9.2 Solving Systems of Linear Equations by Substitution; Applications 1. Solve systems of linear equations using substitution. 2.

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Transcript 9.2 Solving Systems of Linear Equations by Substitution; Applications 1. Solve systems of linear equations using substitution. 2.

9.2
Solving Systems of Linear Equations by
Substitution; Applications
1. Solve systems of linear equations using substitution.
2. Solve applications involving two unknowns using a
system of equations.
Solve the system of equations using substitution.
4 x  5 y  8

x  1 y
Step 1: Isolate a variable in one equation.
The second equation is already solved for x.
Step 2: Substitute 1 – y for x in the first equation.
4x  5 y  8
4(1  y )  5 y  8
4 x  5 y  8

x  1 y
Step 3: Solve for y.
4(1  y)  5 y  8
4  4y  5y  8
4 y 8
y4
Step 4: Solve for x by substituting 4 for y in one of
the original equations.
x  1 y
x  1 4
x  3
The solution is (3, 4).
Solving Systems of Two Equations Using Substitution
1. Isolate one of the variables in one of the
equations.
2. Substitute the expression you found in step 1
for that variable in the other equation.
3. Solve this new equation. (It will now have only
one variable.)
4. Substitute this solution for the variable in one of
the original equations and solve for the other
variable. The solution is an ordered pair.
5. Check the solution in the original equations.
Solve the system of equations using substitution.
2 x  y  7

 x  y  1
Step 1: Isolate a variable in one equation. Use either
equation. (But choose “smart”.)
2x + y = 7
y = 7 – 2x
Step 2: Substitute y = 7 – 2x for y in the second
equation.
x  y  1
x  (7  2 x)  1
Step 3: Solve for x.
x  (7  2 x)  1
2 x  y  7

 x  y  1
x  7  2 x  1
3x  7  1
3x  6
x2
Step 4: Solve for y by substituting 2 for x in either
of the original equations.
2x  y  7
2(2)  y  7
4 y  7
y3
The solution is (2, 3).
Solve the system.
x  y  4

2 x  y  6
a) (10, 3)
b) (1, 4)
c) (10, 14)
d) (3, 6)
9.2
Copyright © 2011 Pearson Education, Inc.
Slide 5- 7
Solve the system.
x  y  4

2 x  y  6
a) (10, 3)
b) (1, 4)
c) (10, 14)
d) (3, 6)
9.2
Copyright © 2011 Pearson Education, Inc.
Slide 5- 8
Solve the system.
 x  3 y  16

7 x  4 y  10
a) (2, 3)
b) (2, 6)
c) (2, 6)
d) (3, 2)
9.2
Copyright © 2011 Pearson Education, Inc.
Slide 5- 9
Solve the system.
 x  3 y  16

7 x  4 y  10
a) (2, 3)
b) (2, 6)
c) (2, 6)
d) (3, 2)
9.2
Copyright © 2011 Pearson Education, Inc.
Slide 5- 10
Inconsistent Systems of Equations
The system has no solution because the graphs of
the equations are parallel lines.
You will get a false statement such as 3 = 4.
Consistent Systems with Dependent Equations
The system has an infinite number of solutions
because the graphs of the equations are the same
line.
You will get a true statement such as 8 = 8.
Copyright © 2011 Pearson Education, Inc.
Solve the system of equations using substitution.
3x  y  4

 y  3x  5
Substitute y = 3x – 5 into the first equation.
3x  y  4
3x  (3x  5)  4
3 x  3 x  5  4
5  4
False statement.
The system is inconsistent and has no solution.
Solve the system of equations using substitution.
 y  4  3x

6 x  2 y  8
Substitute y = 4 – 3x into the second equation.
6 x  2 y  8
6 x  2(4  3x)  8
6 x  8  6 x  8
6 x  6 x  8  8
8  8
True statement.
The system is consistent with dependent equations.
There are an infinite number of solutions.