Week 2, Day 2 Solving linear equations and application ... · 4 D D 3 4 = D 4 3 = 4D 3 Week 2, Day 2 Solving linear equations and application problemsSeptember 8th, 2014 8 / 1. We

Week 2, Day 2Solving linear equations and application problems

September 8th, 2014

Week 2, Day 2 Solving linear equations and application problemsSeptember 8th, 2014 1 / 1

Linear equations in one variable

A linear equation in one variable, x say, is an equation that equates twoexpressions that are sums and differences of multiples of x and constants.The set of values of x that satisfy the equation is called the solution set.There are three possibilities for solution sets

The most common situation: there is exactly one solution

5x + 7 = 32 ⇒ solution set = {5}

The equation is an identity

7x + 9 = 7x + 9 ⇒ solution set = R

The equation has no solutions

2x + 4 = 2x + 7 ⇒ solution set = ∅


Solving linear equations in one variable

To solve linear equations

First clear any fractions : Multiply the equation by a commondenominator (a common multiple of the denominators in theequation).

Distribute any parentheses.

Combine like items

Isolate the variable ; add or subtract multiples of the variable andconstants from the appropriate sides of the equation so as to leaveonly multiples of x on one side of the equation and constants on theother

If the equation isn’t either an identity or a contradiction, divide by thecoefficient of x to obtain x = the solution

Check the answer you got satisfies the equation


Example 1

Solve the linear equation

6(x + 2)− 3x = 3(x + 1) + 9

First we use the distributive property

6x + 12− 3x = 3x + 3 + 9

Combining like terms on each side of the equals sign

3x + 12 = 3x + 12

This is true for any value of x so the solution set is the set of all realnumbers

Solution set = R


Example 1


6(x + 2)− 3x = 3(x + 1) + 9


6x + 12− 3x = 3x + 3 + 9


3x + 12 = 3x + 12


Solution set = R


Example 1


6(x + 2)− 3x = 3(x + 1) + 9


6x + 12− 3x = 3x + 3 + 9


3x + 12 = 3x + 12


Solution set = R


Example 2

Solve the equation

1

2(5x − 4)− 5

3(x − 2) =

2

5(3x + 1)

First clear the fractions. A common denominator is 30. Multiply theequation by this gives

15(5x − 4)− 50(x − 2) = 12(3x + 1)

Now distribute75x − 60− 50x + 100 = 36x + 12

Combining like terms gives

25x + 40 = 36x + 12


Example 2

Solve the equation

1

2(5x − 4)− 5

3(x − 2) =

2

5(3x + 1)

First clear the fractions.

A common denominator is 30. Multiply theequation by this gives

15(5x − 4)− 50(x − 2) = 12(3x + 1)



25x + 40 = 36x + 12


Example 2

Solve the equation

1

2(5x − 4)− 5

3(x − 2) =

2

5(3x + 1)


15(5x − 4)− 50(x − 2) = 12(3x + 1)



25x + 40 = 36x + 12


Example 2

Solve the equation

1

2(5x − 4)− 5

3(x − 2) =

2

5(3x + 1)


15(5x − 4)− 50(x − 2) = 12(3x + 1)



25x + 40 = 36x + 12


Example 2

Solve the equation

1

2(5x − 4)− 5

3(x − 2) =

2

5(3x + 1)


15(5x − 4)− 50(x − 2) = 12(3x + 1)



25x + 40 = 36x + 12


Subtract 25x and 12 from both sides of the equation

28 = 11x

Dividing by the coefficient of x , i.e. 11, gives the solution

x =28

11= 2

6

11

Now substitute this into the original equation

1

2

(5× 28

11− 4

)− 5

3

(28

11− 2

)=

2

5

(3× 28

11+ 1

)1

2

(140

11− 44

11

)− 5

3

(28

11− 22

11

)=

2

5

(84

11+

11

11

)1

2

(96

11

)− 5

3

(6

11

)=

2

5

(95

11

)⇒ 48

11− 10

11=

38

11


Subtract 25x and 12 from both sides of the equation

28 = 11x

Dividing by the coefficient of x , i.e. 11, gives the solution

x =28

11= 2

6

11

Now substitute this into the original equation

1

2

(5× 28

11− 4

)− 5

3

(28

11− 2

)=

2

5

(3× 28

11+ 1

)1

2

(140

11− 44

11

)− 5

3

(28

11− 22

11

)=

2

5

(84

11+

11

11

)1

2

(96

11

)− 5

3

(6

11

)=

2

5

(95

11

)⇒ 48

11− 10

11=

38

11


General problem solving strategy

1 Read the problem carefully

2 Define the problem i.e identify what you are required to find

3 Assign the variables

4 Translate the problem into an equation

5 Solve the equation

6 Check your answer

7 Answer the original question


A uniform motion problem

Landon can climb a certain hill at a rate 2.5 mph slower than his ratedescending the hill. It takes him 2 hours to climb the hill and 45 minutesto come down the hill. What was his rate coming down the hill?

