HOW MANY SOLUTIONS
SEC 5.1.3
Intersection of a circle and a
parabola
Can you think of any other way?
WHAT IS A SYSTEM OF EQUATIONS?
• Answer: A system of equation just means 'more than
1 equation.'. A system of linear equations is just
more than 1 line, see the picture:
WHAT IS THE SOLUTION OF A SYSTEM OF
EQUATIONS?
• Answer: The solution is where the equations 'meet'
or intersect. The red point on the graph is the
solution of the system.
HOW MANY SOLUTIONS CAN SYSTEMS OF
LINEAR EQUATIONS HAVE?
• There can be zero solutions, 1 solution or infinite
solutions--each case is explained in detail in the
following slide.
• Note: Although systems of linear equations can have 3 or more equations, we
are going to refer to the most common case--a stem with exactly 2 lines.
Case I: 1 Solution
This is the most common situation
and it involves lines that intersect
exactly 1 time.
Case 2: No Solutions
This only happens when the lines
are parallel. As you can see,
parallel lines are not going to ever
meet.
Case 3: Infinite Solutions
This is the rarest case and only
occurs when you have the same
line.
☺5-33A; PAGE 229
• Solve the following system.
𝒚 = −𝟑𝒙 + 𝟓
Check the graph.
Check the table.
𝒚 = −𝟑𝒙 − 𝟏
☺ 5-33B; PAGE 229
• Solve the following system.
𝒚=
𝟏
𝒙
𝟐
+𝟏
Check the graph.
Check the table.
𝒚 = 𝟐𝒙 − 𝟏
*** 5-33C; PAGE 229
• Solve the following system.
𝒚𝟐 = 𝒙
𝒚=𝒙−𝟐
Check the graph.
Check the table.
☺ 5-33D; PAGE 229
• Solve the following system.
𝟒𝒙 − 𝟐𝒚 = 𝟏𝟎
Check the graph.
Check the table.
𝒚 = 𝟐𝒙 − 𝟓
OTHER SYSTEMS
• Now consider the system of equations that consists
of a line and a parabola i.e. a linear and a
quadratic function.
• Generate a table similar to the one we just created,
see slide #7.
• Next repeat the process for systems that consist of a
two parabolas.
• Repast the process for systems that consist of a
hyperbola and a circle.
*****5-34; PAGE
• Consider the following system:
𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓
𝒚 = 𝒙𝟐
− 𝟏𝟑
Have many solutions are possible?
The next few slides will display how to use
your calculators to solve the system of
equations graphically.
Go to slide number 20 to see the steps for
algebraic method.
GRAPHING CIRCLES ON YOUR
CALCULATOR
• A circle is not a function and
cannot be graphed in the regular
y=screen.
• To graph a circle in the regular y=
screen, you have to graph it as
two functions on the y= screen.
GRAPHING CIRCLES ON YOUR
CALCULATOR
First solve the equation of circle in terms
of y.
𝑥 2 + 𝑦 2 = 25
𝑦 2 = 25 − 𝑥 2
𝑦 = 25 − 𝑥 2
GRAPHING CIRCLES ON YOUR
CALCULATOR
Remember a square root can be positive
or negative.
In line 1 of y= screen graph what you've
been graphing and then graph the same
equation in line 2 but with a negative in
front of the equation.
You'll get something that looks like an oval
since the calculator screen is rectangular.
To make it look more circular (both parts
aren't going to connect), press zoom and
then select #5 (square).
GRAPHING CIRCLES ON YOUR
CALCULATOR
First solve the equation of circle in terms of y.
Remember a square root can be positive or
negative.
In line 1 of y= screen graph what you've been
graphing and then graph the same equation in
line 2 but with a negative in front of the
equation.
You'll get something that looks like an oval since
the calculator screen is rectangular. To make it
look more circular (both parts aren't going to
connect), press zoom and then select #5
(square).
Choose choice
#5 ZSquare
The two parts
will not
connect
ENTER THE EQUATION OF PARABOLA
Now use the Intersect key to find all points of
intersection.
Hint… both shape are symmetrical about the
y axis.
Careful when you use the Table of values…
as you have three equations.
ALGEBRAIC METHOD
𝑥 2 = 25 − 𝑦 2
25 − 𝑦 2 = 𝑦 + 13𝑥 2
= 𝑦 + 13
𝑦 2 + 𝑦 − 12 = 0
𝑦 = −4 𝑜𝑟 𝑦 = 3
Final points of
intersection:
(-4,3) (4,3) (-3,-4) (3,-4)
1. Rearrange both
equations
2. Use equal values
method.
3. Rearrange and solve
the quadratic.
4. Sub y values into
original equations
and solve for x.
ON YOUR OWN:
• Review your
notes. Rewrite
and fortify them if
needed.
• Update your
vocab list, if
needed.
• Review and Preview
• Page 230
• # 37-43; 39 is a
milestone problem