Download Ch4-Sec 4.2

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4.2 – Slide 1
Chapter 4
Systems of Linear Equations
and Inequalities
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 2
4.2
Solving Systems of Linear Equations
by Substitution
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 3
4.2 Solving Systems of Linear Equations
by Substitution
Objectives
1.
2.
3.
Solve linear systems by substitution.
Solve special systems by substitution.
Solve linear systems with fractions and
decimals by substitution.
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4.2 – Slide 4
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 1 Solve the system by the substitution method.
5x + 2y = 2
y = – 3x
The second equation is already solved for y. This equation says that
y = – 3x. Substituting – 3x for y in the first equation gives
5x + 2y = 2
5x + 2(– 3x) = 2
5x + – 6x = 2
– 1x = 2
x = –2
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Let y = – 3x.
Multiply.
Combine like terms.
Multiply by – 1.
4.2 – Slide 5
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 1 (cont.) Because x = – 2, we find y from the equation
y = – 3x by substituting – 2 for x.
y = – 3x = – 3(– 2) = 6
Let x = – 2.
Check that the solution of the given system is (– 2, 6) by
substituting – 2 for x and 6 for y in both equations.
y = – 3x
5x + 2y = 2
5(– 2) + 2(6) = 2
2 = 2
6 = – 3(– 2)
?
True
6 = 6
?
True
The solution set of the system is {(– 2, 6)}.
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4.2 – Slide 6
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 2 Solve the system 3x + 4y = 4
by the substitution method.
x = 2y + 18
The second equation gives x in terms of y. Substitute 2y + 18 for x
in the first equation.
3x + 4y = 4
3(2y + 18) + 4y = 4
6y + 54 + 4y = 4
10y + 54 = 4
10y = – 50
y = –5
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Let x = 2y + 18.
Distributive property
Combine like terms.
Subtract 54.
Divide by 10.
4.2 – Slide 7
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 2 (continued) Because y = – 5, we find x from the
equation x = 2y + 18 by substituting – 5 for y.
x = 2y + 18 = 2(– 5) + 18 = 8
Let y = – 5.
Check that the solution of the given system is (8, – 5) by
substituting 8 for x and – 5 for y in both equations..
3x + 4y = 4
3(8) + 4(– 5) = 4
4 = 4
x = 2y + 18
8 = 2(– 5) + 18
?
True
8 = 8
?
True
The solution set of the system is {(8, – 5)}.
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4.2 – Slide 8
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Solving a Linear System by Substitution
Step 1 Solve one equation for either variable. If one of
the variables has a coefficient of 1 or – 1, choose it,
since it usually makes the substitution 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.
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4.2 – Slide 9
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 3 Use substitution to solve the system.
3x + 2y = 26
(1)
5x – y = 13
(2)
Step 1 For the substitution method, we must solve one of the
equations for either x or y. Because the coefficient of y in
equation (2) is – 1, we choose equation (2) and solve for y.
5x – y = 13
5x – y – 5x = 13 – 5x
– y = 13 – 5x
y = – 13 + 5x
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(2)
Subtract 5x.
Combine like terms.
Multiply by – 1.
4.2 – Slide 10
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 3 (continued) Use substitution to solve the system.
3x + 2y = 26
(1)
5x – y = 13
(2)
Step 2 Now substitute – 13 + 5x for y in equation (1).
3x + 2y = 26
3x + 2(– 13 + 5x) = 26
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(1)
Let y = – 13 + 5x.
4.2 – Slide 11
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 3 (continued) Use substitution to solve the system.
3x + 2y = 26
(1)
5x – y = 13
(2)
Step 3 Now solve the equation from Step 2.
3x + 2(– 13 + 5x) = 26
3x – 26 + 10x = 26
13x – 26 = 26
13x – 26 + 26 = 26 + 26
13x = 52
x = 4
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From Step 2.
Distributive property
Combine like terms.
Add 26.
Combine like terms.
Divide by 13.
4.2 – Slide 12
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 3 (continued) Use substitution to solve the system.
3x + 2y = 26
(1)
5x – y = 13
(2)
Step 4 Since y = – 13 + 5x and x = 4, y = – 13 + 5(4) = 7.
Step 5 Check that (4, 7) is the solution.
5x – y = 13
5(4) – 7 = 13
3x + 2y = 26
3(4) + 2(7) = 26
12 + 14 = 26
26 = 26
?
?
True
20 – 7 = 13
13 = 13
?
?
True
Since both results are true, the solution set of the system is {(4, 7)}.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 13
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4
Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 1 To use the substitution method, we must solve one of the
equations for one of the variables. We choose equation (2)
and solve for x.
5x – 4y = 6
5x – 4y + 4y = 6 + 4y
5x = 6 + 4y
6
4
x =
+
y
5
5
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(2)
Add 4y.
Combine like terms.
Divide by 5.
4.2 – Slide 14
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 2 Now substitute
6
4 for x in equation (1).
+
y
5
5
3x – 2y = 0
3 6 + 4 y
5
5
– 2y = 0
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(1)
Let x =
6
4 .
+
y
5
5
4.2 – Slide 15
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 3 Now solve the equation from Step 2.
3 6 + 4 y
5
5
– 2y = 0
From Step 2.
18
12 – 2y = 0
+
y
5
5
Distributive property
18
2
+
y = 0
5
5
Combine like terms.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 16
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 3 (continued)
18
2
+
y
5
5
18 + 2 y – 18
5
5
5
2 y
5
= 0
From previous slide.
= 0 – 18
5
Subtract 18 .
5
= – 18
5
Combine like terms.
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4.2 – Slide 17
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 3 (continued)
2 y
5
= – 18
5
y = –9
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From previous slide.
Multiply by 5 .