To find Landons rate of descent we need to find the distance to the top ofthe hill.

Let the distance to the top of the hill be D

Time(hours) Distance(miles) Rate=Distance/Time

Ascending 2 D D2

Descending 34 D D ÷ 3

4 = D × 43 = 4D

3







Ascending 2 D D2


4 = D × 43 = 4D

3







Ascending 2 D D2


4 = D × 43 = 4D

3


We are told that the rate of ascent is 2.5 mph slower than the rate ofdescent. Translating this into an equation gives

4D

3− 2.5 =

D

2

To clear the equation of fractions we need to multiply the equation by thelowest common denominator,which in this case is 6.

6

(4D

3− 2.5

)= 6

(D

2

)Distributing

24D

3− 15 =

6D

2

Simplifying the fractions gives

8D − 15 = 3D



4D

3− 2.5 =

D

2

To clear the equation of fractions we need to multiply the equation by thelowest common denominator,

which in this case is 6.

6

(4D

3− 2.5

)= 6

(D

2

)Distributing

24D

3− 15 =

6D

2


8D − 15 = 3D



4D

3− 2.5 =

D

2


6

(4D

3− 2.5

)= 6

(D

2

)

Distributing24D

3− 15 =

6D

2


8D − 15 = 3D



4D

3− 2.5 =

D

2


6

(4D

3− 2.5

)= 6

(D

2

)Distributing

24D

3− 15 =

6D

2


8D − 15 = 3D



4D

3− 2.5 =

D

2


6

(4D

3− 2.5

)= 6

(D

2

)Distributing

24D

3− 15 =

6D

2


8D − 15 = 3D


Adding 15 to both sides of the equation gives

8D = 3D + 15

Now subtracting 3D from both sides of the equation

5D = 15

Dividing by 5 gives5D

5=

15

5D = 3 miles

So Landon’s rate of descent will be

Rate of descent =4D

3=

4× 3

3=

12

3= 4 mph



8D = 3D + 15


5D = 15


5=

15

5D = 3 miles


Rate of descent =4D

3=

4× 3

3=

12

3= 4 mph



8D = 3D + 15


5D = 15


5=

15

5D = 3 miles


Rate of descent =4D

3=

4× 3

3=

12

3= 4 mph



8D = 3D + 15


5D = 15


5=

15

5D = 3 miles


Rate of descent =4D

3=

4× 3

3=

12

3= 4 mph


Now checking that the answer is reasonable.

The rate of ascent is given by

Rate of ascent =D

2=

3

2= 1.5 mph

andRate of descent− 2.5 = 4− 2.5 = 1.5 = Rate of ascent

so the solution we have found

Rate of descent = 4 mph

satisfies the question that was asked.


Group Work 9/8/2014

1 Solve the linear equation

(3x + 6) + 2x + 2 = 5(x − 1) + 10

2. Solve the linear equation

3

5(2x − 4) + x − 8 = −7

2(x − 3)

3. Manuel traveled 3 hours nonstop to Mexico, a total of 112 miles. Hetook a train part of the way, which averaged 50 mph. He then took a busfor the remaining distance, which averaged 30 mph. How long was Manuelon the bus?


Group Work Solutions

1 This is a contradiction it has solution set = ∅.

2. The solution set is {113 }.

3. Let T be the time Manuel spent on the bus then

Distance on bus = 30T Distance on train = 50(3− T )

Total distance = 112 = 30T + 50(3− T )

This has solution T = 1910 = 1 9

10 hours.


Documents

Week 2, Day 2 Solving linear equations and application ... · 4 D D 3 4 = D 4 3 = 4D 3 Week 2, Day 2 Solving linear equations and application problemsSeptember 8th, 2014 8 / 1. We