Download 6. Systems of Equations and Inequalities

6. Systems of Equations
and Inequalities
6.1. LINEAR AND NONLINEAR SYSTEMS OF
EQUATIONS
What You Should Learn
• Use the method of substitution to solve systems of
linear equations in two variables.
• Use the method of substitution to solve systems of
nonlinear equations in two variables.
• Use a graphical approach to solve systems of
equations in two variables.
• Use systems of equations to model and solve real-life
problems.
Introduction
Most problems have involved either a function of one
variable or a single equation in two variables.
However, many problems in science, business, and engineering
involve two or more equations in two or more variables.
To solve such problems, you need to find solutions of a system
of equations.
Introduction
Here is an example of a system of two equations in two unknowns.
2x + y = 5
3x – 2y = 4
Equation 1
Equation 2
A solution of this system is an ordered pair that satisfies each
equation in the system.
Finding the set of all solutions is called solving the system of
equations.
Introduction
For instance, the ordered pair (2, 1) is a solution of this system. To check
this, you can substitute 2 for x and 1 for y in each equation.
Check (2, 1) in Equation 1 and Equation 2:
2x + y = 5
2(2) + 1 ≟ 5
Write Equation 1.
Substitute 2 for x and 1 for y.
•
4+1=5
3x – 2y = 4
3(2) – 2(1) ≟ 4
6–2=4
Solution checks in Equation 1.
Write Equation 2.
Substitute 2 for x and 1 for y.
Solution checks in Equation 2.
The Method of Substitution
We will study four ways to solve systems of equations,
beginning with the method of substitution.
Method
Type of System
1. Substitution
Linear or nonlinear, two variables
2. Graphical method
Linear or nonlinear, two variables
3. Elimination
Linear, two variables
4. Gaussian elimination
Linear, three or more variables
The Method of Substitution
Example 1 – Solving a System of Equations by
Substitution
Solve the system of equations.
x+y=4
Equation 1
x–y=2
Equation 2
Example 1 – Solution
Begin by solving for y in Equation 1.
y=4–x
Solve for y in Equation 1.
Next, substitute this expression for y into Equation 2 and
solve the resulting single variable equation for x.
x–y=2
Write Equation 2.
Example 1 – Solution
x – (4 – x) = 2
cont’d
Substitute 4 – x for y.
x–4+x=2
Distributive Property
2x = 6
Combine like terms.
x=3
Divide each side by 2.
Finally, you can solve for y by back-substituting x = 3 into
the equation y = 4 – x, to obtain
y=4–x
Write revised Equation 1.
y=4–3
Substitute 3 for x.
Example 1 – Solution
y = 1.
cont’d
Solve for y.
The solution is the ordered pair (3, 1). You can check this
solution as follows.
Check:
Substitute (3, 1) into Equation 1:
x+y=4
3+1≟4
4=4
Write Equation 1.
Substitute for x and y.
Solution checks in Equation 1.
Example 1 – Solution
cont’d
Substitute (3, 1) into Equation 2:
x–y=2
3–1≟2
2=2
Write Equation 2.
Substitute for x and y.
Solution checks in Equation 2.
Because (3, 1) satisfies both equations in the system, it is a
solution of the system of equations.
The Method of Substitution
The term back-substitution implies that you work
backwards.
First you solve for one of the variables, and then you
substitute that value back into one of the equations in the
system to find the value of the other variable.
Nonlinear Systems of Equations
The equations in Example 1 is linear.
The method of substitution can also be used to solve systems in
which one or both of the equations are nonlinear.
Example 3 – Substitution: Two-Solution Case
Solve the system of equations.
3x2 + 4x – y = 7
Equation 1
2x – y = –1
Equation 2
Example 3 – Solution
Begin by solving for y in Equation 2 to obtain y = 2x + 1.
Next, substitute this expression for y into Equation 1 and
solve for x.
