Download Section 20-1: Solving Systems of Linear Equations and Inequalities

Section 20-1: Solving Systems of Linear Equations and Inequalities Graphically
Learning Outcome 1
Solve the system of equations by graphing: x + 2y = −4, 3x + y = 3
Find the x- and y-intercepts for x + 2y = −4
x-intercept: y = 0
x + 2(0) = −4
x = −4
(−4,0)
y-intercept: x = 0
x + 2y = −4
0 + 2y = −4
2y = −4
y = −2
(0,−2)
Write the equation 3x + y = 3 in slope-intercept form.
3x + y = 3
y = −3x + 3
m = −3 b = 3
The solution of the system is (2, –3) or x = 2 and y = –3.
Learning Outcome 2
Graph the system of inequalities and show the solution by shading: x + y ≥ 5, x − y < 3
Graph each equation and shade the appropriate side for the inequality. Find the x- and y-intercepts
for each equation.
x+y=5
x−y=3
(0,5) (5,0)
(3,0) (0,−3)
Graph the lines. Then test points to determine which side to shade for the inequality x + y ≥ 5. Test
the point (0,0).
0+0≥5?
0≥5
False
(0,0) makes a false statement, so shade the side of the line that does not include the point.
Determine which side to shade for the inequality x − y < 3. Test the point (0,0).
0−0<3?
0<3
True
(0,0) makes a true statement, so shade the side of the line that includes the point. The solution is the
portion of the graph that contains the overlapped shading. It includes the portion of the y-axis above
(0,5).
Section 20-2: Solving Systems of Equations Using the Addition Method
Learning Outcome 1
Solve the system of equations by addition: x + 2y = 4, 2x − 2y = 5
x + 2y = 4
Add the two equations.
2x − 2y = 5
3x = 9
Solve for x.
x=3
x + 2y = 4
Substitute to find y.
3 + 2y = 4
2y = 1
y = 1/2
(3,1/2)
Solution.
Learning Outcome 2
Solve the system by addition: 2x − y = 5, x + 3y = −1
6x − 3y = 15
Multiply the first equation by 3 to form opposites.
x + 3y = −1
7x = 14
x=2
2x − y = 5
Substitute in the first equation to find y.
2(2) − y = 5
4−y=5
Sort terms.
−y = 5 − 4
Combine terms.
−y = 1
Multiply by −1.
y = −1
(2, −1)
Solution.
Learning Outcome 3
Solve the system: x + 2y = −3, −x − 2y = 5
x + 2y = −3
Add the equations.
−x − 2y = 5
0+0 = 2
0 = 2
False.
When both variables add to zero and the resulting equation is false, the lines that graph the equation
are parallel and the system has no solution.
Solve the system: 5x − y = 8, −5x + y = −8
5x − y = 8
Add the equations.
−5x + y = −8
0 + 0=0
0=0
True.
When both variables add to zero and the resulting equation is true, the lines coincide and the system
has an infinite number of solutions.
Section 20-3: Solving Systems of Linear Equations Using the Substitution
Method
Learning Outcome 1
Use substitution to solve the system: x + 2y = 3, 2x − y = 11
x + 2y = 3
Solve the first equation for x.
x = −2y + 3
Substitute −2y + 3 in second equation.
2(−2y + 3) − y = 11
Solve for y.
−4y + 6 − y = 11
−5y + 6 = 11
−5y = 5
y = −1
x = −2y + 3
Substitute in solved equation above or one of the original equations.
x = −2(−1)+ 3
x=2+3
x=5
(5,−1)
Solution.
Section 20-4: Problem Solving Using Systems of Equations
Learning Outcome 1
Three pounds of cashews and 4 pounds of pretzels cost $6.73. The difference in the cost of 5 pounds
of cashews and 3 pounds of pretzels is $3.58. Write two equations and solve the system to find the
cost of 1 pound of cashews and the cost of 1 pound of pretzels.
3c + 4p = $6.73
Multiply by 3.
5c − 3p = $3.58
Multiply by 4.
9c + 12p = 20.19
20c − 12p = 14.32
29c = 34.51
29c 34.51
=
29
29
c = 1.19
3c + 4p = 6.73
3(1.19) + 4p = 6.73
3.57 + 4p = 6.73
4p = 3.16
Substitute to find p.
4 p 3.16
=
4
4
p = 0.79
One pound of cashews costs $1.19 and one pound of pretzels costs $0.79.