Download Grade 10 Academic Math

Grade 10 Academic Math
Chapter 1 – Linear
Systems
Elimination and Substitution
Days 3 through 9
Agenda – Day 3
1. Elimination
2. Types of Word Problems
Learning Goal
By the end of the lesson(s)…
… students will be able to…
linear systems using algebra
Curriculum Expectations
• Students will: - solve systems of two
linear equations involving two variables,
using the algebraic method of substitution
or elimination
• Ontario Catholic School Graduate
Expectations: The graduate is expected
to be… a self-directed life long learner
who CGE4f applies effective… problem
solving… skills
Mathematical Process
Expectations
• Selecting Tools and Computational
Strategies – select and use a variety of…
computational strategies to investigate
mathematical ideas and to solve
problems
Unidentified student after first
day of math after summer
holidays…
How can we solve this?
• A family goes to a fish
store to buy some fish.
They get a new fish tank
and 20 small fish for a
total of $95.
Later that same day
another customer buys
the same fish tank and 12
of the same small fish for
a total of $83.
How much does a fish
tank cost? How much do
the small fish cost?
How can we solve this?
• Thinking
• Communication
How can we solve this?
• At a local building supply store:
One customer has ordered 75
pieces of lumber (2x4s) and 2
large buckets of screws.
The wood and screws weigh a
total of 442.5 lbs.
Another customer ordered 25
pieces of lumber (2x4s) and
one large bucket of screws.
The total weight of that order is
152.5 lbs.
How much does a bucket of
screws weigh? How much
does one piece of lumber
weigh?
How can we solve this?
• Thinking
• Communication
Elimination
The elimination method is the process of
eliminating one of the variables in a
system of equations using addition in
conjunction with multiplication or division
and solving the system of equations
Example 1 – Easiest case…
2x + y = 9
3x – y = 16
5
25
x
5
5
(#1)
(#2)
Example 1 – Easiest case…
2x + y = 9
(#1)
3x – y = 16
(#2)
-------------------------5x + 0y = 25
(#3) Add (#1) & (#2)
5x = 25
Simplify
5
25
x
5
5
x=5
Example 1 – Cont’d
Take your answer of x=5 and substitute it back
into equation (#1) or (#2), whichever is easier…
2(5) + y = 9
10 + y = 9
y = 9 – 10
y = -1
Check your answer…
• So your answer is (5,-1)
• To check your answer
2x+y=9 (#1)
3x-y=16 (#2)
2(5)+(-1)=9
3(5)-(-1)=16
10-1=9
15+1=16
9=9
16=16
True
True
Example 2 – x’s or y’s same but
same sign…
x – 2y = –9 (#1)
x + 3y = 16 (#2)
Example 2 – x’s or y’s same but
same sign…
x – 2y = –9 (#1)………
x + 3y = 16 (#2) mult by -1
Now add (#1 & #3)
x – 2y = –9 (#1)
-x - 3y = -16 (#3)
----------------------0x – 5y = -25 (#4)
-5y = -25
5
25
y
5
5
y= 5
Example 2 – Cont’d
Take your answer of y=5 and substitute it back
into equation (#1) or (#2), whichever is easier…
x – 2(5) = –9 (#1)
x – 10 = -9
x = -9 +10
x=1
Check your answer…
• So your answer is (1,5)
• To check your answer
x-2y=-9 (#1)
x+3y=16 (#2)
1-(2)(5)=-9
1+(3)(5)=16
1-10=-9
1+15=16
-9=-9
16=16
True
True
Example 3 – What do I do when my
equations look different?
2x +1 = y (#1)
y = 3x
(#2)
Example 3 – What do I do when my
equations look different?
2x +1 = y (#1)………
2x +1 = y (#1)
y = 3x
(#2)……… rearrange… -3x = -y (#3)
Now add (#1 & #3)
-----------------------x + 1 = 0 (#4)
-x = -1
1x 1

1 1
x=1
Example 3 – Cont’d
Take your answer of x=1 and substitute it back
into equation (#1) or (#2), whichever is easier…
y = 3(1) (#2)
y=3
Don’t forget to check!!!
Example 4 – What do I with
decimals?
0.06x +0.9y = 0.42 (#1)
0.3x - 0.5y = 0.1 (#2)
100 y 40

100 100
Example 4 – What do I with
decimals?
0.06x +0.9y = 0.42 (#1) x 100…6x + 90y = 42 (#3)
0.3x - 0.5y = 0.1 (#2) x 10…3x - 5y = 1 (#4)
Then,
(#4) x -2…-6x +10y = -2 (#5)
Now, add (#3) and (#5)
0x +100y=40 (#6)
100 y 40

