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8.1 Solving Systems of Linear
Equations by Graphing
• To solve by graphing, graph both linear equations.
This gives an approximate solution. Algebraic
methods are more exact (next 2 sections).
• If the graphs intersect at one point the system is
consistent and the equations are independent.
8.1 Solving Systems of Linear
Equations by Graphing
• If the graphs are parallel lines, there is no solution and the
solution set is . The system is inconsistent.
• If the graphs represent the same line, there are an infinite
number of solutions. The equations are dependent.
8.2 Solving Systems of Linear
Equations by Substitution
•
Solving by substitution:
1. Solve for a variable
2. Substitute for that variable in the other
equation
3. Solve this equation for the remaining variable
4. Put your solution back into either of the
original equations to solve for the other
variable
5. Check your solution with the other equation
8.2 Solving Systems of Linear
Equations by Substitution
• Example:
2x  y  7
3 x  y  13
From the first equation we get y=2x-7, so
substituting into the second equation:
3 x  2 x  7  13
5 x  7  13  5 x  20  x  4
24  y  7  8 y  7
 y  1  y  1
8.2 Solving Systems of Linear
Equations by Substitution
• If when using substitution both variables drop out
and you get something like: 10=6
The system inconsistent and there is no solution
(parallel lines)
• If when using substitution both variables drop out
and you get something like: 10=10
The system dependent and every solution of one
line is also on the other (same lines)
8.3 Solving Systems of Linear
Equations by Elimination
•
Solving systems of equations by elimination:
1. Write equations in standard form (variables line up)
2. Multiply one of the equations to get coefficients of
one of the variables to be opposites
3. Add (or subtract) equations – so that one variable
drops out
4. Solve for the remaining variable.
5. Plug you solution back into one of the original
equations and solve for the other variable.
8.3 Solving Systems of Linear
Equations by Elimination
• Example:
2x  3y  5
4 x  y  17
• Multiply the second equation by 3 to get:
2x  3y  5
12 x  3 y  51
• Adding equations you get:
14 x  56  x  4
4  4  y  17  y  17  16  1
8.3 Solving Systems of Linear
Equations by Elimination
• If when using elimination both variables drop out
and you get something like: 10=6
The system inconsistent and there is no solution
(parallel lines)
• If when using elimination both variables drop out
and you get something like: 10=10
The system dependent and every solution of one
line is also on the other (same lines)
8.4 Linear Systems of Equations in
Three Variables
• Linear system of equation in 3 variables:
Ax  By  Cz  D
Ex  Fy  Gz  H
Ix  Jy  Kz  L
• Example:
4x  8 y  z  2
x  7 y  3z  14
2x  3y  2z  3
8.4 Linear Systems of Equations in
Three Variables
•
Graphs of linear systems in 3 variables:
1. Single point (3 planes intersect at a point)
2. Line (3 planes intersect at a line)
3. No solution (all 3 equations are parallel
planes)
4. Plane (all 3 equations are the same plane)
8.4 Linear Systems of Equations in
Three Variables
•
Solving linear systems in 3 variables:
1. Eliminate a variable using any 2 equations
2. Eliminate the same variable using 2 other
equations
3. Eliminate a different variable from the
equations obtained from (1) and (2)
8.4 Linear Systems of Equations in
Three Variables
•
Solving linear systems in 3 variables:
4. Use the solution from (3) to substitute into 2
of the equations. Eliminate one variable to
find a second value.
5. Use the values of the 2 variables to find the
value of the third variable.
6. Check the solution in all original equations.
8.5 Applications of Linear
Systems of Equations
•
Solving an applied problem by writing a system
of equations:
1. Determine what you are to find – assign variables
2. Draw a diagram, figure or make a chart of
information.
3. Write the system of equations
4. Solve the system using substitution or elimination
5. Answer the question from the problem.
8.5 Applications of Linear
Systems of Equations
• Mixture problem: How many ounces of a 5%
solution must be added to a 20% solution to get 10
ounces of 12.5% solution.
Let x = # ounces of 5% solution
Let y = # ounces of 20% solution
8.5 Applications of Linear
Systems of Equations
• Solution to mixture problem in 2 variables:
x  y  10
5% x  20% y  12.5%  10
.05 x  .210  x   .12510
.05 x  2.0  .2 x  1.25
 .15 x  1.25  2.0  .75
x  5  y  10  5  5