Transcript Thinking Mathematically by Robert Blitzer

Systems of
Nonlinear Equations
in Two Variables
Systems of Nonlinear Equations
and Their Solutions
A system of two nonlinear equations in two variables contains at least one
equation that cannot be expressed in the form Ax + By = C. Here are two
examples:
x2 = 2y + 10
3x – y = 9
y = x2 + 3
x2 + y2 = 9
A solution to a nonlinear system in two variables is an ordered pair of real
numbers that satisfies all equations in the system. The solution set to the
system is the set of all such ordered pairs.
Text Example
Solve by the substitution method:
x–y=3
(x – 2)2 + (y + 3)2 = 9
The graph is a line.
The graph is a circle.
Solution Graphically, we are finding the intersection of a line and a circle
whose center is at (2, -3) and whose radius measures 2.
Step 1 Solve one of the equations for one variable in terms of the other.
We will solve for x in the linear equation - that is, the first equation. (We could
also solve for y.)
x–y=3
This is the first equation in the given system.
x=y+3
Add y to both sides.
Text Example cont.
Solution
Step 2 Substitute the expression from step 1 into the other equation. We
substitute y + 3 for x in the second equation.
x=y +3
( x – 2)2 + (y + 3)2 = 4
This gives an equation in one variable, namely
(y + 3 – 2)2 + (y + 3)2 = 4.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing one variable.
(y + 3 – 2)2 + (y + 3)2 = 4
This is the equation containing one variable.
(y + 1)2 + (y + 3 )2 = 4
Combine numerical terms in the first parentheses.
y2 + 2y + 1 + y2 + 6y + 9 = 4
Square each binomial.
2y2 + 8y + 10 = 4
Combine like terms on the left.
2y2 + 8y + 6 = 0
Subtract 4 from both sides and set the quadratic
equation equal to 0.
Solution
Text Example cont.
y2 + 4y + 3 = 0
(y + 3)(y + 1) = 0
y + 3 = 0 or y + 1 = 0
y = -3
or y = -1
Simplify by dividing both sides by 2.
Factor.
Set each factor equal to 0.
Solve for y.
Step 4 Back-substitute the obtained values into the
equation from step 1. Now that we have the y-coordinates of the solutions, we back-substitute -3 for y and -1 for
y in the equation x = y + 3.
If y = -3: x = -3 + 3 = 0, so (0, -3) is a solution.
If y = -1: x = -1 + 3 = 2, so (2, -1) is a solution.
Step 5 Check the proposed solution in both of the
system's given equations. Take a moment to show that
each ordered pair satisfies both equations. The solution
set of the given system is {(0, -3), (2, -1)}.
7
6
5
x – y =4 3
3
2
1
-5 -4 -3 -2 -1
-1
(2, -1)
1
2 3 4
5
6
7
-2
(0, -3) -3
-4
-5
-6
-7
(x – 2)2 + (y + 3)2 = 4
Text Example
Solve the system:
4x2 + y2 = 13
x2 + y2 = 10
Equation 1.
Equation 2.
Solution We can use the same steps that we did when we solved linear systems
by the addition method.
Step 1 Write both equations in the form Ax2 + By2 = C. Both equations
are already in this form, so we can skip this step.
Step 2 If necessary, multiply either equation or both equations by
appropriate numbers so that the sum of the x2-coefficients or the sum of
the y2-coefficients is 0. We can eliminate y by multiplying Equation 2 by -1.
4x2 + y2 = 13
x2 + y2 = 10
No change.
Multiply by -1.
4x2 + y2 = 13
-x2 – y2 = -10
Solution
Steps 3 and 4
Add.
Text Example cont.
Add equations and solve for the remaining variable.
4x2 + y2
-x2 – y2
3x2
x2
x
=
=
=
=
=
13
-10
3
1
+1
Step 5 Back-substitute and find the values for the other variables. We
must back-substitute each value of x into either one of the original equations.
Let's use x2 + y2 = 10, Equation 2. If x = 1,
12 + y2 = 10
Replace x with 1 in Equation 2.
y2 = 9
Subtract 1 from both sides.
y = ±3
Apply the square root method.
(1, 3) and (1, -3) are solutions. If x = -1,
(-1)2 + y2 = 10
Replace x with -1 in Equation 2.
y2 = 9
The steps are the same as before.
y = ±3
(-1, 3) and (-1, -3) are solutions.
Text Example cont.
Solution
Step 6 Check. Take a moment to show that each of the four ordered pairs
satisfies Equation 1 and Equation 2. The solution set of the given system is
{(1, 3), (1, -3), (-1, 3), (-1, -3)}.
7
4x2 + y2 = 13
6
(-1, 3)
5
4
(1, 3)
3
2
x2 + y2 = 10
1
-5 -4 -3 -2 -1
-1
1
2 3 4
-2
(-1, -3)
-3
-4
-5
-6
-7
(1, -3)
5
6
7
Text Example
Solve the system:
y = x2 + 3 Equation 1 (The graph is a parabola.)
x2 + y2 = 9
Equation 2 (The graph is a circle.)
Solution We could use substitution because Equation 1 has y expressed in
terms of x, but this would result in a fourth-degree equation. However, we
can rewrite Equation 1 by subtracting x2 from both sides and adding the
equations to eliminate the x2-terms.
Add.
-x2 + y
= 3
x2
+ y2 = 9
y + y2 = 12