100 100
y = 0.4
Example 4 – Cont’d
Take your answer of y=0.4 and substitute it back
into equation (#1) or (#2), whichever is easier…
0.3x - 0.5(0.4) = 0.1 (#2)
0.3x – 0.2 = 0.1
0.3x = 0.3
x=1
Example 5 – What do I do with
fractions?
x y 1
  (#1)
4 3 2
2x y
  8(# 2)
5 2
Example 5 – What do I do with
fractions?
x y 1
  (#1)
4 3 2
2x y
  8(# 2)
5 2
Multiply each equation by its LCD &
simplify…
(#1) x 12-> 3x+4y=6 (#3)
(#2) x 10-> 4x+5y=80 (#4)
Then, multiply each equation by
number necessary to get a variable
opposite but equal
(#3) x 4-> 12x+16y=24 (#5)
(#2) x -3->-12x-15y=-240 (#6)
(#5)+(#6)
y=-216
Example 5 – Cont’d
Take your answer of y=-216 and substitute it back
into equation (#1) or (#2), OR (#3) or (#4)
whichever is easier…
3x+4y=6 (#3)
3x+4(-216)=6
3x-864=6
3x=6+864
3x=870
x=290
Don’t forget to check!!!
Day 3 Practice
• all using elimination
– Text p.101, #2,3,6odd
Day 4 Practice
• all using elimination
– Text p.51, #11,13,15-21,23,24,26,28,33
– Text p.61, #12,15
– Text p.84, #15-17
– Text p.93, #14
– Text p.102, #5
– Text p.103, #14-18
Agenda – Day 4
1. Substitution
How can we solve this?
• A store in the mall
has a sale on shirts
and sweaters. All
shirts cost $30 less
than the sweaters.
Together, 2 shirts and
2 sweaters cost $360.
• How much does one
shirt cost?
How can we solve this?
• Thinking
• Communication
How can we solve this?
• At a different store,
5 shirts and a pair
of pants cost $233.
Pants cost $23
more than shirts.
How much does
one shirt cost at
this store?
How can we solve this?
• Thinking
• Communication
Definition of Substitution
• a person or thing acting or serving in place of
another…
• In linear relations, the method of solving "by
substitution" works by solving one of the
equations (you choose which one) for one of the
variables (you choose which one), and then
plugging this back into the other equation,
"substituting" for the chosen variable and solving
for the other. Then you back-solve for the first
variable.
Substitution example 1
• y = x - 1 (#1)
• y = 2x - 3 (#2)
• You substitute the x – 1 from (#1) into y in
(#2)
• This gives you x – 1 = 2x-3 (#3)
Substitution example 1
• y = x - 1 (#1)
• y = 2x - 3 (#2)
Substitution example 1 (cont’d)
x – 1 = 2x-3 (#3)
-1 + 3 = 2x - x
2= x
x=2
Now solve for x
Collect like terms
Simplify & solve
Switch sides
Substitution example 1 (cont’d)
• Now take your answer of x=2 and plug it
back into either equation (1) or (2)… pick
whichever is the easiest
y = x - 1 (1)
y= 2 - 1
y=1
Check your answer…
• So your answer is (2,1)
• To check your answer
y = x – 1 (#1)
y = 2x - 3 (#2)
1=2–1
1 = 2(2) -3
1=1
1=1
True
True
Substitution example 2
• x+4y=6 (#1)
• 2x-3y=1 (#2)
Substitution example 2
• x+4y=6 (#1)….. Rearrange to x=6-4y (#3)
• 2x-3y=1 (#2)
• You substitute the 6-4y from (#3) into x in
(#2)
• This gives you 2(6-4y)-3y=1 (#4)
Substitution example 2 (cont’d)
2(6-4y)-3y=1 (#4) Now solve for y
12-8y-3y=1
Use distributive law
-11y=1-12
Collect like terms
-11y=-11
Simplify
11
11
y
11
11
y=1
Mental health break…
Substitution example 2 (cont’d)
• Now take your answer of y=1 and plug it
back into either equation (1) or (2)… pick
whichever is the easiest
x+4y=6 (1)
x+4(1)=6
x+4=6
x=6-4
x=2
Check your answer…
• So your answer is (2,1)
• To check your answer
x+4y=6 (#1)
2x-3y=1 (#2)
2+4(1)=6
2(2)-3(1)=1
6=6
4-3=1
True
1=1
True
Humour Break
Day 5 Practice
• Text p.93, #1-4(all (f) only),
•
6,7f,8f,9,10,12bfi
Day 6 Practice
• all using substitution
– Solve p.93, #13,14
– Solve p.94, #16,18, p.95, #22
– Solve p.51, #10,12,14,22,25,32
Humour Break