### 3.2 Solving Systems of Equations Algebraically The Substitution

```3.2 Solving Systems of Equations Algebraically
The Substitution Method:
1. Solve one of the equations for one of its variables. (coefficient of one on variable)
2. Substitute this expression into the other equation and solve for the other variable.
3. Substitute the value from #2 above in the revised first equation and solve.
4. Check the solution in each of the original equations.
Solve the following problems by substitution.
Ex. 1
4x − y = 9
x − 3 y = 16
The Elimination Method:
1. Arrange the equations with like terms in columns. Equations should look like standard
form.
2. Obtain coefficients for x (or y) that have matching coefficients with opposite signs. If
none match, multiply one or both to make a match.
3. Add so that the x column or y column is eliminated.
4. Solve for the remaining variable.
5. Substitute the value from #4 above into one of the original equations and solve for the
other variable.
6. The solution should give an ordered pair, all reals, or empty set.
Solve each problem below using elimination.
Ex. 1
5a + 4b = 11
3a − 5b = −23
Ex. 2
5x − 3y = −3
2x + 6y = 0
heck
6 Solve the system using intersection.
a graphing calculator. y ! x2 " 2
y ! "x
Solve using any algebraic method:
d 3 to find the second
1
x =1
4
y = x − 4Skill and Word Problem Practice.
For more
exercises, see2 Extra
Intersection
x + 2y = −8
3x − 4y = −24
y+
Sare (1, 4) and (4, 1).
X=4
oblem Solving
Y=1
aphing calculator. y ! x2 " 2
! "x by graphing. Find the number of solutions for each system.
mple
Solve eachy system
nd 2
753)
1. y ! x2 # 1
y!x#1
2. y ! x2 # 4
y ! 4x
3. y ! x2 " 5x " 4
y ! "2x
Application:
4. y ! x2 # 2x # 4
5. y ! x2 # 2x # 5
6. y ! 3x # 4
A
hair
salon
a
shipment
of
84
bottles
of
hair
conditioner
to
use
and
exercises, see
Extra
y!
x # 1Skill and Word Problem
y ! "2xPractice.
#1
y !sell
"xto2
customers. The two types of conditioners received are type A, which is used for regular
hair, and type B, which is used for frizzy hair. Type A costs \$6.50 per bottle and type B
Solve
each
using
ple 3
costs
\$8.25
per system
bottle. The
hairelimination.
salon’s invoice for the conditioner is \$588. How many of
each
type
of
conditioner
are
in
the
shipment? 2
753)
7. y ! "x # 3
y!x #1
8. y ! x
y!x#2
ing. Find the number
of solutions for each system.
2
2. y ! x2 # 4 2
10. y ! x # 11
y ! 4x y ! "12x
3. y ! x2 " 5x " 4
11. y ! 5x " 20
y ! "2x y ! x2 " 5x # 5
9. y ! "x " 7
y ! x2 " 4x " 5
12. y ! x2 " x " 90
y ! x # 30
5. y !Solving
x2 # Linear
6. yEquations
! 3x #Algebraically:
4
each
system
using
substitution.
2
ple 4 y ! Solve
"2x # 1
y ! "x
754)
Ex.
13. y ! x2 " 2x " 6
mination. y ! 4x # 10
14. y ! 3x " 20
y ! "x2 # 34
15. y ! x2 # 7x # 100
y # 10x ! 30
17. "
3x7" y ! " 2
8. y ! 16.
x2 "x2 " x # 19 !9.y y ! "x
! y5
y ! x #x2! y # 80
y ! x2 "2x
4x2 "
18. y ! 3x2 # 21x " 5
"10x # y ! "1
11. y ! 5x " 20
y ! x2 " 5x # 5
Practice:
bstitution.
12. y ! x2 " x " 90
y ! x # 30
Lesson NY-6
Systems of Linear and Quadratic Equations
14. y ! 3x " 20
y ! "x2 # 34
15. y ! x2 # 7x # 100
y # 10x ! 30
17. 3x " y ! " 2
2x2 ! y
18. y ! 3x2 # 21x " 5
"10x # y ! "1
H.W. Pg. 164-165 #’s 4,6,10,16,20,22,28,30,32,36,55
Systems of Linear and Quadratic Equations
NY 755
NY 755
```