Subtract x2 from both sides of Equation 1.
This is Equation 2.
Add the equations.
Solution
Text Example cont.
We now solve this quadratic equation.
y + y2 = 12
y2 + y – 12 = 0
Subtract 12 from both circles and get the quadratic
(y + 4)(y – 3) = 0
y + 4 = 0 or y – 3 = 0
y = -4
or
y=3
equation equal to 0.
Factor.
Set each factor equal to 0.
Solve for y.
To complete the solution, we must back-substitute each value of y into either
one of the original equations. We will use y = x2 + 3, Equation 1. First, we
substitute -4 for y.
-4 = x2 + 3
-7 = x2
Subtract 3 from both sides.
Text Example cont.
Solution
Because the square of a real number cannot be negative, the equation x2 = -7
does not have real-number solutions. Thus, we move on to our other value for
y, 3, and substitute this value into Equation 1.
y = x2 + 3
3 = x2 + 3
0 = x2
0=x
This is Equation 1.
7
6
Back-substitute 3 for y.
Subtract 3 from both sides.
y=
x2 +
3
(0, 3)
3
2
Solve for x.
We showed that if y = 3, then x = 0. Thus, (0, 3) is
the solution. Take a moment to show that (0, 3)
satisfies Equation 1 and Equation 2. The solution
set of the given system is {(0, 3)}.
5
4
1
-5 -4 -3 -2 -1
-1
1
2 3 4
5
6
7
-2
-3
-4
-5
-6
-7
x2 + y2 = 9
Example
Solve the following system of equations
x y=-12
x-2y+14 = 0
Solution:
so x=-14+2y
and (-14+2y)y=-12
(-14+2y)y=-12
-14y+2y2=-12
2y2-14y+12=0
Example
Solve the following system of equations
x y=-12
x-2y+14 = 0
Solution:
y=6
y=1
2y2-14y+12=0
x=-14+2y
x=-14+2y
2(y2-7y+6)=0
2(y-6)(y-1)=0
=-14+2(6)
=-14+2(1)
(y-6)=0, (y-1)=0
=-14+12
=-14+2
y=6, y=1
=-2
=-12
(-2, 6) and (-12,1)
Systems of Linear
Equations in Three
Variables
Systems of Linear Equations in
Three Variables and Their Solutions
An equation such as x + 2y - 3z = 9 is called a linear equation in three
variables. In general, any equation of the form
Ax + By + Cz = D
where A, B, C, and D are real numbers such that A, B, and C are not all 0, is a
linear equation in the variables x, y, and z. The graph of this linear equation
in three variables is a plane in three-dimensional space.
Text Example
Show that the ordered triple (-1, 2, -2) is a solution of the system:
x + 2y – 3z = 9
2x – y + 2z = -8
-x + 3y – 4z = 15.
Solution Because -1 is the x-coordinate, 2 is the y-coordinate, and -2 is
the z-coordinate of (-1, 2, -2), we replace x by -1, y by 2, and z by -2 in each
of the three equations.
x + 2y – 3z = 9
?
-1 + 2(2) – 3(-2) = 9
?
-1 + 4 + 6 = 9
9 = 9 true
2x – y + 2z = -8
?
2(-1) – 2 + 2(-2) = -8
?
-2 – 2 – 4 = -8
-8 = -8 true
-x + 3y – 4z = 15
?
-(-1) + 3(2) – 4(-2) = 15
?
1 + 6 + 8 = 15
15 = 15 true
The ordered triple (-1, 2, -2) satisfies the three equations. Thus, the solution
set is {(-1, 2, -2)}.
Solving Linear Systems in Three
Variables by Eliminating Variables
• Reduce the system to two equations in two variables. This is
usually accomplished by taking two different pairs of equations
and using the addition method to eliminate the same variable
from each pair.
• Solve the resulting system of two equations in two variables
using addition or substitution. The result is an equation in one
variable that gives the value of that variable.
• Back-substitute the value of the variable found in step 2 into
either of the equations in two variables to find the value of the
second variable.
• Use the values of the two variables from steps 2 and 3 to find
the value of the third variable by back-substituting into one of
the original equations.
• Check the proposed solution in each of the original equations.
Text Example
Solve the system:
5x – 2y – 4z = 3
3x + 3y + 2z = -3
-2x + 5y + 3z = 3
Equation 1
Equation 2
Equation 3
Solution There are many ways to proceed. Because our initial goal is to reduce
the system to two equations in two variables, the central idea is to take two
different pairs of equations and eliminate the same variable from each pair.
Step 1 Reduce the system to two equations in two variables. We choose any
two equations and use the addition method to eliminate a variable. Let's
eliminate z from Equations 1 and 2. We do so by multiplying Equation 2 by 2.
Then we add equations.
Equation 1
Equation 2
5x – 2y – 4z = 3
3x + 3y + 2z = -3
No change.
5x – 2y – 4z = 3
6x + 6y + 4z = -6
Add: 11x + 4y
= -3
Multiply by 2.
Equation 4
Solution
Text Example cont.
Now we must eliminate the same variable from another pair of
equations. We can eliminate z from Equations 2 and 3. First, we multiply
Equation 2 by –3. Next, we multiply Equation 3 by 2. Finally, we add
equations.
Equation 2 3x + 3y + 2z = -3
Equation 3 -2x + 5y + 3z = 3
-9x – 9y – 6z = 9
-4x + 10y + 6z = 6
