Download POWERPOINT - Solving Systems of Equations

Solving Systems of
Equations
Modeling real-world problems
Profi
t
Loss
In business, the point at which income equals
expenses is called the break-even point.
When starting a business, people want to know
the point a which their income equals their
expenses, that’s the point where they start to
make a profit. In the example above the values
of y on the blue line represent dollars made and
the value of y on the dotted red line represent
dollars spent.
A system of equations is a set of two or more
equations that have variables in common.
The common variables relate to similar quantities.
You can think of an equation as a condition
imposed on one or more variables, and a system as
several conditions imposed simultaneously.
Remember, when solving systems of equations, you
are looking for a solution that makes each
equation true.
In earlier chapters, you learned to solve an
equation for a specified variable, graph the
equation, and find how pairs of lines
represented by linear equations in two
variables are related.
When solving a system of equations, you look for a
solution that makes each equation true. There
are several strategies you can use. To begin with
we will be using tables and graphs. Let’s look at
the following example and work through the
problem step-by-step to find a solution.
Edna leaves the trailhead at dawn to hike 12
miles toward the lake, where her friend Maria
is camping. At the same time, Maria starts
her hike toward the trailhead. Edna is walking
uphill so she averages only 1.5 mi/hr, while
Maria averages 2.5 mi/hr walking downhill.
When and where will they meet?
Here’s what we need to complete to solve this
example.
• Define variables for time (x) and for distance (y)
from the trailhead.
• Write a system of two equations to model this
situation.
• Solve this system by creating a table and finding
values for the variables that make both equations
true. Then locate this solution on a graph.
• Check your solution and explain its real-world
meaning.
Let x represent the
time in hours. Both
women hike the same
amount of time. Let
y represent the
distance in miles
from the trailhead.
When Edna and
Maria meet they will
both be the same
distance from the
trailhead, although
they will have hiked
different distances.
The system of equations that models this
situation is grouped in a brace.
Edna starts at the trailhead so she increases
her distance from it as she hikes 1.5 mi/hr for
x hours. Maria starts 12 miles from the
trailhead and reduces her distance from it as
she hikes 2.5 mi/hr for x hours.
 y  1.5 x

 y  12  2.5 x
Create a table from the equations. Fill in the
times and calculate each distance. The table
shows the x-value that gives equal y-values for
both equations. When x = 3, both y-values are
4.5. So the solution is the ordered pair (3, 4.5).
We say these values “satisfy” both equations.
Hiking Times and Distances
X
0
1
2
3
4
5
y = 1.5x
0
1.5
3
4.5
6
7.5
y = 12 – 2.5x
12
9.5
7
4.5
2
-0.5
Let’s create the data on your graphing
calculator. Enter the equations in y = on your
calculator and create a table.
Do both tables model this situation?
What do you notice about y when x increases
in each equation?
Why are the values different?
Hiking Times and Distances
X
0
1
2
3
4
5
y = 1.5x
0
1.5
3
4.5
6
7.5
y = 12 – 2.5x
12
9.5
7
4.5
2
-0.5
On the graph this solution is the point where
the two lines intersect. You can use trace
function or calculate function on your
calculator to approximate the coordinates of
the solution point, though sometimes you’ll get
an exact answer as in our example here.
Solving Systems of
Equations
Graphing Method
A system of linear equations is a set of two
or more equations with the same variable. The
solution of a system in x and y is any ordered
pair (x, y) that satisfies each of the equations
in the system.
The solution of a system of equations is the
intersection of the graphs of the equations.
If you can graph a
straight line, you can
solve systems of
equations graphically!
The process is very
easy. Simply graph the
two lines and look for
the point where they
intersect (cross).
Remember using the
graphing method many
times only
approximates the
solution, so sometimes
it can be unreliable.
Solving Systems of Equations by
Graphing.
To solve a system of equations graphically, graph both equations and
see where they intersect. The intersection point is the solution.
4x – 6y = 12
4x = 6y + 12
4x – 12 = 6y
6y = 4x – 12
6 6
6
y = 2/3x – 2
slope = 2/3
y-intercept = -2
2x + 2y = 6
2y = -2x + 6
2
2 2
y = -x + 3
slope = -1/1
y-intercept = 3
Graph the equations.
The slope-intercept
method of graphing was
used in this example.
The point of intersection
of the two lines (3, 0) is
the solution to the system
of equations.
This means that (3, 0),
when substituted into
either equation, will make
them both true.
Use a graph to solve the system of equations
below. Graph both equations on the same
coordinate plane.
Graph x + y = 5 using the intercepts: (5, 0) and (0, 5)
Graph y = 2x – 1 using the slope-intercept method.
x  y  5

 y  2x 1
Locate the point where the lines intersect.
