Download System of equations

Chapter 4
Systems of
Linear
Equations and
Inequalities
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
1
CHAPTER
4
Systems of Equations and
Inequalities
4.1
4.2
4.3
Solving Systems of Linear Equations in
Two Variables
Solving Systems of Linear Equations in
Three Variables
Solving Applications Using Systems of
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4.1
Solving Systems of Linear
Equations in Two Variables
1. Determine whether an ordered pair is a
solution for a system of equations.
2. Solve systems of linear equations graphically
and classify systems.
3. Solve systems of linear equations using
substitution.
4. Solve systems of linear equations using
elimination.
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System of equations: A group of two or more
equations.
 x y 5

3x  4 y  8
(Equation 1)
(Equation 2)
Solution for a system of equations: An ordered set of
numbers that makes all equations in the system true.
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Checking a Solution to a System of Equations
To verify or check a solution to a system of equations,
1. Replace each variable in each equation with its
corresponding value.
2. Verify that each equation is true.
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Example
Determine whether each ordered pair is a solution to
the system of equations.
x  y  7

 y  3x  2
a. (3, 2)
Solution
a. (3, 2)
(Equation 1)
(Equation 2)
b. (3, 4)
x+y=7
y = 3x – 2
3 + 2 = 7
2 = 3(3) – 2
1 = 7
2 = 11
False
False
Because (−3, 2) does not satisfy both equations, it is not a
solution for the system.
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continued
b. (3, 4)
x+y=7
3+4=7
7=7
True
y = 3x – 2
4 = 3(3) – 2
4=7
False
Because (3, 4) satisfies only one equation, it is not a
solution to the system of equations.
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A system of two linear equations in two variables can
have one solution, no solution, or an infinite number of
solutions.
x  y  5

 y  2x  4
The lines intersect
at a single point.
There is one
solution.
x  2 y  4

2 x  4 y  8
 y  3x  1

 y  3x  2
The equations have
the same slope, the
lines are parallel.
There is no solution.
The lines are
identical. There are
an infinite number of
solutions.
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Solving Systems of Equations Graphically
To solve a system of linear equations graphically,
1. Graph each equation.
a. If the lines intersect at a single point, then the
coordinates of that point form the solution.
b. If the lines are parallel, there is no solution.
c. If the lines are identical, there are an infinite
number of solutions, which are the coordinates of
all the points on that line.
2. Check your solution.
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Example
Solve the system of equations graphically.
y  2  x

2 x  4 y  12
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines intersect at a single
point, (2, 4).
We can check the point in each
equation to verify and we
leave that to you.
(2, 4)
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2x + 4y = 12
y=2–x
10
Example
Solve the system of equations graphically.
3

y  x 3
4


3x  4 y  4
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines appear to be parallel,
which we can verify by
comparing the slopes. 3x  4 y  4
The slopes are the
same, so the lines
are parallel.
4 y  3x  4
3
y  x 1
4
The system has
no solution.
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Example
Solve the system of equations graphically.
4 x  2 y  2

 y  2x 1
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines appear to be identical.
4x  2 y  2
2 y  2  4 x
y  1  2 x
y  2x 1
The equations are identical. All
ordered pairs along the line are
solutions.
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Consistent system of equations: A system of
equations that has at least one solution.
Inconsistent system of equations: A system of
equations that has no solution.
Dependent linear equations in two unknowns:
Equations with identical graphs.
Independent linear equations in two unknowns:
Equations that have different graphs.
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Classifying Systems of Equations
To classify a system of two linear equations in two
unknowns, write the equations in slope-intercept form
and compare the slopes and y-intercepts.
Consistent system with
independent equations:
The system has a single
solution at the point of
intersection.
Consistent system with
dependent equations: The
system has an infinite
number of solutions.
The graphs are different.
They have different slopes.
The graphs are identical.
They have the same slope
and same y-intercept.
Inconsistent system: The
system has no solution.
The graphs are parallel
lines. They have the same
slope, but different yintercepts.
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Example
For each of the systems of equations in the example,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
(Equation 1)
2a.  y  2  x

2 x  4 y  12
The graphs intersected
at a single point.
The system is
consistent.
(Equation 2)
The equations are
independent (different
graphs), and the
system has one
solution: (2, 4).
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Example
For each of the systems of equations in the example,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
2b.  y  3 x  3
(Equation 1)
4


3x  4 y  4
(Equation 2)
The graphs were parallel lines. The system is
inconsistent and has no solutions.
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Example
For each of the systems of equations in the example,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
2c. 4 x  2 y  2
(Equation 1)

 y  2x 1
(Equation 2)
The graphs coincide. The system is consistent (it
has a solution) with dependent equations (same
graph) and has an infinite number of solutions.
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Solving Systems of Two Linear Equations Using
Substitution
To find the solution of a system of two linear equations
using the substitution method,
1. If necessary, isolate one of the variables in one of the
equations.
2. In the other equation, substitute the expression you
found in step 1 for that variable.
3. Solve this new equation. (It now has only one
variable.)
4. Using one of the equations containing both variables,
substitute the value you found in step 3 for that
variable and solve for the value of the other variable.
5. Check the solution in the original equations.
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Example
Solve the system of equations using substitution.
4 x  5 y  8

