Transcript Solve linear systems by substitution.

4.2 Solving Systems of Linear Equations by Substitution
1
Solve linear systems by substitution.
2
Solve special systems by substitution.
3
Solve linear systems with fractions and decimals by substitution.
Objective 1
Solve linear systems by
substitution.
Slide 4.2-3
Solve linear systems by substitution.
Graphing to solve a system of equations has a serious drawback. It is
difficult to find an accurate solution, such as  1 ,  5  , from a graph.
6
3
One algebraic method for solving a system of equations is the
substitution method.
This method is particularly useful for solving systems in which one
equation is already solved, or can be solved quickly, for one of the
variables.
Slide 4.2-4
EXAMPLE 1 Using the Substitution Method
Solve the system by the substitution method.
2 x  7 y  12
x  2 y
Solution:
2  2 y   7 y  12
4 y  7 y  12
3 y 12

3
3
y  4
x  2 y
x  2  4
x 8
8, 4
The solution set found by the substitution method will be the same as the
solution found by graphing. The solution set is the same; only the method
is different. A system is not completely solved until values for both x
and y are found.
Slide 4.2-5
Solve linear systems by substitution. (cont’d)
Solving a Linear System by Substitution
Step 1: Solve one equation for either variable. If one of the
variables has coefficient 1 or −1, choose it, since it usually
makes the substitution method easier.
Step 2: Substitute for that variable in the other equation. The
result should be an equation with just one variable.
Step 3: Solve the equation from Step 2.
Step 4: Substitute the result from Step 3 into the equation
from Step 1 to find the value of the other variable.
Step 5: Check the solution in both of the original equations. Then
write the solution set.
Slide 4.2-6
EXAMPLE 2 Using the Substitution Method
Solve the system by the substitution method.
2 x  7 y  12
x  3 2y
Solution:
2  3  2 y   7 y  12
6  4 y  7 y  12
6  3 y  6  12  6
3 y 18

3
3
y  6
x  3  2  6
x  3 12
x  15
15, 6
Be careful when you write the ordered-pair solution of a system. Even though we
found y first, the x-coordinate is always written first in the ordered pair.
Slide 4.2-7
EXAMPLE 3 Using the Substitution Method
Use substitution to solve the system.
x  1  4 y
2 x  5 y  11
Solution:
x  1  1  4 y  1
x  4 y  1
x  4  1  1
x  4 1
x3
2  4 y 1  5 y  11
8 y  2  5 y  2  11  2
13 y 13

13 13
y  1
 3, 1
Slide 4.2-8
Objective 2
Solve special systems by
substitution.
Slide 4.2-9
Solve special systems by substitution.
Recall that systems of equations with graphs that are parallel lines
have no solution. Systems of equations with graphs that are the same
line have an infinite number of solutions.
Slide 4.2-10
EXAMPLE 4 Solving an Inconsistent System by Substitution
Use substitution to solve the system.
y  8x  4
16 x  2 y  8
Solution:
16 x  2 8x  4  8
16x 16x  8  8
8  8
Since the statement is false, the solution set is Ø.
It is a common error to give “false” as the solution of an inconsistent system.
The correct response is Ø.
Slide 4.2-11
EXAMPLE 5 Solving a System with Dependent Equations by Substitution
Solve the system by the substitution method.
x  3 y  7
4 x  12 y  28
Solution:
x  3 y  3 y  7  3 y
x  7  3 y
4  7  3 y   12 y  28
28  12 y  12 y  28
28  28
Since the statement is true every solution of one equations is also
a solution to the other, so the system has an infinite number of
solutions and the solution set is {(x,y)|x + 3y = −7}.
It is a common error to give “true” as the solution of a system of dependent
equations. Remember to give the solution set in set-builder notation using
the equation in the system that is in standard form with integer coefficients
that have no common factor (except 1).
Slide 4.2-12
Objective 3
Solve linear systems with
fractions and decimals by
substitution.
Slide 4.2-13
EXAMPLE 6 Using the Substitution Method with Fractions as Coefficients
Solve the system by the substitution method.
1
1
1
x
y 
2
3
3
1
x  2 y  2
2
1 
1
1
 6  x  y     6
3 
3
2
3x  2 y  2
3  4  4 y   2 y  2
Solution:
12  12 y  2 y  12  2  12
1

2
x

2
y
 
10 y
10
  2  2 
2



10
10
x  4 y  4 y  4  4 y
y  1
x  4  4 y
x  4  4  1
x  4  4
x0
 0, 1
Slide 4.2-14
EXAMPLE 7 Using the Substitution Method with Decimals as Coefficients
Solve the system by the substitution method.
0.2 x  1.2 y  1
0.1x  0.3 y  0.1
Solution:
10  0.2 x  1.2 y   10  1
10  0.2 x   10 1.2 y   10  1
2 x  12 y  10
2(1  3 y )  12 y  10
2  6 y  12 y  10
6 y  12
y  2
10  0.1x  0.3 y   10  0.1
10  0.1x   10  0.3 y   10 .01
x  3y  1
x  1 3y
x  1  3  2 
x7
 7, 2
Slide 4.2-15