Add: -13x + y
= 15 Equation 5
Multiply by -3.
Multiply by 2.
Equations 4 and 5 give us a system of two equations in two variables.
Step 2 Solve the resulting system of two equations in two variables. We
choose any two equations and use the addition method to eliminate a variable.
Let's eliminate z from Equations 1 and 2. We do so by multiplying Equation 2
by 2. Then we add equations.
Equation 4 11x + 4y = -3
Equation 5 -13x + y = 15
No change.
11x + 4y = -3
52x – 4y = -60
Add: 63x
= -63
x
= -1
Multiply by -4.
Divide both sides by 63.
Text Example cont.
Solution
Step 3 Use back-substitution in one of the equations in two variables to
find the value of the second variable. We back-substitute -1 for x in either
Equation 4 or 5 to find the value of y.
-13x + y = 15
Equation 5
-13(-1) + y = 15
Substitute -1 for x.
13 + y = 15
Multiply.
y=2
Subtract 13 from both sides.
Step 4 Back-substitute the values found for two variables into one of the
original equations to find the value of the third variable. We can now use
any one of the original equations and back-substitute the values of x and y to
find the value for z. We will use Equation 2.
3x + 3y + 2z = -3
3(-l) + 3(2) + 2z = -3
Equation 2
Substitute -1 for x and 2 for y.
Solution
Text Example cont.
3 + 2z = -3
2z = -6
z = -3
Multiply and then add.
Subtract 3 from both sides.
Divide both sides by 2.
With x = -1, y = 2, and z = -3, the proposed solution is the ordered triple (1,2,-3).
Step 5 Check. Check the proposed solution, (-1, 2, -3), by substituting the
values for x, y, and z into each of the three original equations. These
substitutions yield three true statements. Thus, the solution set is {(-1, 2, -3)}.
Text Example
Solve the system:
x +
z = 8
x + y + 2z = 17
x + 2y + z = 16
Equation 1
Equation 2
Equation 3
Solution
Step 1 Reduce the system to two equations in two variables. Because
Equation 1 contains only x and z, we could eliminate y from Equations 2 and
3. This will give us two equations in x and z. To eliminate y from Equations 2
and 3, we multiply Equation 2 by -2 and add Equation 3.
Equation 2
Equation 3
x + y + 2z = 17
x + 2y + z = -3
Multiply by -2.
No change.
-2x – 2y – 4z = -34
x + 2y + z = 16
Add: -x
– 3z = -18
Equation 4
Equation 4 and the given Equation 1 provide us with a system of
two equations in two variables.
Text Example cont.
Solution
Step 2 Solve the resulting system of two equations in two variables. We
will solve Equations 1 and 4 for x and z.
x
-x
Add:
+ z
– 3z
-2z
z
= 8
= -18
= -10
= 5
Equation 1
Equation 4
Divide both sides by -2.
Step 3 Use back-substitution in one of the equations in two variables
to find the value of the second variable. To find x, we back-substitute 5
for z in either Equation 1 or 4. We will use Equation 1.
x + z = 8 Equation 1
x + 5 = 8 Substitute 5 for z.
Subtract 5 from both sides.
x=3
Text Example cont.
Solution
Step 4 Back-substitute the values found for two variables into one of
the original equations to find the value of the third variable. To find y,
we back-substitute 3 for x and 5 for z into Equation 2 or 3. We can't use
Equation 1 because y is missing in this equation. We will use Equation 2.
x + y + 2z = 17
3 + y + 2(5) = 17
y + 13 = 17
y=4
We found that z = 5, x = 3, and y = 4. Thus, the proposed solution is the
ordered triple (3, 4, 5).
Step 5 Check. Substituting 3 for x, 4 for y, and 5 for z into each of the
three original equations yields three true statements. Consequently, the
solution set is {(3,4,5)}.
Example
Solve the system:
x - 3y + z = 1
2x
-3z = -8
3x +8y +2z = 1
Solution:
Reduce the system to 2X2
-2(x - 3y + z = 1)
2x
-3z = -8
-3(x - 3y + z = 1)
3x +8y +2z = 1
-2x+6y-2z= -2
2x
-3z= -8
6y - 5z = -10
-3x +9y -3z = -3
3x +8y +2z = 1
17y - z = -2
Example cont.
• 6y - 5z = -10
-5(17y - z = -2)
6y - 5z = -10
-85y +5z = 10
-79y = 0
y=0
•6y - 5z = -10
17y - z = -2
6(0) - 5z = -10
-5z = -10
z=2
x - 3(0) + 2 = 1
x+2 = 1
x = -1
Solution: (-1, 0, 2)
Partial Fractions
Steps in Partial Fraction Decomposition
1.
2.
3.
4.
5.
6.
Set up the partial fraction decomposition with the
unknown constants A, B, C, etc., in the numerator of
the composition.
Multiply both sides of the resulting equation by the
least common denominator.
Simplify the right-hand side of the equations.
Write both sides in descending powers, equate
coefficients of like powers of x, and equate constant
terms.
Solve the resulting linear system for A, B, C, etc.
Substitute the values for A, B, C, etc., into the
equation in the first step.
Text Example
x  18
Find the partial fraction decomposition of
2
x(x

