Download Lesson 3: Using Linear Combinations to Solve a System of Equations

Unit 4: Systems of Equations
Lesson 3: Using Linear Combinations to Solve a System of Equations
Steps for Using Linear Combinations to Solve a System of Equations
1. __________________________________________________________________________
2. __________________________________________________________________________
3. __________________________________________________________________________
4. __________________________________________________________________________
5. __________________________________________________________________________
Example 1
Solve the following system using the linear combinations method.
2x – 3y = 12 & 4x + 3y = 6
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
Example 2
Solve the following system using the linear combinations method.
y = 4x - 2 & 8x – 2y = -12
Example 3
Solve the following system using the linear combinations method.
2x – 3y = 3 & 6x – 9y = 9
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
Example 4
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
Lesson 3: Using Linear Combinations Method
1. Which of the following ordered pairs is a solution to the system of equations?
4x +3y = 5
2x – 3y = 7
A.
B.
C.
D.
(5,7)
(2, -1)
(-1,2)
None of the Above.
2. Use Linear Combinations to find the solution to the following system of equations.
3x +2y = 8
x – 2y = 4
3. Which statement best describes the solution to the following system of equations?
5x – y = -2
10x – 2y = -4
A.
B.
C.
D.
(-2,-4) is the solution.
There is no solution, the lines are parallel.
There are an infinite number of solutions, they are the same line.
None of the Above.
4. Solve the following system of equations using linear combinations.
x – 2y = -10
3x – y = 0
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
5. Which ordered pair is the solution to the following system of equations?
3x +2y = 10
2x +5y = 3
A.
B.
C.
D.
(10,3)
(-1,4)
(4,-1)
None of the Above
6. Describe the graph of the solution for the following system of equations.
x – 2y = 10
2x – 4y = 20
7. Find the solution to the following system of equations. Explain what the solution means.
3x +2y = 8
4x – 3y = 5
8. Find the solution to the following system of equations.
4x – 2y = 2
3x – 3y = 9
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
Directions: Solve each system of equations using the
Linear Combinations method.
Each problem is worth 3 points.
1. 2a+8b = 2
-2a- 4b = 6
2. x + 3y = 18
2x+y = 11
3. 2x -4y = -4
3x + 3y = 3
4. 5x –y = -2
10x – 2y = 8
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
Lesson 3: Using Linear Combinations Method – Answer Key
1. Which of the following ordered pairs is a solution to the system of equations?
4x +3y = 5
2x – 3y = 7
A.
B.
C.
D.
(5,7)
(2, -1)
(-1,2)
None of the Above.
4x +3y = 5
We already have opposite terms so we can add!
2x – 3y = 7
6x
= 12
6x = 12
6
6
Step 2: Substitute 2 for x into either equation.
2x –3y = 7
2(2) – 3y = 7
x= 2
4 – 3y = 7
4 – 4 – 3y = 7 – 4
-3y = 3
-3 -3
Y = -1
Subtract 4 from both sides
(2,-1) is the solution
2. Use Linear Combinations to find the solution to the following system of equations.
3x +2y = 8
x – 2y = 4
3x +2y = 8
We already have opposite terms, so we can add.
x – 2y = 4
4x
= 12
4x = 12
4
4
x=3
Step 2: Substitute 3 for x into either equation.
x – 2y = 4
3 – 2y = 4
3 -3 – 2y = 4 – 3
-2y = 1
-2 -2
Y = -1/2
Substitute 3 for x.
Subtract 3 from both sides
Simplify: 4-3 = 1
Copyright© 2009 Algebra-class.com
The solution to the system is (3, -1/2)
Unit 4: Systems of Equations
3. Which statement best describes the solution to the following system of equations?
5x – y = -2
10x – 2y = -4
-2[5x – y = -2]
10x – 2y = -4
-10x +2y = 4
10x – 2y = -4
0 + 0 =0
We don’t have opposite terms, so I multiplied
the first equation by -2, to make the y term 2y.
0 = 0 does make sense.
After adding the two equations, I am left with 0 = 0. This statement does make sense. Therefore, there
is an infinite number of solutions to this system. The lines are equivalent.
A.
B.
C.
D.
(-2,-4) is the solution.
There is no solution, the lines are parallel
There are an infinite number of solutions, they are the same line.
None of the Above.
4. Solve the following system of equations using linear combinations.
x – 2y = -10
3x – y = 0
-3[x – 2y = -10]
3x – y = 0
-3x +6y = 30
3x – y = 0
5y = 30
5y = 30
5
5
We did not have terms that were opposites, so I multiplied the
first equation by -3 to make the x term, -3x.
Add the equation together and you get 5y = 30.
Solve for y. Divide by 5 on both sides.
Y=6
Step 2: Substitute 6 for y into either equation.
3x – y = 0
I chose equation #2 because y didn’t have a coefficient. It doesn’t matter which
equation you choose. You’ll end up with the same answer!
3x – 6 = 0
3x -6 +6 = 0 +6
Add 6 to both sides.
3x = 6
Simplify: 0 +6 = 6
3 3
x=2
The solution to this system of equations is: (2, 6)
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
5.
A.
B.
C.
D.
Which ordered pair is the solution to the following system of equations?
3x +2y = 10
2x +5y = 3
(10,3)
(-1,4)
(4,-1)
None of the Above
-2[3x +2y = 10]
3[2x +5y = 3]
11y = -11
11
11
-6x -4y = -20 We don’t have opposite terms and we can’t multiply one
6x + 15y = 9 equation by a number to create opposite terms. Therefore, we
11y = -11 need to multiply both equations by a number and either make the
x terms opposites (6x and -6x) or make the y terms opposites (10y
and -10y. I chose to multiply the first equation by -2 and the second
equation by 3 to make the x terms (-6x and 6x).
