Transcript Linear and Nonlinear Systems of Equations 7.1

Linear and Nonlinear
Systems of Equations
7.1
JMerrill, 2010
Example: Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
y= x–1
x+y=7
Step 1 If necessary, solve one equation for one variable. The first
equation is already solved for y.
Step 2 Substitute the expression into the other equation.
x+y=7
x + (x – 1) = 7
2x – 1 = 7
2x = 8
x=4
Substitute (x –1) for y in the other equation.
Combine like terms.
Example Continued
Step 3 Substitute the x-value into one of the
original equations to solve for y.
y=x–1
y = (4) – 1
Substitute x = 4.
y=3
The solution is the ordered pair (4, 3).
Example Continued
Check A graph or table supports your answer.
Example: Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
2y + x = 4
3x – 4y = 7
Method 1 Isolate y.
2y + x = 4
y=
++2
3x – 4y = 7
3x –4
+2 +2
=7
3x + 2x – 8 = 7
5x – 8 = 7
Method 2 Isolate x.
2y + x = 4
First equation.
x = 4 – 2y
Isolate one variable.
Second equation.
Substitute the
expression into
the second equation.
3x – 4y= 7
3(4 – 2y)– 4y = 7
12 – 6y – 4y = 7
Combine like terms.
5x = 15
x=3
First part of the solution
12 – 10y = 7
–10y = –5
Example Continued
Substitute the value into one of the original
equations to solve for the other variable.
Method 1
2y + (3) = 4
Method 2
Substitute the value to
solve for the other
variable.
2
+x=4
1+x=4
2y = 1
Second part of the solution
By either method, the solution is
.
x=3
You Try
Use substitution to solve the system of equations.
5x + 6y = –9
2x – 2 = –y
(3, –4)
You can also solve systems of equations by
elimination. With elimination, you get rid of one of
the variables by adding or subtracting equations. This
is actually 7.2, but if it’s easier, do it!
Example: Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
3x + 2y = 4
4x – 2y = –18
Step 1 Add the equations together to solve for one variable.
3x + 2y = 4
+ 4x – 2y = –18
7x
The y-terms have opposite coefficients.
= –14
Add the equations to eliminate y.
x = –2
First part of the solution
Example Continued
Step 2 Substitute the x-value into one of the original equations to
solve for y.
3(–2) + 2y = 4
2y = 10
y=5
Second part of the solution
The solution to the system is (–2, 5).
Systems may have one solution, no solutions, or infinitely many
solutions. When you try to solve these systems algebraically, the
result will be an identity or a contradiction.
Remember!
An identity, such as 0 = 0, is always true and indicates infinitely
many solutions. A contradiction, such as 1 = 3, is never true and
indicates no solution.
Example: Solving Systems with No Solution
Determine the solution, if it exists.
3x + y = 1
2y + 6x = –18
Isolate y.
3x + y = 1
y = 1 –3x
2(1 – 3x) + 6x = –18
2 – 6x + 6x = –18
2 = –18
Solve the first equation for y.
Substitute (1–3x) for y in the second equation.
Distribute.
Simplify.
Because 2 is never equal to –18, the equation is a contradiction.
Therefore, the system is inconsistent and has no solution.
Example: Solving Systems with Infinitely Many Solutions
Determine the solution, if it exists.
56x + 8y = –32
7x + y = –4
Isolate y.
7x + y = –4
y = –4 – 7x
Solve the second equation for y.
Substitute (–4 –7x) for y in the first
equation.
Distribute.
56x + 8(–4 – 7x) = –32
56x – 32 – 56x = –32
–32 = –32

Simplify.
Because –32 is equal to –32, the equation is an identity. The system is
consistent, dependent and has infinite number of solutions.
You Try
Determine the solution, if it exists.
6x + 3y = –12
2x + y = –6
Because –18 is never equal to –12, the equation is a
contradiction. Therefore, the system is inconsistent and has no
solutions.
Example: Zoology Application
A veterinarian needs 60 pounds of dog food that is 15%
protein. He will combine a beef mix that is 18% protein
with a bacon mix that is 9% protein. How many pounds of
each does he need to make the 15% protein mixture?
Let x present the amount of beef mix in the mixture.
Let y present the amount of bacon mix in the mixture.
Example Continued
Write one equation based on the amount of dog food:
Amount of
beef mix
plus
x
+
amount of
bacon mix
equals
=
y
Write another equation based on the amount of protein:
Protein of
beef mix
0.18x
plus
+
protein of
bacon mix
0.09y
equals
=
protein in
mixture.
0.15(60)
60.
60
Example Continued
x + y = 60
Solve the system.
0.18x +0.09y = 9
x + y = 60
y = 60 – x
0.18x + 0.09(60 – x) = 9
0.18x + 5.4 – 0.09x = 9
0.09x = 3.6
x = 40
First equation
Solve the first equation for y.
Substitute (60 – x) for y.
Distribute.
Simplify.
Example Continued
Substitute x into one of the original equations to solve for y.
40 + y = 60
y = 20
Substitute the value of x into one
equation.
Solve for y.
The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix.
Nonlinear Systems
When graphing nonlinear systems, you
may get:
 2 solutions:


Or no real solutions:
Solving Nonlinear Systems

The process is exactly the same:
◦ Solve for one variable
◦ Substitute that equation into the other
equation
◦ Factor and solve