3)
Solution
Step 1 Set up the partial fraction decomposition with the unknown
constants. Because the linear factor x – 3 is repeated twice, we must
include one fraction with a constant numerator for each power of x – 3.
x  18
A
B
C


2 
2
x(x  3)
x x  3 (x  3)
Text Example cont.
Solution
Step 2 Multiply both sides of the resulting equation by the least common
denominator. We clear fractions, multiplying both sides by x(x – 3)2, the least
common denominator.
 x  18 

B
C 
2 A

 

x(x  3) 

2  x(x  3)
2
x(x  3) 
x x  3 (x  3) 
2
We use the distributive property on the right side.
A
B
C
2
2
x  18  x(x  3)   x(x  3) 
 x(x  3) 
2
x
x 3
(x  3)
2
Dividing out common factors in numerators and denominators, we obtain
x – 18 = A(x – 3)2 + Bx(x – 3) + Cx.
Text Example cont.
Solution
Step 3 Simplify the right side of the equation. Square x – 3. Then apply the
distributive property.
x – 18 = A(x2 – 6x + 9) + Bx(x – 3) + Cx
Square x – 3.
x – 18 = Ax2 – 6Ax + 9A + Bx2 – 3Bx + Cx
Apply the distributive property.
Step 4 Write both sides in descending powers, equate coefficients of like
powers of x, and equate constant terms. Square x – 3. Then apply the
distributive property.
x – 18 = Ax2 + Bx2 – 6Ax – 3Bx + Cx + 9A
Express both sides in the same form.
0x2 + 1x – 18 = (A + B)x2 + (-6A – 3B + C)x + 9A.
Text Example cont.
Solution
Step 5 Solve the resulting system for A, B, and C. Dividing both sides of
the last equation by 9, we obtain A = -2. Substituting –2 for A in the first
equation, A + B = 0, gives –2 + B = 0 or B = 2. We find C by substituting –2 for
A and 2 for B in the middle equation, -6A – 3B + C = 1. We obtain C = -5.
Step 6 Substitute values of A, B, and C and write the partial fraction
decomposition. With A = -2, B = 2, and C = -5, the required partial fraction
decomposition is
x  18
A
B
C
2
2
5




2 
2  
2
x(x  3)
x x  3 (x  3)
x x  3 (x  3)
Text Example
Find the partial fraction decomposition of
3x 2  17x  14
(x  2)(x 2  2x  4)
Solution
Step 1 Set up the partial fraction decomposition with the unknown
constants. We put a constant (A) over the linear factor and a linear
expression (Bx + C) over the prime quadratic factor.
3x 2  17x  14
A
Bx  C

 2
2
(x  2)(x  2x  4) x  2 x  2x  4
Text Example cont.
Solution
Step 2 Multiply both sides of the resulting equation by the least
common denominator. We clear fractions, multiplying both sides by (x - 2)
(x2 + 2x + 4), the least common denominator.
2

3x  17x  14 
2
(x  2)(x  2x  4)

2

(x  2)(x  2x  4) 

