Download Solving Systems of Equations

Solve Systems
of Equation
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.
Example 1
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.
Define the variables
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.
Write equations
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
Use a graph to solve each system of equations. If
the system has no solution, write none.
 y  x 1

y  5  x
 y  x  5

 y  x 1
y  x  5

y  x  3
 y  14 x  4

1
y


2 x2

 2 x  y  1

 2 x  y  2
x  y  3

y  x  7
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.
Use the substitution method to solve each
system of equations. Check your answers.
 y  5 x  1

 y  4
 y  x  4

 2 x  y  5
2 x  y  5

x  2 y  4
 2 x  3 y  1

 y  3 x  1
2 x  3 y  6

 y  2x  5
 15 x  3  y

 3 x  y  5
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
Write a system of
equations and solve.
At a baseball game,
Jose bought five hot
dogs and three sodas
for $17. At the same
time, Allison bought
two hot dogs and four
sodas for $11. Find the
cost of one hot dog
and one soda.
Solving Systems
of Equations
Elimination method
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.
Step 1
If you add the two equations
together, the +6y and -6y
cancel
each other out because
5x – 6y = -32
of the Property of Additive
Inverse
3x + 6y = 48
Eliminate y because the sum of the coefficients of
y is zero
5x – 6y = -32 Addition Property of Equality
3x + 6y = 48 Solve for x
8x + 0 = 16
x = 2
Elimination Method
Step 2
Solve for the eliminated variable y using either of
the original equations.
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.
Elimination Method
Since x = 2 and y = 7, the solution is (2, 7)
Check
5x – 6y = -323x + 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.
Elimination Method
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.
Elimination Method
Let a = number of adult tickets
Let s = number of student tickets
total number of ticket
a + s = 292
total number of sales
3a + 1s = 470
Solve by elimination (get rid of s) because the
difference of the coefficients of s is zero.
That means you must subtract
the two equations so,
-3a – a = -470 is what you
must subtract.
a + s = 292
3a + s = 470
-2a + 0 = -178
a = 89
This is the number
of adult tickets
sold.
Next Step
Elimination Method
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.
Elimination Method
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
We can add these two equations
We can add these two
the variables.
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?
Elimination Method
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
Elimination Method
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).
Elimination Method-Practice
Solve by elimination.
-2x + 5y = -32
7x – 5y = 17
3x – 10y = -25
4x + 40y = 20
2x – 3y = 61
2x + y = -7
Ask:
Is one coefficient
a factor of the
other coefficient
for the same
variable?
Elimination Method
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
Elimination Method
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)
To prepare to eliminate y, multiply
the first equation by 3 and the
second equation by 2.
12X + 6Y = 42
14X – 6Y = -16
26X + 0 = 26
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).
Practice
Practice and Problem
Solving
Solve by elimination:
1) 2x + 5y = 17
6x + 5y = -9
2) 7x + 2y = 10
-7x + y = -16
3) 2x – 3y = 61
2x + y = -7
4) 24x + 2y = 52
6x – 3y = -36
5) y = 2x
y=x–1
You choose what
method you what
to use to solve
question 5 thru
10.
6) 9x + 5y = 34
8x – 2y = -2
Word Problems
Practice (con’t)
7) The sum of two numbers is 20. Their difference
is 4. Write and solve a system of equations.
8) Your school sold 456 tickets for a school play.
An adult ticket cost $3.50. A student ticket cost
$1. Total ticket sales equaled $1131. Let a = adult
tickets sold, and s = student tickets sold. How
many tickets of each were sold?
9) Suppose the band sells cans of popcorn for $5
each and mixed nuts for $8 each. The band sells
a total of 240 cans and makes a total of $1614.
Find the number of cans of each sold.
One more problem,
my favorite.
Practice (con’t)
10) A farmer raises chicken and cows. He
has a total of 34 animals in his barnyard. His
six-year son came in one day all excited
saying, “Daddy, daddy, did you know all
your animals have a total of 110 legs.”
Write a system of equations to represent this
situation.
How many chickens and how many cows
does the farmer have?
Helpful Hints
When you solve systems using elimination,
plan a strategy. Here are a few hints to help
you decide how to eliminate a variable.
Can I eliminate a variable by
adding or subtracting the
given equations?
Answer these questions.
Do It.
YES
Or
NO
Can I multiply one of
the equations by a
number, and then
add or subtract the
equations?
YES
or
NO
Do It.
Multiply both equations
by different numbers.
then add or subtract the
equations
When to use which?
Graphing: Used to estimate the
solution, since graphing usually
does not give an exact solution.
Y = 2x - 3
y=x-1
Substitution: If one of the
variables in either equation has a
coefficient of 1 or –1.
3y + 2x = 4
-6x + y = -7
Elimination Using Addition: If one
of the variables has opposite
coefficients in the two equations.
5x – 6y = -32
3x + 6y = 48
Summary
Elimination Using Subtraction: 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)