3x2 + 4x – (2x + 1) = 7
3x2 + 2x – 1 = 7
Substitute 2x + 1 for y in Equation 1.
Simplify.
Example 3 – Solution
3x2 + 2x – 8 = 0
(3x – 4)(x + 2) = 0
cont’d
Write in general form.
Factor.
Solve for x.
Back-substituting these values of x to solve for the
corresponding values of y produces the solutions
and
Graphical Approach to Finding Solutions
A system of two equations in two unknowns can have
exactly one solution, more than one solution, or no solution.
By using a graphical method, you can gain insight about
the number of solutions and the location(s) of the
solution(s) of a system of equations by graphing each of
the equations in the same coordinate plane.
The solutions of the system correspond to the points of
intersection of the graphs.
Graphical Approach to Finding Solutions
For instance, the two equations in Figure 6.1 graph as two
lines with a single point of intersection; the two equations in
Figure 6.2 graph as a parabola and a line with two points of
intersection; and the two equations in Figure 6.3 graph as a
line and a parabola that have no points of intersection.
One intersection point
Two intersection points
Figure 6.1
Figure 6.2
No intersection points
Figure 6.3
𝑥
2
The above graphs indicate that the system of equations with + 3𝑦 =
1 and 𝑥 − 𝑦 = 2 has one solution and the solution is the intersection
point (2,0). The system of equations with 𝑦 = 𝑥 2 − 𝑥 − 1 and 𝑦 = 𝑥 −
1 has two solutions (2,1) and (0,-1). And the system of equations −𝑥 +
𝑦 = 4 and 𝑥 2 + 𝑦 = 3 has no solution because the graphs never meet.
Example 5 – Solving a System of Equations Graphically
Solve the system of equations.
y = ln x
x+y=1
Equation 1
Equation 2
Solution:
Sketch the graphs of the two
equations. From the graphs of
these equations, it is clear that
there is only one point of
intersection and that (1, 0) is the
solution point (see Figure 6.4).
Figure 6.4
Example 5 – Solution
You can check this solution as follows.
Check (1, 0) in Equation 1:
y = ln x
Write Equation 1.
0 = ln 1
Substitute for x and y.
0=0
Solution checks in Equation 1.
Check (1, 0) in Equation 2:
x+y=1
1+0=1
1=1
Write Equation 2.
Substitute for x and y.
Solution checks in Equation 2.
cont’d
Graphical Approach to Finding Solutions
Example 5 shows the value of a graphical approach to
solving systems of equations in two variables.
Notice what would happen if you tried only the substitution
method in Example 5.
You would obtain the equation x + ln x = 1. It would be
difficult to solve this equation for x using standard algebraic
techniques.
Applications
The total cost C of producing x units of a product typically
has two components—the initial cost and the cost per unit.
When enough units have been sold so that the total
revenue R equals the total cost C, the sales are said to
have reached the break-even point.
You will find that the break-even point corresponds to the
point of intersection of the cost and revenue curves.
Example 6 – Break-Even Analysis
A shoe company invests $300,000 in equipment to produce a
new line of athletic footwear. Each pair of shoes costs $5 to
produce and is sold for $60. How many pairs of shoes must
be sold before the business breaks even?
Solution:
The total cost of producing x units is
C = 5x + 300,000.
Equation 1
Example 6 – Solution
cont’d
The revenue obtained by selling x units is
R = 60x.
Equation 2
Because the break-even point occurs when R = C, you
have C = 60x, and the system of equations to solve is
C = 5x + 300,000
C = 60x
.
Example 6 – Solution
cont’d
Solve by substitution.
60x = 5x + 300,000
Substitute 60x for C in Equation 1.
55x = 300,000
Subtract 5x from each side.
x ≈ 5455
Divide each side by 55.
So, the company must sell about 5455 pairs of shoes to
break even.
Applications
Another way to view the solution in Example 6 is to
consider the profit function
P = R – C.
The break-even point occurs when the profit is 0, which is
the same as saying that R = C.