 A
Bx  C 
(x  2)(x  2x  4)
 2
x  2 x  2x  4 

2
We use the distributive property on the right side.
Dividing out common factors in numerators and denominators, we obtain
3x2 + 17x + 14 = A(x2 + 2x + 4) + (Bx + C)(x - 2).
Text Example cont.
Solution
Step 3 Simplify the right side side of the equation. We multiply on the
right side by distributing A over each term in parentheses and multiplying
(Bx + C)(x – 2) using the FOIL method.
3x2 + 17x + 14 = Ax2 + 2Ax + 4A + Bx2 – 2Bx + Cx – 2C.
Step 4 Write both sides in descending powers, equate coefficients of like
powers of x, and equate constant terms. The left side, 2x2 + 17x + 14, is in
descending powers of x. We write the right side in descending powers of x
3x2 + 17x + 14 = Ax2 + Bx2 + 2Ax – 2Bx + Cx + 4A – 2C.
And express both sides in the same form.
3x2 + 17x + 14 = (A + B)x2 + (2A – 2B + C) x + (4A – 2C).
Text Example cont.
Solution
Equating coefficients of like powers of x and constant terms results in the
following system of linear equations.
A+B=3
2A – 2B + C = 17
4A – 2C = 14
Step 5 Solve the resulting system for A, B, and C. Because the first
equation involves A and B, we can obtain another equation in A and B by
eliminating C from the second and third equations. Multiply the second
equation by 2 and add equations. Solving in this manner, we obtain A = 5,
B = -2, and C = 3.
Text Example cont.
Solution
Step 6 Substitute the values of A, B, and C and write the partial
fraction decomposition. With A = 5, B = -2, and C = 3, the required partial
fraction decomposition is
3x 2  17x  14
A
Bx  C

 2
2
(x  2)(x  2x  4) x  2 x  2x  4
5
2x  3

 2
x  2 x  2x  4
Example
• Find the partial fraction decomposition of
3x  1
x2  9
Solution
3x  1
3x  3

2
x  9 ( x  3)(x  3)
A
B
3x  3


x  3 x  3 ( x  3)(x  3)
A( x  3)  B( x  3)  3 x  3
Ax  3 A  Bx  3B  3 x  3
Ax  Bx  3 x, 3 A  3B  3
A  B  3, 3 A  3B  3
A  3  B, 3(3  B)  3B  3
A
B
3x  3


x  3 x  3 ( x  3)(x  3)
A( x  3)  B( x  3)  3x  3
A  3  B, 3(3  B)  3B  3
9  3B  3B  3
9  6B  3
 6 B  6
B 1
A  3  (1)  2
Example Cont.
• Find the partial fraction decomposition of
Solution
3x  1
2
x 9
3x  3
A
B