From the graph, the solution appears to (2, 3).
Check to be sure that (2, 3) is the solution,
substitute 2 for x and 3 for y into each
equation.
x+y=5
y = 2x – 1
2+3=5
3 = 2(2) - 1
Summary of Solutions of
Systems of Linear Equations
The lines
intersect so
there is one
solution.
The lines are
parallel so there
are no solutions.
y – 2x = 7
Y = 2x + 3
x + 2y = 7
x = y + 4
-3x = 5 – y
2y = 6x + 10
The lines are
the same so
there are
infinitely many
solutions.
Note
Some systems of equations may
be very difficult to solve using
the graphing method. The exact
solution would be hard to
determine from a graph because
the coordinates are not integers.
Solving a system algebraically is
better than graphing when you
need an accurate solution.
Take for example the system of
equations:
3x + 2y = 12
x–y=3
x = 3/8 and y = 15/4
Solving Systems of
Equations Algebraically
Substitution Method
The substitution
method is used to
eliminate one of the
variables by
replacement when
solving a system of
equations.
Think of it as
“grabbing” what one
variable equals from
one equation and
“plugging” it into the
other equation.
Solve this system of equations using the
substitution method.
Step 1
3y – 2x = 11
Y + 2x = 9
Solve one of the equations for either “x” or “y”.
In this example it is easier to solve the second
equation for “y”, since it only involves one step.
Y = 9 – 2x
Step 2
Replace the “y” value in the first equation by
what “y” now equals (y = 9 – 2x). Grab the “y” value
and plug it into the other equation.
3y – 2x = 11
3(9 – 2x) – 2x = 11
Step 3
Solve this new equation for “x”.
3(9 – 2x) – 2x = 11
27 – 6x – 2x = 11
27 – 6x – 2x = 11
27 – 8x = 11
-8x = -16
x=2
Step 4
Now that we know the “x” value (x = 2), we
place it into either of the ORIGINAL
equations in order to solve for “y”. Pick the easier
one to work with!
Y + 2x = 9
y + 2(2) = 9
y+4=9
y=5
Step 5
Check: substitute x = 2 and y = 5 into BOTH
ORIGINAL equations. If these answers are
correct BOTH equations will be true!
3y – 2x = 11
3(5) – 2(2) = 11
15 – 4 = 11
11 = 11 True?
Y + 2x = 9
5 + 2(2) = 9
5+4=9
9 = 9 True?
The Substitution Method
• Step 1 Solve one equation for x (or y).
• Step 2 Substitute the expression from Step 1
into the other equation.
• Step 3 Solve for y (or x).
• Step 4 Take the value of y (or x) found in Step 3
and substitute it into one of the original
equations. Then solve for the other variable.
• Step 5 The ordered pair of values from Steps 3
and 4 is the solution. If the system has no
solution, a contradictory statement will result in
either Step 3 or 4.
Solving systems of equations in real-world
problems.
April sold 75 tickets to a school play and collected a
total of $495. If the adult tickets cost $8 each and
child tickets cost $5 each, how many adult tickets and
how many child tickets did she sell?
Solution: Let a represent the adult tickets and c
represent the child tickets.
Individual tickets sold equaled 75, so a + c = 75
All total April sold $495 in tickets, since adult tickets are $8 and
child tickets are $5, so 8a + 5c = 495.
System of Equations
a + c = 75
8a + 5c == 495
Solution
a + c = 75
8a + 5c = 495
a = 75 – c
8(75 – c) + 5c = 495
600 – 8c + 5c = 495
600 - 3c = 495
105 = 3c
35 = c
a + c = 75
a + 35 = 75
a = 40
There were 40 adult tickets and 35 child
tickets sold. 40 + 35 = 75
8(40) + 5(35) = 495
320 + 175 = 495
Solving Systems of
Equations
Elimination method
You can use the Addition and Subtraction
Properties of Equality to solve a system by the
elimination method. You can add or subtract
equations to eliminate (getting rid of) a variable.
If you add the two equations
Step 1
together, the +6y and -6y
each other out because
5x – 6y = -32 cancel
of the Property of Additive
Inverse
3x + 6y = 48
Eliminate y because the sum of the coefficients of y
is zero
5x – 6y = -32
3x + 6y = 48
8x + 0 = 16 Addition Property of Equality
Solve for x
x = 2
Step 2
Solve for the eliminated variable y using
either of the original equations.
Remember
x = 2
3x + 6y = 48
3(2) + 6y = 48
6 + 6y = 48
6y = 42
y = 7
Choose the 2nd equation
Substitute 2 for x
Simplify. Then solve for y.
