REVIEW: 6.1 Solving by Graphing:
Remember:
To graph a line we use the slope intercept
form:
y = mx +b
Slope =
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
𝑹𝒊𝒔𝒆 ⍙𝒚
=
𝑹𝒖𝒏 ⍙𝒙
STARING POINT
(The point
where it crosses
the y-axis)
System Solution: The point where the two
lines intersect (cross):
(1, 3)
Remember: What are the requirements
for this to happen?
REVIEW: 6.2: Solving by Substitution:
0): THINK - Which variable is the easiest to
isolate?
1): Isolate a variable
2): Substitute the variable into the other
equation
3): Solve for the variable
4): Go back to the original equations,
substitute, solve for the second variable
5): Check
6.3: Solving by Elimination:
0): THINK: Which variable is easiest to
eliminate.
1): Pick a variable to eliminate
2): Add the two equations to Eliminate a
variable
3): Solve for the remaining variable
4): Go back to the original equation,
substitute, solve for the second variable.
5): Check
NOTE:
We can solve system of equations
using a graph, the substitution or
eliminations process.
The best method to use will depend
on the form of the equations and how
precise we want the answer to be.
CONCEPT SUMMARY:
METHOD
Graphing
WHEN TO USE
When you want a visual
display of the equations, or
when you want to estimate
the solution.
http://player.discoveryeducation.com/index.cfm?guidAs
setId=8A6198F2-B782-4C69-8F6D8CD683CAF9DD&blnFromSearch=1&productcode=US
YOU TRY IT:
Solve the system by Graphing:
−2𝑥 + 𝑦 = 2
6𝑥 + 2𝑦 = 14
YOU TRY IT: (SOLUTION)
−2𝑥 + 𝑦 = 2
6𝑥 + 2𝑦 = 14
→ 𝐘 = 𝟐𝐗 + 𝟐
→ 𝒀 = −𝟑𝑿 + 𝟕
(1,4)
CONCEPT SUMMARY:
METHOD
WHEN TO USE
Substitution When one equation is
already solved:
y=mx+b or x= ym+b .
 2 x  7  2

y  x2
http://www.khanacademy.org/mat
h/algebra/systems-of-eq-andineq/fast-systems-ofequations/v/solving-linearsystems-bysubstitution?exid=systems_of_equ
ations
http://player.discoveryeducation.com/index.cfm?guidAssetId=A9199767-40AB-4AD1-9493-9391E75638D0
YOU TRY IT:
Solve the system by Substitution:
−2𝑥 + 𝑦 = 2
6𝑥 + 2𝑦 = 14
YOU TRY IT:(SOLUTION)
−𝟐𝒙 + 𝒚 = 𝟐 → 𝐲 = 𝟐𝐱 + 𝟐
6𝑥 + 2𝑦 = 14
6𝑥 + 2(2𝑥 + 2) = 14
6𝑥 + 4𝑥 + 4 = 14
10𝑥 = 10  x = 1
(𝟏, 𝟒)
y=2 1 +2 →4
CONCEPT SUMMARY: (continue)
METHOD
WHEN TO USE
Elimination When the equations are in
Ax +By = C form or the
coefficients of one variable
are the same and/or opposites
 2 x  5 y  17

6 x  5 y  9
http://www.khanacademy.org/mat
h/algebra/systems-of-eq-andineq/fast-systems-ofequations/v/solving-systems-ofequations-by-elimination
http://player.discoveryeducation.com/index.cfm?guidAssetId=02B482AE-EB9F-4960-BC5C7D2360BDEE66
YOU TRY IT:
Solve the system by Elimination:
−2𝑥 + 𝑦 = 2
6𝑥 + 2𝑦 = 14
YOU TRY IT: (SOLUTION)
−𝟐𝒙 + 𝒚 = 𝟐 → −𝟐(−𝟐𝒙 + 𝒚 = 𝟐)
𝟔𝒙 + 𝟐𝒚 = 𝟏𝟒
𝟒𝒙 − 𝟐𝒚 = −𝟒
+ 𝟔𝒙 + 𝟐𝒚 = 𝟏𝟒
𝟏𝟎𝒙
= 10
𝟔𝒙 +
𝟐𝒚𝟏=+𝟏𝟒
𝟔
𝟐𝒚 =𝟔𝟏𝟒
+ 𝟐𝒚 =
𝟏𝟒 𝟐𝒚 = 𝟖
 x=1
 y=4
ADDITIONALLY: System of equations
help us solve real world problems.
http://player.discoveryeducation.com/index.
cfm?guidAssetId=A9199767-40AB-4AD19493-9391E75638D0
VIDEO-Word Prob.
NOTE:
We can solve system of equations
using a graph, the substitution or
eliminations process.
The best method to use will depend
on the form of the equations and how
precise we want the answer to be.
6.4 Application of Linear Systems:
Break-Even Point: The point for business is
where the income equals the expenses.
GOAL:
MODELING PROBLEMS: Systems of
equations are useful to for solving and
modeling problems that involve
mixtures, rates and Break-Even points.
Ex:
A puzzle expert wrote a new
sudoku puzzle book. His initial costs are
$864. Binding and packaging each book
costs $0.80. The price of the book is
$2.00. How many books must be sold to
break even?
SOLUTION:
1) Write the system of equations
described in the problem.
Let x = number of books sold
Let y = number of dollars of expense
or income
Expense: y = $0.80x + 864
Income: y = $2x
SOLUTION: (Continue)
2) Solve the system of equations for the
break-even point using the best method.
To break even we want:
Expense = Income
$0.80x + 864 = $2x
864 = 2x -0.80x
864 = 1.2x
720 = x
There should be 720 books sold for the
puzzle expert to break-even.
YOU TRY IT:
Ex:
A fashion designer makes and sells
hats. The material for each hat costs
$5.50. The hats sell for $12.50 each. The
designer spends $1400 on advertising.
How many hats must the designer sell
to break-even?
SOLUTION:
1) Write the system of equations
described in the problem.
Let x = number of hats sold
Let y = number of dollars of expense
or income
Expense: y = $5.50x + $1400
Income: y = $12.50x
SOLUTION: (Continue)
2) Solve the system of equations for the
break-even point using the best method.
To break even we want:
Expense = Income
$5.50x + $1400 = $12.50x
1400 = 12.5x -5.50x
1400 = 7x
200 = x
There should be 200 hats sold for the
fashion designer to break-even.
VIDEOS:
Special Linear
Equations
https://www.khanacademy.org/math/algebra/syst
ems-of-eq-and-ineq/fast-systems-ofequations/v/special-types-of-linear-systems
CLASSWORK:
Page 386-388
Problems: As many as needed
to master the
concept.
SUMMARY:
http://www.bing.com/
videos/search?q=SYSTE
M+OF+EQUATIONS+&v
iew=detail&mid=2CFE6
3B47EDB353AFDCF2CF
E63B47EDB353AFDCF&
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