MAT 150 Module 10 – Systems of
Equations
Lesson 1 – Systems
of Linear Equations
Systems of Linear Equations
A system of linear equations is
a set of two linear equations in
two variables, x and y.
A solution to the system is the
point (x, y) that satisfies both
equations in the system.
Graphically, the point (x, y) is
the point where the lines
formed by the two equations
intersect.
Systems of Linear Equations
Systems of Linear Equations
How do we decide whether a system has a
unique solution, no solution, or infinitely
many solutions?
To decide, we want to put both equations in
slope-intercept form (y = mx + b). Then it
is easy to tell if the equations are different
lines, parallel lines, or the same line.
Example 1
Without solving, decide whether the system have a
unique solution, no solution, or many solutions.
A. 6x - y = 10
x + 2y = 6
B. 2x + 3y = 6
4x + 6y = 12
C. 2x - y = 5
- 4x + 2y = -18
Example 1 - Solution
Without solving, decide whether the system have a
unique solution, no solution, or many solutions.
a. 6x - y = 10
x + 2y = 6
6x – y = 10
-y = -6x + 10
x + 2y = 6
2y = -x + 6
y = 6x - 10
y = -½x + 3
Example 1 - Solution
Without solving, decide whether the system have a
unique solution, no solution, or many solutions.
b. 2x + 3y = 6
4x + 6y = 12
2x + 3y = 6
3y = -2x + 6
y = -⅔x + 2
4x + 6y = 12
6y = -4x + 12
y = -⅔x + 2
Example 1 - Solution
Without solving, decide whether the system have a
unique solution, no solution, or many solutions.
C. 2x - y = 5
- 4x + 2y = -18
2x - y = 5
-4x + 2y = -18
-y = -2x + 5
y = 2x -5
2y = 4x -18
y = 2x -9
Solving systems of linear equations
There are three methods used to solve systems of
linear equations:
Graphing
Substitution
Elimination
Any of these three methods will work on any
problem. However, your goal is to use the method
that makes the most sense for that problem.
Solving systems by graphing
To solve a system by graphing, graph both lines and find
the point of intersection, if there is one.
Example 2
Solve the system of equations by graphing:
-x + 3y = 6
2x + 3y = 15
Example 2 - Solution
Solve the system of
equations by graphing:
-x + 3y = 6
2x + 3y = 15
The solution is the point
(3, 3).
Solving systems by substitution
To solve a system by substitution:
Solve one equation
for either x or y.
Solve for the
variable you are
left with.
Then substitute for
either or y into the
other equation.
Plug back in to find
the other variable.
Example 3
Solve the system by substitution:
2x – y = 14
x=y+8
Example 3 - Solution
Solve the system by substitution:
Since we have x = y+8,
2x – y = 14
2x –y = 14 becomes
Substitute y+8 in for x
2(y+8) – y = 14
into
the
first
equation.
x=y+8
Distribute
2y+16 – y = 14
Combine y terms
y+ 16 = 14
Subtract 16
y = -2
Plug into x = y+8
to find x
X = -2 + 8 = 6
The solution is
(6, -2)
Solving systems by elimination
Elimination is a good method when the equations
are written in general form, Ax + By = C. In this
method, our goal is to cancel out either x or y,
leaving an equation with only one variable.
Multiply one or
both equations by
a constant
Then add the two
equations to
cancel out x or y
So that either the x terms or the y
terms in both equations have the
same coefficient with opposite signs.
This will result in an
equation you can solve
for x or y.
Like 2x and -2x
or -4y and 4y
Plug that solution back
in to solve for the other
variable.
Example 4
Solve the systems by elimination:
a. 5x + 3y = 6
-3x – 2y = 2
b. 4x – 6y = 12
2x – 3y = 6
c. 7x = 8 – 14y
x + 2y = 5
Example 4 - Solution
Solve the systems by elimination:
a. 5x + 3y = 6
-3x – 2y = 2
2(5x +3y = 6)
3(-3x -2y = 2)
5(18)+3y = 6
Want 6y and -6y
10x + 6y = 12
+ -9x - 6y = 6
x
= 18
90+3y = 6
3y = -84
y = -84/3
=-28
Example 4 - Part a Solution
Solve the systems by elimination:
a. 5x + 3y = 6
-3x – 2y = 2
To check the answer, plug in x = 19 and y = -28 for x
and y into BOTH equations.
5(18) +3(-28) = 6?
-3(18) -2(-28) = 2?
Example 4 – Part b solution
Solve the systems by elimination:
b. 4x – 6y = 12
2x – 3y = 6
4x - 6y = 12
-2(2x - 3y = 6)
Want 4x and -4x
4x - 6y = 12
+ -4x + 6y = -12
0 =0
Same line –
infinitely many
solutions
Example 4 – Part b solution
Example 4 – Part C Solution
Solve the systems by elimination:
c. 7x = 8 – 14y
x + 2y = 5
7x + 14y = 8
-7(x + 2y = 5)
7x + 14y = 8
x + 2y = 5
7x + 14y = 8
-7x - 14y = -35
0 = -27
No Solution
Example 4 – Part C solution
Applications of systems of linear
equations
NEXT UP:
Lesson 2
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