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Grade 10 Academic Math Chapter 1 – Linear
SystemsElimination 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 (#1)
3x – y = 16 (#2)
5 25
5 5x
Example 1 – Easiest case…2x + y = 9 (#1)
3x – y = 16 (#2)
--------------------------
5x + 0y = 25 (#3) Add (#1) & (#2)
5x = 25 Simplify
x=5
5 25
5 5x
Example 1 – Cont’dTake 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 – 2y = –9 (#1)
x + 3y = 16 (#2) mult by -1 -x - 3y = -16 (#3)
Now add (#1 & #3) -----------------------
0x – 5y = -25 (#4)
-5y = -25
5 25
5 5y
y= 5
Example 2 – Cont’dTake 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 = -11 1
1 1
x
x = 1
Example 3 – Cont’dTake 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 40
100 100
y
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 40
100 100
y
y = 0.4
Example 4 – Cont’dTake 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?
1(#1)
4 3 22
8(#2)5 2
x y
x y
Example 5 – What do I do with fractions?
1(#1)
4 3 22
8(#2)5 2
x y
x y
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’dTake 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) Now solve for x
-1 + 3 = 2x - x Collect like terms
2= x Simplify & solve
x = 2 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
y=1
11 11
11 11y
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