Download Systems: Special Cases

Name:____________________
April 9, 2014
Algebra 2
Systems: Special Cases
Systems that do not have solutions
Sometimes a linear system will not have any solutions. Here’s a system that has no
solutions:
x+ y= 4
2x + 2y = 10
What happens in our system solving methods, if the system actually has no solution?
Substitution Method:
Here’s what happens for the example problem.
First equation solved for y:
Substitute into second equation:
Distribute and simplify:
y=4–x
2x + 2(4– x) = 10
2x + 8 – 2x = 10
8 = 10
We end up with an always-false equation, which tells us that there’s no solution.
Elimination Method:
Here’s what happens for the example problem.
First equation multiplied by 2:
Second equation:
Subtract:
2x + 2y = 8
2x + 2y = 10
0x + 0y = 2
0 =2
We end up with an always-false equation, which tells us that there’s no solution.
In general, using either of the methods, the indicator that a system has no solutions is that
you reach an equation that is always false, such as a number equaling a different number.
Since the equation is false no matter what values x and y have, that means the system has
no solution.
You try it
1. a. Solve by substitution:
2x + 4y = 20
3x + 6y = 50
b. Solve by elimination:
2x + 4y = 20
3x + 6y = 50
Name:____________________
April 9, 2014
Algebra 2
Systems that have infinitely many solutions
Sometimes a linear system will have infinitely many solutions. Here’s such a system:
x+ y= 4
2x + 2y = 8
What happens in our system solving methods, if the system has infinitely many
solutions?
Substitution Method:
Here’s what happens for the example problem.
First equation solved for y:
Substitute into second equation:
Distribute and simplify:
y=4–x
2x + 2(4– x) = 8
2x + 8 – 2x = 8
8=8
We get an always-true equation, which tells that there are infinitely many
solutions.
Elimination Method:
Here’s what happens for the example problem.
First equation multiplied by 2:
Second equation:
Subtract:
2x + 2y = 8
2x + 2y = 8
0x + 0y = 0
0 =0
We get an always-true equation, which tells that there are infinitely many
solutions.
In general, using either of the methods, the indicator that a system has infinitely many
solutions is that you reach an equation that is always true, such as a number equaling the
same number. When that happens, any of the infinitely many (x, y) pairs that makes one
of the equations true also makes the other equation true.
You try it
2. a. Solve by substitution:
2x + 4y = 20
3x + 6y = 30
b. Solve by elimination:
2x + 4y = 20
3x + 6y = 30
Name:____________________
April 9, 2014
Algebra 2
More Problems:
Directions: Solve these systems using your choice of substitution or elimination. If the
system has a solution, check it.
3. 4x – 2y = 5
–8x + 4y = –10
4.
–3x – y = 6
6x + 2y = 12
5.
2x + 5y = 20
–4x + 10y = 40
Name:____________________
April 9, 2014
Algebra 2
x  13 y  2
3x  2 y  6
6.
1
2
7.
0.3x – 0.6y = 1
–3x + 6y = –10
8. a. Fill in the ____’s with numbers so that this system would have no solutions.
2x + 7y
= _______
4x + ____y =
20
b. Fill in the ____’s with numbers so that this system would have infinitely many
solutions.
2x + 7y
= _______
4x + ____y =
20
c. Fill in the ____’s with numbers so that this system would have just a single
solution.
2x + 7y
= _______
4x + ____y =
20
Name:____________________
April 9, 2014
9. system: 4x – 8y = 20, –2x + 4y = –10
a. Solve this system using the elimination method.
b. Solve this system again using the substitution method
Algebra 2