lesson 5: solving linear systemslesson 5: solving linear systems quiz solutions algebra 1 © 2009...
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Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 7
Show all of your work in order to receive full credit. Attach graph paper for graphs.
1. Decide which of the following ordered pairs are solutions to the system. Answer yes or no.
4 11
7 19
x y
x y
− = −
+ =
a) (-4, -5)
( ) ( )?
?
?
4 11
4 4 5 11
16 5 11
11 11
x y− = −
− − − = −
− + = −
− = − �
( ) ( )
?
?
?
7 19
4 7 5 19
4 35 19
39 19
x y+ =
− + − =
− − =
− = �
No
b) (-2, 3)
( )?
?
?
4 11
4 2 3 11
8 3 11
11 11
x y− = −
− − = −
− − = −
− = − �
( ) ( )
?
?
?
7 19
2 7 3 19
2 21 19
19 19
x y+ =
− + =
− + =
= �
Yes
2. Solve the system graphically. Attach graph paper.
3 4 8
5
x y
x y
− =
+ =
3 4 8
4 3 8
32
4
x y
y x
y x
− =
− = − +
= −
5
5
x y
y x
+ =
= − +
See next page for graph.
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 7
The graphs intersect at the point ( )4,1 .
Check:
( ) ( )?
?
?
3 4 8
3 4 4 1 8
12 4 8
8 8
x y− =
− =
− =
= �
( ) ( )?
?
5
4 1 5
5 5
x y+ =
+ =
= �
The solution is ( )4,1 .
3. Solve the following system by substitution. 2x + y = 9
3x - 4y = 8
2 9
2 9
x y
y x
+ =
= − +
Substitute this into the second equation:
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 7
( )
3 4 8
3 4 2 9 8
3 8 36 8
11 44
4
x y
x x
x x
x
x
− =
− − + =
+ − =
=
=
Substitute this into an equation to find y.
( )
2 9
2 4 9
8 9
1
y x
y
y
y
= − +
= − +
= − +
=
Check:
( ) ( )
?
?
?
2 9
2 4 1 9
8 1 9
9 9
x y+ =
+ =
+ =
= �
( ) ( )
?
?
?
3 4 8
3 4 4 1 8
12 4 8
8 8
x y− =
− =
− =
= �
The solution is ( )4,1 .
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 7
4. Solve the following system by elimination.
4x + 3y = 5
5x + 2y = 8
( )
( )
4 3 5
5 2 8
2 4 3 5
3 5 2 8
8 6 10
15 6 24
7 14
2
x y
x y
x y
x y
x y
x y
x
x
+ =
+ =
− + =
+ =
− − = −
+ =
=
=
Substitute this into one of the equations to find y:
( )
4 3 5
4 2 3 5
8 3 5
3 3
1
x y
y
y
y
y
+ =
+ =
+ =
= −
= −
Check:
( ) ( )?
?
?
4 3 5
4 2 3 1 5
8 3 5
5 5
x y+ =
+ − =
− =
= �
( ) ( )
?
?
?
5 2 8
5 2 2 1 8
10 2 8
8 8
x y+ =
+ − =
− =
= �
The solution is ( )2, 1− .
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 7
5. Solve the following systems algebraically.
a) 2 7
10 5 35
x y
x y
+ =
= − +
2 7
2 7
x y
y x
+ =
= − +
Substitute this into the second equation:
( )
10 5 35
10 5 2 7 35
10 10 35 35
0 0
x y
x x
x x
= − +
= − − + +
= − +
=
This means that the two equations are the same. Therefore, any
pair of numbers ( ),x y that satisfies the equation 2 7x y+ = will
also satisfy the equation 10 5 35x y= − +
The solution is the set of numbers ( ),x y such that 2 7x y+ = .
b) 2 9
4 2 2
x y
x y
− = −
+ =
( )
2 9
4 2 2
2 2 9
4 2 2
4 2 18
4 2 2
8 16
2
x y
x y
x y
x y
x y
x y
x
x
− = −
+ =
− = −
+ =
− = −
+ =
= −
= −
Substitute this into one of the equations to find y:
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 6 of 7
( )
2 9
2 2 9
4 9
5
5
x y
y
y
y
y
− = −
− − = −
− − = −
− = −
=
Check:
( ) ( )
?
?
?
2 9
2 2 5 9
4 5 9
9 9
x y− = −
− − = −
− − = −
− = − �
( ) ( )
?
?
?
4 2 2
4 2 2 5 2
8 10 2
2 2
x y+ =
− + =
− + =
= �
The solution is ( )2,5− .
c) 3 2 5
6 7 4
x y
x y
+ =
= −
( )
3 2 5
6 7 4
3 2 5
6 4 7
2 3 2 5
6 4 7
6 4 10
6 4 7
0 3
x y
x y
x y
x y
x y
x y
x y
x y
+ =
= −
+ =
+ =
− + =
+ =
− − = −
+ =
= −
No solution!
Lesson 5: Solving Linear Systems Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 7 of 7
6. Solve the following system graphically. Attach graph paper.
2 10
3 4
x y
y x
− ≥
> − +
2 10
2 10
15
2
x y
y x
y x
− ≥
− ≥ − +
≤ −
3 4y x> − +
The solution is the set of points that lie in the purple region on the
graph.