9.3 Solving Systems of Linear Equations by Elimination; Applications 1. Solve systems of linear equations using elimination. 2.

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Transcript 9.3 Solving Systems of Linear Equations by Elimination; Applications 1. Solve systems of linear equations using elimination. 2.

9.3
Solving Systems of Linear Equations by
Elimination; Applications
1. Solve systems of linear equations using elimination.
2. Solve applications using elimination.
Solve the system of equations
using elimination.
x  y  9

3x  y  7
One equation with one unknown.
Look for additive inverses.
x y 9
3x  y  7
4 x  0  16
4 x  16
x4
Solve for y.
x+y=9
4+y=9
y=5
The solution is (4, 5).
Solve the system of equations
using elimination.
2 x  3y  8

4 x  3y  16
Look for additive inverses.
2x  3y  8
4 x  3y  16
6 x  24
x 4
24   3y  8
8  3y  8
 3y  0
y0
The solution is (4, 0).
Solve the system of equations. -34  x  y  8

3x  4 y  11

Look for additive inverses.
4 x  y  8
4x  4y  32
4 x  4 y  32
3x  4 y  11
7 x  0  21
7 x  21
x3
Solve for y.
x+y=8
3+y=8
y=5
The solution
is (3, 5).
Solve the system of equations.
20 x  4 y  104
7x  4y  4
27 x  108
x 4
7 x  4 y  4

45 x  y  26
54   y  26
20  y  26
y6
The solution is (4, 6).
Solve the system of equations.
24 x  5 y
-3
 19

-5
4 3x  2 y  9
Look for additive inverses.
4 x  5 y  19
3x  2 y  9
Multiply by 2.
Multiply by 5.
Solve for y.
4x – 5y = 19
4(1) – 5y = 19
4 – 5y = 19
–5y = 15
y = 3
8 x  10 y  38
15 x  10 y  45
7 x  0  7
7 x  7
x 1
The solution is (1, –3).
Solve the
5 x  6 y  11
system. 
2 x  4 y  2
a) (1, 1)
b) (1, 5)
c) (–1, 1)
d) no solution
9.3
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Slide 5- 7
Solve the
5 x  6 y  11
system. 
2 x  4 y  2
a) (1, 1)
b) (1, 5)
c) (–1, 1)
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 8
Solve the system of equations.
0.03x  0.02 y  0.03

2
2
4
x y 

5
5
5
To clear the decimals in Equation 1, multiply by 100.
To clear the fractions in Equation 2, multiply by 5.
0.03 x  0.02 y  0.03
4
2
2
x y 
5
5
5
3x  2 y  3
-1 4 x  2 y  2
Multiply by 100.
Multiply by 5.
3x  2 y  3
4x  2 y  2
3x  2 y  3
-1
4 x  2 y  2
0.03x  0.02 y  0.03

2
2
4
x y 

5
5
5
continued
3x  2 y  3
4 x  2 y  2
x  0  1
x  1
x  1
Substitute to find y.
The solution is (1, 3).
3x  2 y  3
3(1)  2 y  3
3  2 y  3
2 y  6
y  3
Solving Systems of Two Linear Equations Using Elimination
1. Write the equations in standard form (Ax + By = C).
2. Use the multiplication principle to clear fractions or
decimals (optional).
3. If necessary, multiply one or both equations by a number so
that they have a pair of terms that are additive inverses.
4. Add the equations. The result should be an equation in
one variable.
5. Solve the equation from step 4.
6. Using an equation containing both variables, substitute the
value you found in step 5 for the corresponding variable and
solve for the value of the other variable.
7. Check your solution in the original equations.
Copyright © 2011 Pearson Education, Inc.
Solve the system.
x  3y  7

 x  5 y  13
a) (2, 3)
b) (7, 0)
c) (–2, 3)
d) (5, 5)
9.3
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Slide 5- 12
Solve the system.
x  3y  7

 x  5 y  13
a) (2, 3)
b) (7, 0)
c) (–2, 3)
d) (5, 5)
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 13
Inconsistent Systems and Dependent Equations
When both variables have been eliminated and the
resulting equation is false, such as 0 = 5, there is no
solution. The system is inconsistent.
When both variables have been eliminated and the
resulting equation is true, such as 0 = 0, the equations
are dependent. There are an infinite number of
solutions.
Solve the system of equations.
 3x  y  4

3x  y  5
3x  y  4
3x  y  5
0  9
False statement.
The system is inconsistent
and has no solution.
Solve the system of equations.
2  3x  y  4

6 x  2 y  8
To eliminate y, multiply the first equation by 2.
3x  y  4
6 x  2 y  8
Multiply by 2.
6x  2 y  8
6 x  2 y  8
00
True statement. The equations are dependent.
There are an infinite number of solutions.
Solve the
5 x  5 y  50
system. 
 x  y  2.5
a) (0, 10)
b) (-2, -3)
c) infinite number of solutions
d) no solution
9.3
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Slide 5- 17
Solve the
5 x  5 y  50
system. 
 x  y  2.5
a) (0, 10)
b) (-2, -3)
c) infinite number of solutions
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 18