Transcript 12_04 - Solving Systems by Elimination - 1

Solve the following system using the addition method.
3x – 2y = – 8
5x + 2y = – 8
Notice that the y terms are opposites.
Add the two equations together.
8x = – 16
Solve the equation for x.
x =–2
Replace x with – 2 in either
equation from the system.
5x + 2y = – 8
5(– 2) + 2y = – 8
→ – 10 + 2y = – 8
→ 2y = 2
Solution:
x =–2
and
→ y = 1
y = 1
12.04
Solving
Systems of Equations
by
Elimination
(1 Multiplier)
Another method that can be used to solve a system
of equations is called the elimination method.
Some systems do not contain opposites so
adding the two equations together will not
cancel out any variable terms.
Examine the following system.
2x + y = 3
2x + 3y = 5
If the two equations are added the new equation is:
4x + 4y = 8
Neither variable term cancelled out.
If a system does not have opposites, we have to
create them by multiplying one of the equations by
some factor (number) in order to create opposites.
To create opposites, examine the variable terms
to see which equation most easily can be
changed in order to create opposites.
Remember opposite terms have the same
numbers and variables with different signs.
Choose an equation and multiply that
equation by some number that will create
opposites in the system.
When the system contain opposite terms, add
the equations and solve as usual.
Solve the following system using the elimination method.
2x – 5y = 9
2x – 3y = 11
Notice that the x and y terms
do not contain opposites.
If the 2nd equation is multiplied
by – 1, the x terms will
become opposites.
2x – 5y = 9
– 2x + 3y = – 11
– 2y = – 2
y = 1
Add the two equations and solve.
2x – 5y = 9
2x – 5(1) = 9
Solve for x.
→ 2x – 5 = 9
→ 2x = 14
Solution:
x = 7
and
→ x = 7
y = 1
Solve the following system using the elimination method.
5x – 2y = 13
2x + y = 7
Notice that the x and y terms
do not contain opposites.
If the 2nd equation is multiplied
by 2, the y terms will become
opposites.
5x – 2y = 13
4x + 2y = 14
9x = 27
Add the two equations and solve.
x =3
2x + y = 7
Solve for y.
2(3) + y = 7 → 6 + y = 7
Solution:
x = 3
and
→ y = 1
y = 1
Solve the following system using the elimination method.
x + y = 2
3x + 3y = 6
– 3x – 3y = – 6
3x + 3y = 6
0 = 0
Notice that the x and y terms
do not contain opposites.
If the 1st equation is multiplied
by – 3, the x terms will
become opposites.
Add the two equations and solve.
The variable terms have cancelled
and the statement is true.
There are infinite solutions to the system.
Solve the following system using the elimination method.
2x + 4y = 8
x + 2y = 5
Notice that the x and y terms
do not contain opposites.
2x + 4y = 8
– 2x – 4y = – 10
If the 2nd equation is
multiplied by – 2, the x terms
will become opposites.
0 =–2
Add the two equations and solve.
The variable terms have cancelled
and the statement is false.
There are no solutions to the system.
Try This:
Solve the following system using the elimination method.
4x + y = 7
6x + 2y = 10
– 8x – 2y = – 14
6x + 2y = 10
– 2x = – 4
x = 2
4x + y = 7
4(2) + y = 7
Solution:
Notice that the x and y terms
do not contain opposites.
If the 1st equation is multiplied
by – 2, the x terms will
become opposites.
Add the two equations and solve.
Solve for y.
→ 8 + y =7
x = 2
and
→ y = –1
y =–1