Since x = 2 and y = 7, the solution is (2, 7)
Check
5x – 6y = -32
3x + 6y = 48
5(2) – 6(7) = -32 3(2) + 6(7) = 48
10 – 42 = -32
6 + 42 = 48
-32 = -32
48 = 48
True
True
Remember, the order pair (2, 7) must make both
equations true.
Suppose your community center sells a
total of 292 tickets for a basketball
game. An adult ticket cost $3. A
student ticket cost $1. The sponsors
collected $470 in ticket sales. Write
and solve a system to find the number of
each type of ticket sold.
Let a = number of adult tickets
Let s = number of student tickets
total number of ticket total number of sales
a + s = 292
3a + 1s = 470
Solve by elimination (get rid of s) because the
difference of the coefficients of s is zero.
a + s = 292
=(3a + s = 470)
This is the number
That means you must subtract
of adult tickets
-2a
+
0
=
-178
the two equations so,
sold.
-3a – a = -470 is what you
a = 89
must subtract.
Next Step
Solve for the eliminated variable using either of
the original equations.
a + s = 292
89 + s = 292
s = 203
There were 89 adult tickets sold and 203 student
tickets sold.
Is the solution reasonable? The total number of tickets is
89 + 203 = 292. The total sales is $3(89) + $1(203) =
$470. The solution is correct.
If you have noticed in the last few examples
that to eliminate a variable its coefficients
must have a sum or difference of zero.
Sometime you may need to multiply one or both of the
equations by a nonzero number first so that you can then
add or subtract the equations to eliminate one of the
We can add these two equations
variables.
We can add these two
together to eliminate the y variable.
2x + 5y = 17
6x – 5y = -19
equations together to eliminate
the x variable.
7x + 2y = 10
-7x + y = -16
2x + 5y = -22
10x + 3y = 22
If you notice the systems of equations above, two of them have
something in common. The third doesn’t.
What are we going to do with these
equations, can’t eliminate a variable the
way they are written?
Multiplying One
Equation
Be careful when you
subtract. All the signs
in the equation that is
being subtracted
change.
-10x – 3y = -22
Solve by Elimination
2x + 5y = -22
10x + 3y = 22
Start with the
given system.
Step 1
2x + 5y = -22
10x + 3y = 22
To prepare for
eliminating x, multiply
the first equation by 5.
5(2x + 5y = -22)
10x + 3y = 22
Ask:
Is one coefficient
a factor of the
other coefficient
for the same
variable?
Subtract the equations to
eliminate x.
10x + 25y = -110
-(10x + 3y = 22)
0 + 22y = -132
y = -6
NEXT
Step 2
Solve for the eliminated variable using either
of the original equations.
2x + 5y = -22
2x +5(-6) = -22
2x – 30 = -22
2x = 8
x=4
Choose the first equation.
Substitute -6 for y.
Solve for x.
The solution is (4, -6).
Multiplying Both
Equations
To eliminate a variable, you may need to multiply
both equations in a system by a nonzero number.
Multiply each equation by values such that when
you write equivalent equations, you can then add
or subtract to eliminate a variable.
In these two equations you
cannot use graphing or
substitution very easily.
However ever if we multiply
the first equation by 3 and
the second by 2, we can
eliminate the y variable.
4x + 2y = 14
7x + 3y = -8
4 x 7 = 28
2x3=6
Find the least common
multiple LCM of the
coefficients of one
variable, since working
with smaller numbers
tends to reduce the
likelihood of errors.
NEXT
Add the equations to
eliminate y.
4X + 2Y = 14
7X – 3Y = - 8
Start with the given
system.
3(4X + 2Y = 14)
2(7X – 3Y = -8)
12X + 6Y = 42
14X – 6Y = -16
26X + 0 = 26
To prepare to eliminate y, multiply
the first equation by 3 and the
second equation by 2.
26X = 26
X = 1
Solve for the eliminated variable y using either of the
original equations.
4x + 2y = 14
4(1) + 2y = 14
4 + 2y = 14
2y = 10
y= 5
The solution is (1, 5).
Solving Systems of Equations
The Best Time to Use Which Method
Graphing:
Substitution:
Elimination
Using
Addition:
Elimination
Using
Subtraction:
Used to estimate the solution, since graphing usually does
not give an exact solution.
Y = 2x - 3
y=x-1
If one of the variables in either equation has a
coefficient of 1 or –1
3y + 2x = 4
-6x + y = -7
If one of the variables has opposite coefficients in the
two equations 5x – 6y = -32
3x + 6y = 48
If one of the variables has the same coefficient in the
two equations
2x + 3y = 11
2x + 9y = 1
Elimination
Using
Multiplication: If none of the coefficients are 1 or –1 and neither of the
variables can be eliminated by simply adding and
subtracting the equations.
5(2x + 5y = -22)
10x + 3y = 22
3(4x + 2y = 14)
2(7x = 3y = -8)