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

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