Transcript Section 7.1, Example 1

example 1
Solve the system
Solution by Elimination
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
Chapter 7.1
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
1. If necessary, interchange two equations or use multiplication to make the
coefficient of x in the first equation a 1.
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
E1  E2
x  y  2 z  3
2 x  3 y  z  1
3x  y  z  9
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
2 x  3 y  z  1
3x  y  z  9
-2 R1 + R2  R2
 2x  2 y  4z  6
2 x  3 y  z  1
x  y  2 z  3
 y  3z  5
3x  y  z  9
 y  3z  5
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
2 x  3 y  z  1
3x  y  z  9
-2 E1 + E2  E2
 2x  2 y  4z  6
2 x  3 y  z  1
x  y  2 z  3
 y  3z  5
3x  y  z  9
 y  3z  5
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
2 x  3 y  z  1
3x  y  z  9
-2 E1 + E2  E2
 2x  2 y  4z  6
2 x  3 y  z  1
x  y  2 z  3
 y  3z  5
3x  y  z  9
 y  3z  5
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
 y  3z  5
3x  y  z  9
-3 R1 + R3  R3
3 x  3 y  6 z  9
3x  y  z  9
x  y  2 z  3
 y  3z  5
4 y  7 z  18
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
 y  3z  5
3x  y  z  9
-3 E1 + E3  E3
3 x  3 y  6 z  9
3x  y  z  9
x  y  2 z  3
 y  3z  5
4 y  7 z  18
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
 y  3z  5
3x  y  z  9
-3 E1 + E3  E3
3 x  3 y  6 z  9
3x  y  z  9
x  y  2 z  3
 y  3z  5
4 y  7 z  18
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
2. Add a multiple of the first equation to each of the following equations so
that the coefficients of x in the second and third equations become 0.
x  y  2 z  3
 y  3z  5
3x  y  z  9
-3 E1 + E3  E3
3 x  3 y  6 z  9
3x  y  z  9
x  y  2 z  3
 y  3z  5
4 y  7 z  18
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
3. Multiply (or divide) both sides of the second equation by a number that
makes the coefficient of y in the second equation equal to 1.
x  y  2 z  3
 y  3z  5
4 y  7 z  18
-1 R2  R2
x  y  2 z  3
y  3 z  5
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
3. Multiply (or divide) both sides of the second equation by a number that
makes the coefficient of y in the second equation equal to 1.
x  y  2 z  3
 y  3z  5
4 y  7 z  18
-1 E2  E2
x  y  2 z  3
y  3 z  5
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
3. Multiply (or divide) both sides of the second equation by a number that
makes the coefficient of y in the second equation equal to 1.
x  y  2 z  3
 y  3z  5
4 y  7 z  18
-1 E2  E2
x  y  2 z  3
y  3 z  5
4 y  7 z  18
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
4. Add a multiple of the (new) second equation to the (new) third equation
so that the coefficient of y in the newest third equation becomes 0.
x  y  2 z  3
y  3 z  5
4 y  7 z  18
-4 R2 + R3  R3
4 y  12 z  20
4 y  7 z  18
x  y  2 z  3
y  3 z  5
 19 z  38
 19 z  38
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
4. Add a multiple of the (new) second equation to the (new) third equation
so that the coefficient of y in the newest third equation becomes 0.
x  y  2 z  3
y  3 z  5
4 y  7 z  18
-4 E2 + E3  E3
4 y  12 z  20
4 y  7 z  18
x  y  2 z  3
y  3 z  5
 19 z  38
 19 z  38
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
4. Add a multiple of the (new) second equation to the (new) third equation
so that the coefficient of y in the newest third equation becomes 0.
x  y  2 z  3
y  3 z  5
4 y  7 z  18
-4 E2 + E3  E3
4 y  12 z  20
4 y  7 z  18
x  y  2 z  3
y  3 z  5
 19 z  38
 19 z  38
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
4. Add a multiple of the (new) second equation to the (new) third equation
so that the coefficient of y in the newest third equation becomes 0.
x  y  2 z  3
y  3 z  5
4 y  7 z  18
-4 E2 + E3  E3
4 y  12 z  20
4 y  7 z  18
x  y  2 z  3
y  3 z  5
 19 z  38
 19 z  38
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
5. Multiply (or divide) both sides of the third equation by a number that
makes the coefficient of z in the third equation equal to 1. This gives the
solution for z in the system of equations.
x  y  2 z  3
y  3 z  5
 19 z  38
 191 E3  E3
x  y  2 z  3
y  3 z  5
z  2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
5. Multiply (or divide) both sides of the third equation by a number that
makes the coefficient of z in the third equation equal to 1. This gives the
solution for z in the system of equations.
x  y  2 z  3
y  3 z  5
 19 z  38
 191 E3  E3
x  y  2 z  3
y  3 z  5
z  2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
6. Use the solution for z to solve for y in the second equation. Then
substitute values for y and z to solve for x in the first equation.
x  y  2 z  3
y  3 z  5
z  2
y  3  2   5
x  1  2  2   3
y  6  5
x  5  3
y 1
x2
2009 PBLPathways
Solve the system
2 x  3 y  z  1
x  y  2 z  3
3x  y  z  9
Does the solution solve the system?
2  2   3 1   2   1
2  1  2  2   3
3  2   1   2   9
2009 PBLPathways