Download Solve Linear Systems Algebraically Part I

Solve Linear Systems
Algebraically Part I
Chapter 3.2
Solutions of Linear Systems of Equations
• A linear system of equations will always have one of the following as
a solution
• Exactly one solution in x and y (the lines intersect in a single point)
• An infinite number of solutions (the lines coincide and share all points)
• No solution (the lines are parallel and never intersect)
• The next slide shows how graphs of the last two would look
Solutions of Linear Systems of Equations
Solve Linear Systems Algebraically
• Although it is possible to solve a linear system of equations by
graphing, this is seldom the best method
• The reason is that, if the solution is not an ordered pair with integer
coordinates, then the point of intersection has fractional values
• These are usually impossible to read unless the coordinate plane is
broken in the right fractional values
• The best method for solving a linear system of equations is by
algebraic methods
Solve Linear Systems Algebraically
• You will learn about two such methods
• The first is called the substitution method
• The second is called the elimination method (or sometimes it is
called the addition method)
• In today’s lesson you will solve linear systems by the substitution
method
• This method is best used when one or both equations are solved for
either y or for x
The Substitution Method
• Suppose you are to solve a linear system of equations like the one
below
𝑦 = −3𝑥 + 10
𝑥 + 2𝑦 = 0
• Since the solution is the point (𝑥, 𝑦) that is common to both lines, then
the x and y values from the first equation must be the same as the x and
y values from the second equation
• This means that we can substitute the right part of the first equation
into y in the second equation
The Substitution Method
𝑦 = −3𝑥 + 10
𝑥 + 2𝑦 = 0
Substitute this:
𝑦 = −3𝑥 + 10
here:
𝑥 + 2 ___ = 0
and solve for x
The Substitution Method
𝑦 = −3𝑥 + 10
𝑥 + 2𝑦 = 0
𝑥 + 2 −3𝑥 + 10 = 0
𝑥 − 6𝑥 + 20 = 0
−5𝑥 + 20 = 0
−5𝑥 = −20
𝑥=4
So we have part of the solution: (4, )
We need to find y to complete the solution. Do this by substituting for x
in the first equation.
The Substitution Method
𝑦 = −3𝑥 + 10
𝑥 + 2𝑦 = 0
𝑦 = −3 4 + 10
𝑦 = −12 + 10
𝑦 = −2
The solution is (4, −2)
The Substitution Method
• Some linear systems might have both equations solved for y, like the
one shown below
𝑦 = 3𝑥 + 7
𝑦 = −2𝑥 + 2
• The substitution method is the same: replace y in either equation with
the right side of the other equation
The Substitution Method
𝑦 = 3𝑥 + 7
𝑦 = −2𝑥 + 2
Substitute this:
𝑦 = 3𝑥 + 7
here:
___ = −2𝑥 + 2
and solve for x.
The Substitution Method
𝑦 = 3𝑥 + 7
𝑦 = −2𝑥 + 2
3𝑥 + 7 = −2𝑥 + 2
5𝑥 + 7 = 2
5𝑥 = −5
𝑥 = −1
So we have part of the solution: (−1, )
To find y, substitute this value into either equation.
The Substitution Method
𝑦 = 3𝑥 + 7
𝑦 = −2𝑥 + 2
𝑦 = 3 −1 + 7
𝑦=4
The solution is (−1,4).
A System With No Solution
• How would you know when a system of linear equations has no
solution?
• The following example shows what to look for
𝑦 = 3𝑥 − 2
3𝑥 − 𝑦 = 4
Use the substitution method
A System With No Solution
• How would you know when a system of linear equations has no
solution?
• The following example shows what to look for
𝑦 = 3𝑥 − 2
3𝑥 − 𝑦 = 4
• You should get something like 2 = 4, or possibly some other equation
that is false
• When this happens, you conclude that the system has no solution
A System With Infinite Solutions
• How do you know when a system has an infinite number of solutions?
• The next example illustrates
10𝑥 − 2𝑦 = 2
𝑦 = 5𝑥 − 1
• Use the substitution method
A System With Infinite Solutions
• How do you know when a system has an infinite number of solutions?
• The next example illustrates
10𝑥 − 2𝑦 = 2
𝑦 = 5𝑥 − 1
• You should get something like 2 = 2 or possibly 0 = 0
• Both of these are always true, so the system has an infinite number of
solutions
Guided Practice
Solve the following systems of linear equations by the substitution
method.
2𝑥 − 5𝑦 = −10
1.
𝑦 =𝑥+2
𝑦 = −8𝑥 − 3
2.
−24𝑥 − 3𝑦 = −8
𝑦 =𝑥+6
3.
𝑦 = −7𝑥 − 2
𝑦 = −4𝑥 + 15
4.
𝑦 = 3𝑥 − 13
Exercise 3.2a
• Handout