( x  3)(x  3) x  3 x  3
2
1


x 3 x 3
Systems of
Nonlinear Equations
in Two Variables
Systems of Nonlinear Equations
and Their Solutions
A system of two nonlinear equations in two variables contains at least one
equation that cannot be expressed in the form Ax + By = C. Here are two
examples:
x2 = 2y + 10
3x – y = 9
y = x2 + 3
x2 + y2 = 9
A solution to a nonlinear system in two variables is an ordered pair of real
numbers that satisfies all equations in the system. The solution set to the
system is the set of all such ordered pairs.
Text Example
Solve by the substitution method:
x–y=3
(x – 2)2 + (y + 3)2 = 9
The graph is a line.
The graph is a circle.
Solution Graphically, we are finding the intersection of a line and a circle
whose center is at (2, -3) and whose radius measures 2.
Step 1 Solve one of the equations for one variable in terms of the other.
We will solve for x in the linear equation - that is, the first equation. (We could
also solve for y.)
x–y=3
This is the first equation in the given system.
x=y+3
Add y to both sides.
Text Example cont.
Solution
Step 2 Substitute the expression from step 1 into the other equation. We
substitute y + 3 for x in the second equation.
x=y +3
( x – 2)2 + (y + 3)2 = 4
This gives an equation in one variable, namely
(y + 3 – 2)2 + (y + 3)2 = 4.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing one variable.
(y + 3 – 2)2 + (y + 3)2 = 4
This is the equation containing one variable.
(y + 1)2 + (y + 3 )2 = 4
Combine numerical terms in the first parentheses.
y2 + 2y + 1 + y2 + 6y + 9 = 4
Square each binomial.
2y2 + 8y + 10 = 4
Combine like terms on the left.
2y2 + 8y + 6 = 0
Subtract 4 from both sides and set the quadratic
equation equal to 0.
Solution
Text Example cont.
y2 + 4y + 3 = 0
(y + 3)(y + 1) = 0
y + 3 = 0 or y + 1 = 0
y = -3
or y = -1
Simplify by dividing both sides by 2.
Factor.
Set each factor equal to 0.
Solve for y.
Step 4 Back-substitute the obtained values into the
equation from step 1. Now that we have the y-coordinates of the solutions, we back-substitute -3 for y and -1 for
y in the equation x = y + 3.
If y = -3: x = -3 + 3 = 0, so (0, -3) is a solution.
If y = -1: x = -1 + 3 = 2, so (2, -1) is a solution.
Step 5 Check the proposed solution in both of the
system's given equations. Take a moment to show that
each ordered pair satisfies both equations. The solution
set of the given system is {(0, -3), (2, -1)}.
7
6
5
x – y =4 3
3
2
1
-5 -4 -3 -2 -1
-1
(2, -1)
1
2 3 4
5
6
7
-2
(0, -3) -3
-4
-5
-6
-7
(x – 2)2 + (y + 3)2 = 4
Text Example
Solve the system:
4x2 + y2 = 13
x2 + y2 = 10
Equation 1.
Equation 2.
Solution We can use the same steps that we did when we solved linear systems
by the addition method.
Step 1 Write both equations in the form Ax2 + By2 = C. Both equations
are already in this form, so we can skip this step.
Step 2 If necessary, multiply either equation or both equations by
appropriate numbers so that the sum of the x2-coefficients or the sum of
the y2-coefficients is 0. We can eliminate y by multiplying Equation 2 by -1.
4x2 + y2 = 13
x2 + y2 = 10
No change.
Multiply by -1.
4x2 + y2 = 13
-x2 – y2 = -10
Solution
Steps 3 and 4
Add.
Text Example cont.
Add equations and solve for the remaining variable.
4x2 + y2
-x2 – y2
3x2
x2
x
=
=
=
=
=
13
-10
3
1
+1
Step 5 Back-substitute and find the values for the other variables. We
must back-substitute each value of x into either one of the original equations.
Let's use x2 + y2 = 10, Equation 2. If x = 1,
12 + y2 = 10
Replace x with 1 in Equation 2.
y2 = 9
Subtract 1 from both sides.
y = ±3
Apply the square root method.
(1, 3) and (1, -3) are solutions. If x = -1,
(-1)2 + y2 = 10
Replace x with -1 in Equation 2.
y2 = 9
The steps are the same as before.
y = ±3
(-1, 3) and (-1, -3) are solutions.
Text Example cont.
Solution
Step 6 Check. Take a moment to show that each of the four ordered pairs
satisfies Equation 1 and Equation 2. The solution set of the given system is
{(1, 3), (1, -3), (-1, 3), (-1, -3)}.
7
4x2 + y2 = 13
6
(-1, 3)
5
4
(1, 3)
3
2
x2 + y2 = 10
1
-5 -4 -3 -2 -1
-1
1
2 3 4
-2
(-1, -3)
-3
-4
-5
-6
-7
(1, -3)
5
6
7
Text Example
Solve the system:
y = x2 + 3 Equation 1 (The graph is a parabola.)
x2 + y2 = 9
Equation 2 (The graph is a circle.)
Solution We could use substitution because Equation 1 has y expressed in
terms of x, but this would result in a fourth-degree equation. However, we
can rewrite Equation 1 by subtracting x2 from both sides and adding the
equations to eliminate the x2-terms.
Add.
-x2 + y
= 3
x2
+ y2 = 9
y + y2 = 12
Subtract x2 from both sides of Equation 1.
This is Equation 2.
Add the equations.
Solution
Text Example cont.
We now solve this quadratic equation.
y + y2 = 12
y2 + y – 12 = 0
Subtract 12 from both circles and get the quadratic
(y + 4)(y – 3) = 0
y + 4 = 0 or y – 3 = 0
y = -4
or
y=3
equation equal to 0.
Factor.
Set each factor equal to 0.
Solve for y.
To complete the solution, we must back-substitute each value of y into either
one of the original equations. We will use y = x2 + 3, Equation 1. First, we
substitute -4 for y.
-4 = x2 + 3
-7 = x2
Subtract 3 from both sides.
Text Example cont.