x  1 y
Solution
Step 1: Isolate a variable in one equation. The second
equation is solved for x.
Step 2: Substitute x = 1 – y for x in the first equation.
4x  5 y  8
4(1  y )  5 y  8
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4 x  5 y  8

x  1 y
continued
Step 3: Solve for y.
4(1  y )  5 y  8
4  4y  5y  8
4 y 8
y4
Step 4: Solve for x by substituting 4 for y in one of
the original equations.
x  1 y
x  1 4
x  3
The solution is (3, 4).
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Example
Solve the system of equations using substitution.
2 x  y  7

 x  y  1
Solution
Step 1: Isolate a variable in one equation. Use either
equation. 2x + y = 7
y = 7 – 2x
Step 2: Substitute y = 7 – 2x for y in the second
equation.
x  y  1
x  (7  2 x)  1
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2 x  y  7

 x  y  1
continued
Step 3: Solve for x.
x  (7  2 x)  1
x  7  2x  1
3x  7  1
3x  6
x2
Step 4: Solve for y by substituting 2 for x in one of
the original equations.
2x  y  7
2(2)  y  7
4 y  7
The solution is (2, 3).
y 3
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Example
x  y  9

Solve the system using elimination. 3x  y  7
Solution We can add the equations.
x y 9
3x  y  7
4x  0  16
4x  16
x4
Notice that y is eliminated,
so we can easily solve for
the value of x.
Divide both sides by 4 to isolate x.
Now that we have the value of x, we can find y by
substituting 4 for x in one of the original
equations.
x+y=9
4+y=9
The solution is (4, 5).
y=5
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Example
x  y  8
Solve the system of equations. 3x  4 y  11

Solution Because no variables are eliminated if we
add, we rewrite one of the equations so that it has a
term that is the additive inverse of one of the terms
in the other equation.
Multiply the each term of the first equation by 4.
4 x  4 y  32
3 x  4 y  11
7 x  0  21
7 x  21
x 3
Solve for y.
x+y=8
3+y=8
y=5
x y 8
4  x  4  y  4 8
4 x  4 y  32
The solution
is (3, 5).
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Example
4 x  5 y  19
Solve the system of equations. 3x  2 y  9

Solution Choose either variable to eliminate. We can
multiply both equations by numbers that make the
terms additive inverses to cancel them.
Multiply the first equation by 2.
Multiply the second equation by 5.
4 x  5 y  19
3x  2 y  9
Multiply by 2.
Multiply by 5.
Add the equations to
eliminate y.
8 x  10 y  38
15 x  10 y  45
7 x  0  7
7 x  7
x 1
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continued
4 x  5 y  19

3x  2 y  9
Substitute x = 1 into either of the original equations.
4x – 5y = 19
4(1) – 5y = 19
4 – 5y = 19
–5y = 15
y = 3
The solution is (1, –3).
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Example
Solve the system of equations. 0.03 x  0.02 y  0.03

2
2
4
 5 x  5 y  5
Solution
To clear the decimals in Equation 1, multiply by 100.
To clear the fractions in Equation 2, multiply by 5.
0.03 x  0.02 y  0.03
4
2
2
x y 
5
5
5
Multiply by 100.
Multiply by 5.
Multiply equation 2 by 1
then combine the
equations.
3x  2 y  3
4x  2 y  2
3x  2 y  3
4x  2 y  2
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3x  2 y  3
4 x  2 y  2
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continued
3x  2 y  3
4 x  2 y  2
x  0  1
x  1
x  1
Substitute to find y.
3x  2 y  3
3(1)  2 y  3
3  2 y  3
2 y  6
y  3
The solution is (1, 3).
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Solving Systems of Two Linear Equations Using Elimination
To solve a system of two linear equations using the elimination
method,
1. Write the equations in standard form (Ax + By = C).
2. Use the multiplication principle to clear fractions or decimals
(optional).
3. If necessary, multiply one or both equations by a number (or
numbers) so that they have a pair of terms that are additive
inverses.
4. Add the equations. The result is an equation in terms
of one variable.
5. Solve the equation from step 4 for the value of that variable.
6. Using an equation containing both variables, substitute the
value you found in step 5 for the corresponding variable and
solve for the value of the other variable.
7. Check your solution in the original equations.
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Inconsistent Systems and Dependent Equations
When both variables have been eliminated and the
resulting equation is false, such as 0 = 5, there is no
solution. The system is inconsistent.
When both variables have been eliminated and the
resulting equation is true, such as 0 = 0, the equations
are dependent. There are an infinite number of
solutions.
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Example
Solve the system of equations.
 3x  y  4

3 x  y  5
Solution
Notice that the left side of the equations are additive
inverses. Adding the equations will eliminate both
variables.
3 x  y  4
3 x  y   5
0  9
False statement. The
system is inconsistent
and has no solution.
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Example
Solve the system of equations using substitution.
 y  4  3x

6 x  2 y  8
Solution
Substitute y = 4 – 3x into the second equation.
6 x  2 y  8
6 x  2(4  3x)  8
6x  8  6x  8
6x  6x  8  8
8  8
True statement. The
number of solutions is
infinite.
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