2
4.2 – Slide 18
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 4 Find x by substituting – 9 for y in
x =
6
+
5
Copyright © 2010 Pearson Education, Inc. All rights reserved.
6
4
+
y.
5
5
4
(– 9) = – 6
5
4.2 – Slide 19
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems by Substitution
Example 4 (continued) Use substitution to solve the system.
3x – 2y = 0
(1)
5x – 4y = 6
(2)
Step 5 Check that (– 6, – 9) is the solution.
3x – 2y
3(– 6) – 2(– 9)
(– 18) – (– 18)
– 18 + 18
0
=
=
=
=
=
0
0
0
0
0
5x – 4y
5(– 6) – 4(– 9)
(– 30) – (– 36)
(– 30) + 36
6
=
=
=
=
=
6
6
6
6
6
Both results are true, so the solution set of the system is {(– 6, – 9)}.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 20
4.2 Solving Systems of Linear Equations
by Substitution
Solving Special Systems by Substitution
Example 5 Use substitution to solve the system.
y = 2x – 7
3y – 6x = – 6
(1)
(2)
Substitute 2x – 7 for y in equation (2).
3y – 6x = 6
(2)
3(2x – 7) – 6x = – 6
Let y = 2x – 7.
6x – 21 – 6x = – 6
Distributive property
– 21 = – 6
False.
This false result means that the equations in the system have graphs
that are parallel lines. The system is inconsistent and the solution set
is 0 .
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4.2 – Slide 21
4.2 Solving Systems of Linear Equations
by Substitution
Solving Special Systems by Substitution
y
Example 5 (continued)
y = 2x – 7
3y – 6x = – 6.
(1)
(2)
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(2)
(1)
x
4.2 – Slide 22
4.2 Solving Systems of Linear Equations
by Substitution
Solving Special Systems by Substitution
Example 6
Solve the system by the substitution method.
3 = 5x – y
(1)
– 4y – 12 = – 20x
(2)
Begin by solving equation (1) for y to get y = 5x – 3. Substitute
5x – 3 for y in equation (2) and solve the resulting equation.
– 4y – 12 = – 20x
(2)
– 4(5x – 3) – 12 = – 20x
Let y = 5x – 3.
– 20x + 12 – 12 = – 20x
Distributive property
0 = 0
Add 20x; combine terms.
This true result means that every solution of one equation is also a
solution of the other, so the system has an infinite number of solutions.
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4.2 – Slide 23
4.2 Solving Systems of Linear Equations
by Substitution
Solving Special Systems by Substitution
y
Example 6 (continued)
3 = 5x – y
– 4y – 12 = – 20x
(1)
(2)
x
The system has an infinite number of solutions – all the ordered pairs
corresponding to points that lie on the common graph.
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4.2 – Slide 24
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7
Solve the system by the substitution method.
2 x + 5y = 1
(1)
3
1 x –
4
1 y = – 13
8
8
(2)
Clear equation (1) of fractions by multiplying each side by 3.
3
3 2 x
3
2 x + 5y
3
= 3 1
Multiply by 4.
+ 3 5y
= 3 1
Distributive property
2x + 15y = 3
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(3)
4.2 – Slide 25
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7 (cont.) Solve the system by the substitution method.
2 x + 5y = 1
(1)
3
1 x – 1 y = – 13
(2)
4
8
8
Now, clear equation (2) of fractions by multiplying each side by 8.
8 1 x –
4
8 1 x
4
–
1 y = 8 – 13
8
8
8 1y
8
= 8 – 13
8
2x – y = – 13
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Multiply by 8.
Distributive property
(4)
4.2 – Slide 26
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7 (cont.) The given system of equations has been
simplified to the equivalent system.
2 x + 5y = 1
(1)
2x + 15y = 3
3
1 x – 1 y = – 13
(2)
2x – y = – 13
4
8
8
(3)
(4)
To solve this system by substitution, equation (4) can be solved for y.
2x – y = – 13
2x – y – 2x = – 13 – 2x
– y = – 13 – 2x
y = 13 + 2x
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(4)
Subtract 2x.
Combine like terms.
Multiply by – 1.
4.2 – Slide 27
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7 (cont.) The given system of equations has been
simplified to the equivalent system.
2 x + 5y = 1
(1)
2x + 15y = 3
3
1 x – 1 y = – 13
(2)
2x – y = – 13
4
8
8
(3)
(4)
Now substitute 13 + 2x for y in equation (3).
2x + 15y
2x + 15(13 + 2x)
2x + 195 + 30x
32x + 195
=
=
=
=
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3
3
3
3
(3)
Let y = 13 + 2x.
Distributive property
Combine like terms.
4.2 – Slide 28
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7 (cont.) The given system of equations has been
simplified to the equivalent system.
2 x + 5y = 1
(1)
2x + 15y = 3
3
1 x – 1 y = – 13
(2)
2x – y = – 13
4
8
8
(3)
(4)
Now substitute 13 + 2x for y in equation (3).
32x + 195 = 3
32x + 195 – 195 = 3 – 195
32x = – 192
x = –6
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Combine like terms.
Subtract 195.
Combine like terms.
Divide by 32.
4.2 – Slide 29
4.2 Solving Systems of Linear Equations
by Substitution
Solving Linear Systems with Fractions and Decimals
Example 7 (cont.) The given system of equations has been
simplified to the equivalent system.
2 x + 5y = 1
(1)
2x + 15y = 3
3
1 x – 1 y = – 13
(2)
2x – y = – 13
4
8
8
(3)
(4)
Substitute – 6 for x in y = 13 + 2x to get
y = 13 + 2(– 6) = 1.
Check by substituting – 6 for x and 1 for y in both of the original
equations. The solution set is {(– 6, 1)}.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.2 – Slide 30