Solution
Because the square of a real number cannot be negative, the equation x2 = -7
does not have real-number solutions. Thus, we move on to our other value for
y, 3, and substitute this value into Equation 1.
y = x2 + 3
3 = x2 + 3
0 = x2
0=x
This is Equation 1.
7
6
Back-substitute 3 for y.
Subtract 3 from both sides.
y=
x2 +
3
(0, 3)
3
2
Solve for x.
We showed that if y = 3, then x = 0. Thus, (0, 3) is
the solution. Take a moment to show that (0, 3)
satisfies Equation 1 and Equation 2. The solution
set of the given system is {(0, 3)}.
5
4
1
-5 -4 -3 -2 -1
-1
1
2 3 4
5
6
7
-2
-3
-4
-5
-6
-7
x2 + y2 = 9
Example
Solve the following system of equations
x y=-12
x-2y+14 = 0
Solution:
so x=-14+2y
and (-14+2y)y=-12
(-14+2y)y=-12
-14y+2y2=-12
2y2-14y+12=0
Example
Solve the following system of equations
x y=-12
x-2y+14 = 0
Solution:
y=6
y=1
2y2-14y+12=0
x=-14+2y
x=-14+2y
2(y2-7y+6)=0
2(y-6)(y-1)=0
=-14+2(6)
=-14+2(1)
(y-6)=0, (y-1)=0
=-14+12
=-14+2
y=6, y=1
=-2
=-12
(-2, 6) and (-12,1)
Systems of
Inequalities
Graphing a Linear Inequality in Two Variables
1. Replace the inequality symbol with an equal sign
and graph the corresponding linear equation.
Draw a solid line if the original inequality
contains a < or > symbol. Draw a dashed line if
the original inequality contains a < or > symbol.
2. Choose a test point in one of the half-planes that
is not on the line. Substitute the coordinates of
the test point into the inequality
3. If a true statement results, shade the half-plane
containing this test point. If a false statement
results, shade the half-plane not containing this
test point.
Text Example
Graph: 3x – 5y < 15.
Solution
Step 1 Replace the inequality symbol by = and graph the linear
equation. We need to graph 3x – 5y = 15. We can use intercepts to graph this
line.
We set y = 0 to find
the x-intercept:
3x – 5y = 15
3x – 5 • 0 = 15
3x = 15
x=5
We set x = 0 to find
the y-intercept:
3x – 5y = 15
3 – 0 • 5y = 15
-5y = 15
y = -3
Solution
Text Example cont.
The x-intercept is 5, so the line passes through (5, 0). The y-intercept is -3, so
the line passes through (0, -3). The graph is indicated by a dashed line because
the inequality 3x – 5y < 15 contains a < symbol, rather than <. The graph of
the line is shown below.
5
4
3
2
1
-3 -2 -1
-1
1
2 3 4
5
6
7
-2
-3
-4
-5
3x – 5y = 15
Solution
Text Example cont.
Step 2 Choose a test point in one of the half-planes that is not on the
line. Substitute its coordinates into the inequality. The line 3x – 5y = 15
divides the plane into three parts – the line itself and two half-planes. The
points in one half-plane satisfy 3x – 5y > 15. The points in the other half-plane
satisfy 3x – 5y < 15. We need to find which half-plane is the solution. To do
so, we test a point from either half-plane. The origin, (0, 0), is the easiest point
to test.
3x – 5y < 15
This is the given inequality.
Is 3 • 0 – 5 • 0 < 15?
Test (0, 0) by substituting 0 for x and y.
0 – 0 < 15
0 < 15, true
Solution
Text Example cont.
Step 3 If a true statement results, shade the half-plane containing the
test point. Because 0 is less than 15, the test point (0, 0) is part of the solution
set. All the points on the same side of the line 3x - 5y = 15 as the point (0, 0)
are members of the solution set. The solution set is the half-plane that contains
the point (0, 0), indicated by shading this half-plane. The graph is shown
using green shading and a dashed blue line.
5
4
3
2
1
-3 -2 -1
-1
-2
-3
-4
-5
1
2 3 4
5
6
7
Text Example
Graph: x2 + y2 < 9.
Solution
Step 1 Replace the inequality symbol by = and graph the nonlinear
equation. We need to graph x2 + y2 = 9. The graph is a circle of radius 3 with
its center at the origin. The graph is shown below as a solid circle because
equality is included in the < symbol.
5
4
3
2
1
-5 -4 -3 -2
-1
-1
-2
-3
-4
-5
1
2
3 4
5
Text Example cont.
Solution
Step 2 Choose a test point in one of the regions that is not on the circle.
Substitute its coordinates into the inequality. The circle divides the plane
into three parts – the circle itself, the region inside the circle, and the region
outside the circle. We need to determine whether the region inside or outside
the circle is the solution. To do so, we will use the test point (0, 0) from inside
the circle.
x2 + y2 < 9
This is the given inequality.
Is 02 + 02 < 9?
Test (0, 0) by substituting 0 for x and 0 for y.
0+0<9
0 < 9, true
Text Example cont.
Solution
Step 3 If a true statement results, shade the region containing the test
point. The true statement tells us that all the points inside the circle satisfy
x2 + y2 < 9. The graph is shown using green shading and a solid blue circle.
5
4
3
2
1
-5 -4 -3 -2
-1
-1
-2
-3
-4
-5
1
2
3 4
5
Example
• Graph:
2x+y<8
x+y>4
• Solution:
• First graph
2x+y=8
x+y=4
10
8
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Example cont.
• Graph:
2x+y<8
x+y>4
• Solution:
• Test point (0,0)
2x+y<8
2(0) + 0 = 0 <8 true
Example cont.
• Graph:
2x+y<8
x+y>4
• Solution:
• Test point (0,0)
• x+y>4
0+0=0 >4 false
Example cont.
• Graph:
2x+y<8
x+y>4
• Solution:
Text Example
Graph the solution set:
x–y<2
-2 < x < 4
y<3
