we need to use when
finding intersections.
Our equations are as follows:
Y1:
y  9  x2
Y2: y   9  x 2
Y3: y  x 2  3
Y4: y   x 2  3
We need to remember that Y1 and Y2 were produced from the same
original equation. Similarly, Y3 and Y4 were produced from the same
original equation. Thus it will not make sense to look at the
intersection of Y1 and Y2 or the intersection of Y3 and Y4.
We have to figure out which graphs actually intersect.
Notice that Y1 will always be positive. This means that it will not
make sense for it to intersection a function that has negative values.
Notice also that Y3 will also always be positive. This should mean that
the graphs that can be seen above the x-axis must be the graphs Y1
and Y3. These will run into each other twice (once in the quadrant I
and once in quadrant II).
Notice now that Y2 will always be negative. Y4 will also always be
negative. This means that the graphs that can be seen below the xaxis must be the graphs of Y2 and Y4. These will run into each other
twice (once in the quadrant III and once in quadrant IV).
Thus what will make sense to look at is each of the following:
intersection of: Y1 and Y3 (in quadrant I)
intersection of: Y1 and Y3 (in quadrant II)
intersection of : Y2 and Y4 (in quadrant III)
intersection of: Y2 and Y4 (in quadrant IV)
To find the intersection of Y1 and Y3 (in quadrant I) will have to hit
the ENTER button on the screen below
After we hit the ENTER button, we get a screen that will look like
This indicates that we are
currently on Y2 (  9  x 2 ).
This is asking if we want to use this
curve as one of the two we are going to
find the intersection of.
This indicates where the cursor is now.
We don’t want to use Y2, we want to use Y3. To get to Y3, we need to
hit the down arrow button. This will give us the screen:
We are now on Y3.
Notice that there is no y value.
We cannot hit enter here. We must use the right arrow key to move
to an actual point on the graph of Y3 that is in quadrant I. After
hitting the right arrow a few times, I get:
I know that I am on the graph of Y3, since there is a 3 in the top
right corner and there are both an x value and a y value at the bottom
of the screen. I now will need to hit the ENTER button on this
screen. This will produce a screen that looks like
This is the point that we are
looking for.
This is asking us to try to point to
the point that we are looking for.
We need to use the left and/or right arrow keys to move to cursor to
the point we are looking for. Once pretty close, we will hit the ENTER
key. This will give us:
This screen tells us the intersection of
the two equations we chose.
These are the x value and y value of the
point of intersection.
Thus one of the points of intersection is (2.45494897, 1.7320508).
We will complete this process again to find the other 3 points of
intersection.
This gives us the answer for the solution of the system of the
equations:
(2.4494897, 1.7320508)
(-2.44949, 1.7320508)
(2.4494897, -1.732051)
(-2.44949, -1.732051)
When to use the graphing calculator:
Using the graphing calculator is an GRAPHICAL method of solution
and can be used with both linear and non-linear systems of equations.
This is a method that will work if the correct equations are put into
the calculator.
Elimination:
Problem:
Solve the following system of equations using the method of
elimination.
5 x  2 y  21
7 x  3 y  30
Solution:
The first step to solving a system of equations using elimination is to
determine which variable you want to eliminate. I think that I will try
to eliminate y. In order to eliminate y, I need for the coefficient of
the y term in the top equation to be the negative of the coefficient of
the y in the bottom equation. This will cause them to be eliminated
when I add the equations together.
For the top equation, I will multiply the entire equation by 3. This will
give me:
15 x  6 y  63
Since the top equation now has a coefficient of 6 on the y, I need a
coefficient of —6 on the y in the bottom equation. To get a
coefficient of —6 on the y in the bottom equation, I will need to
multiply the entire bottom equation by —2. This will give me:
 14 x  6 y  60
I now have the system of equation of
15 x  6 y  63
 14 x  6 y  60
I now need to add the two equations together. When I add, I get
(15 x  14 x)  (6 y  6 y )  63  60
x  0y  3
x3
Notice the y’s were eliminated as planned. If they do not become
eliminated when adding, I must have made an error.
Now that I have x = 3, I will need to plug that into each of the original
equations and solve for y.
When I plug x = 3 into the original top equation, I get
5(3)  2 y  21
15  2 y  21
2 y  21  15
2y  6
y3
When I plug x = 3 into the original bottom equation, I get
7(3)  3 y  30
21  3 y  30
3 y  30  21
3y  9
y3
I got y = 3 both times. This means that the solution to the system of
equation occurs when x = 3 and when y = 3. Thus the solution to the
system of equations is the point (3, 3).
When to use elimination:
Substitution is an ALGEBRAIC method of solution and can be used
with linear systems of equations.