Y = -1
Step 2: Substitute -1 for y into either equation.
2x +5y = 3
2x +5(-1) = 3
2x -5 = 3
2x -5 +5 = 3 +5
2x = 8
2 2
x=4
The solution to this system of equations is (4, -1)
6. Describe the graph of the solution for the following system of equations.
x – 2y = 10
2x – 4y = 20
-2[x – 2y = 10]
2x – 4y = 20
-2x +4y = -20
2x – 4y = 20
0 = 0
I multiplied the first equation by -2 to make opposite terms (-2x & 2x)
Add the equations together.
We can’t go any further and 0 = 0 is a true statement. Therefore, these two equations are equivalent. They
are the same line lying on top of each other. There are an infinite number of solutions to the system of
equations.
Copyright© 2009 Algebra-class.com
Unit 4: Systems of Equations
7. Find the solution to the following system of equations. Explain what the solution means.
3x +2y = 8
4x – 3y = 5
3[3x +2y = 8]
2[4x – 3y = 5]
9x +6y = 24
8x – 6y = 10
17x
= 34
I multiplied the first equation by 3 and the second equation by 2 to
make the y terms opposites (6y & -6y)
17x = 34
17 17
Step 2: Substitute 2 for x into either equation.
4x – 3y = 5
x=2
4(2) – 3y = 5
8 – 3y = 5
8 -8 – 3y = 5 -8
-3y = -3
-3 -3
y=1
The solution is (2,1)
The solution (2,1) is the point where the two lines will intersect on a graph. It means that (2,1) is a solution
for both equations. This is the only solution that both equations have in common.
8. Find the solution to the following system of equations.
4x – 2y = 2
3x – 3y = 9
-3[4x – 2y = 2]
2[3x – 3y = 9]
-12x + 6y = -6
6x – 6y = 18
-6x
= 12
I multiplied the first equation by -3 and the second equation by 2
in order to make the y terms opposites (6y & -6y).
-6x = 12
-6 -6
Step 2: Substitute -2 for x into either equation and solve for y.
4x – 2y = 2
x = -2
4(-2) – 2y = 2
-8 – 2y = 2
-8 +8 – 2y = 2 +8
-2y = 10
-2 -2
y = -5
Copyright© 2009 Algebra-class.com
Substitute -2 for x.
Simplify: 4(-2) = -8
Add 8 to both sides
Solution is (-2,-5)
Unit 4: Systems of Equations
Directions: Solve each system of equations using the Linear Combinations method.
Each problem is worth 3 points.
1. 2a+8b = 2
-2a- 4b = 6
2. x + 3y = 18
2x+y = 11
The first terms are already opposites so we can go ahead and add:
2a+8b = 2
-2a- 4b = 6
4b = 8
Add the 2 equations. Now solve for b.
4b/4 = 8/4
b=2
Divide by 4 on both sides
b = 2, now we must find a by substituting.
2a + 8b = 2
2a + 8(2) = 2
2a + 16 = 2
2a +16-16 = 2-16
2a = -14
2a/2 = -14/2
a = -7
Choose one of the equation to substitute back into.
Substitute 2 for b.
Simplify: 8(2) = 16
Subtract 16 from both sides
Simplify: 2-16= -14
Divide by 2 on both sides
b = 2 and a = -7
These two equations do not have opposite terms, so we must multiply the
first equation by -2 to create opposite x terms: (-2x & 2x). (Or you could
multiply the second equation by -3 to create opposite y terms: (3y &-3y)
-2(x+3y) = (18)-2
2x + y = 11
Copyright© 2009 Algebra-class.com
-2x – 6y = -36
2x + y = 11
-5y = -25
Multiply all terms by -2
Keep the same
Add the equations
-5y = -25
-5y/-5 = -25 / -5
y=5
Divide by -5 on both sides
y = 5, now solve for x.
2x + y = 11
2x + 5 = 11
2x + 5-5 = 11-5
2x = 6
2x/2 = 6/2
x=3
Use this equation to substitute 5 for y
Substitute 5 for y
Subtract 5 from both sides
Simplify: 11-5 = 6
Divide by 2 on both sides
x = 3 and y =5
Unit 4: Systems of Equations
3. 2x -4y = -4
3x + 3y = 3
We need to create opposites and the only way to do this is to multiply the first
equation by -3 and the second equation by 2. This will create opposite x terms: (6x and 6x). OR you could create opposite y terms by multiplying the first
equation by 3 and the second equation by 4.
-3(2x – 4y) = -3(-4)
2(3x+3y) = 2(3)
4. 5x –y = -2
10x – 2y = 8
-6x + 12y = 12
6x + 6y = 6
18y = 18
Multiply by -3
Multiply by 2
Add the equations
18y = 18
18y/18 = 18/18
y=1
Now we can solve for y.
Divide by 18 on both sides.
y = 1, now we need to solve for x.
3x + 3y = 3
3x +3(1) = 3
3x + 3 = 3
3x +3-3= 3-3
3x = 0
3x/3 = 0/3
x=0
We can use this equation to solve for x.
Substitute 1 for y.
Simplify: 3(1) = 3
Subtract 3 from both sides
Simplify: 3-3 = 0
Divide by 3 on both sides
x = 0 and y = 1
We do not have opposite terms, so we must multiply the first equation by -2 in
order to create opposite terms.
-2(5x – y) = -2(-2)
10x – 2y = 8
-10x + 2y = 4
10x – 2y = 8
0 = 12
Multiply by -2
Keep the same
Add the equations
This statement does not make sense, and we do not have
a variable to solve for. Since 0 is not equal to 12, we know that there are no
solutions to this system of equations. These two lines are parallel and there are
no solutions.
Copyright© 2009 Algebra-class.com