Solution We begin by graphing x - y < 2, the first given inequality. The line
x – y = 2 has an x-intercept of 2 and a y-intercept of -2. The test point (0, 0)
makes the inequality x – y < 2 true, and its graph is shown below.
5
4
3
2
1
-5 -4 -3 -2
-1
-1
-2
-3
-4
-5
1
2
3 4
5
Solution
Text Example cont.
Now let's consider the second given inequality -2 < x < 4. Replacing
the inequality symbols by =, we obtain x = -2 and x = 4, graphed as vertical
lines. The line of x = 4 is not included. Using (0, 0) as a test point and
substituting the x-coordinate, 0, into -2 < x < 4, we obtain the true statement 2 < 0 < 4. We therefore shade the region between the vertical lines.
5
4
3
2
1
-5 -4 -3
-1
-1
-2
-3
-4
-5
1
2
3
5
Solution
Text Example cont.
Finally, let's consider the third given inequality, y < 3. Replacing the
inequality symbol by =, we obtain y = 3, which graphs as a horizontal line.
Because (0, 0) satisfies y < 3 (0 < 3 is true), the graph consists of the halfplane below the line y = 3.
5
4
3
2
1
-5 -4 -3
-1
-1
-2
-3
-4
-5
1
2
3
5
Linear Programming
Objective Functions in Linear
Programming
Many problems involve quantities that must be
maximized or minimized. Businesses are
interested in maximizing profit. An operation in
which bottled water and medical kits are shipped
to earthquake victims needs to maximize the
number of victims helped by this shipment. An
objective function is an algebraic expression in
two or more variables describing a quantity that
must be maximized or minimized.
Text Example
Bottled water and medical supplies are to be shipped to victims of an earthquake by plane. Each container of bottled water will serve 10 people and each
medical kit will aid 6 people. If x represents the number of bottles of water to
be shipped and y represents the number of medical kits, write the objective
function that describes the number of people that can be helped.
Solution Because each bottle of water serves 10 people and each medical kit
aids 6 people, we have
The number of
People helped
is
10 times the number
of bottles of water
plus
6 times the number
of medical kits.
=
10x
+
6y.
Using z to represent the objective function, we have
z = 10x + 6y.
Unlike the functions that we have seen so far, the objective function is an
equation in three variables. For a value of x and a value of y, there is one and
only one value of z. Thus, z is a function of x and y.
Text Example
Planes can carry a total volume for supplies that does not exceed 6000 cubic
feet. Each water bottle is 1 cubic foot and each medical kit also has a volume of
1 cubic foot. With x still representing the number of water bottles and y the
number of medical kits, write an inequality that describes this second constraint.
Solution Because each plane can carry a volume of supplies that does not
exceed 6000 cubic feet, we have
The total volume of
the water bottles
lx
plus
+
Each bottle is
1 cubic foot.
The total volume of
the medical kits
ly
must be less than
or equal to
<
6000
cubic feet.
6000.
Each kit is
1 cubic foot.
The plane's volume constraint is described by the inequality x + y < 6000.
Solving a Linear Programming Problem
•
Let z=ax + by be an objective function that
depends on x and y. Furthermore, z is subject to
a number of constraints on x and y. If a
maximum or minimum value of z exists, it can
be determined as follows:
1. Graph the system of inequalities representing the
constraints
2. Find the value of the objective function at each
corner, or vertex, of the graphed region. The
maximum and minimum of the objective
function occur at one or more of the corner
points.
Example
•
Given the objective function and a system of
linear inequalities. Objective function z = 3x+2y
Constraints x>0, y>0, 2x+y<8, x+y>4
a. Graph the system of inequalities representing the
constraints.
b. Find the value of the objective function at each
corner of the graphed region.
c. Use the values that you found in the prior step to
determine the maximum value of the objective
function and the values of x and y for which the
maximum occurs.
Example cont.
• Graph x=0, y=0
2x+y=8
x+y=4
10
8
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Example cont.
• x>0
Example cont.
• y>0
Example cont.
• x>0
• y>0
Example cont.
• 2x+y<8
Example cont.
• x+y>4
Example cont.
• Corners
• (4,0), (0,4), (0,8)
Example cont.
•
•
•
•
•
Objective function z = 3x+2y
(4,0): z = 3(4) + 2 (0) = 12
(0,4): z = 3(0) + 2(4) = 8
(0,8): z = 3(0) + 2(8) = 16
The maximum value of the objective function is
16 and it occurs when x = 0 and y = 8
Text Example
Find the maximum value of the objective function z = 2x + y subject to the
constraints: x > 0, y > 0, x + 2y < 5, x – y < 2.
Solution We begin by graphing the region in quadrant I
(x > 0, y > 0) formed by the constraints.
5
4
(0, 2.5)
2
(0, 0)
-3 -2 -1
Now we evaluate the objective function at the four vertices
of this region.
Objective Function: z = 2x + y
At (0, 0):
z=2•0+0=0
At (2, 0):
z=2•2+0=4
The maximum value of z.
At (3, 1):
z=2•3+1=7
At (0, 2.5):
z = 2 • 0 + 2.5 = 2.5
x + 2y = 5
(3, 1)
1
-1
-2
-3
1
3 4
5
6
(2, 0)
x–y=2
-4
-5
Thus, the maximum value of z is 7, and this occurs when x = 3 and y = 1.
7