N53-G23 Addition/Subtraction with Linear and Quadratic Denominator

Remember the Algebraic Fractions simplification:

       y  + y - 3

       3       4

 

   = 4 x y + 3 (y - 3).    (Step 1 : Multiply each for same denominator (L.C.M))

       4 x 3 + 3 x4

    = 4y + 3y - 9              (Step2 : Combine into a single fraction)

             12                    (Step 3 : Simplify : Do Order of Operations)

    = y - 9

         12   

<< L2 - NA/O>>

Linear denominator

Similarly, for algebraic expression with variable in the denominator, we simplify the expression accordingly. We can 'treat' the variable as a 'unique' number

Example

a.      2  - y - 3

         y       4

     =    2     -   (x -3)              (Step 1 : Multiply each to get same denominator)

        4 x y      4 x y

     =    2 - x + 3                 (Step2 : Combine to 1 fractional term)

             4y                   (Step 3 : Simplify, - x - = +)

     =    5 - x

             4y

Quadratics Denominator

Example

Simplify  3      +      2   

             y – 3          y

    

     =        3      +      2                  (Step1: Identify common factors)

             y – 3          y                   (Common factor of y, (y – 3) = y x (y – 3))

     =   y x   3      +   2 x (y - 3)      (Step2 : Combine to 1 fractional term)

          y x (y – 3)     y x(y – 3)

     =   3y  +   2 (y - 3)           

              y(y – 3)      

       =   3y  +   2y - 6                       (Step 3 : Simplify, - x - = +)

              y(y – 3)    

       =        5y - 6                      

             y(y – 3)    

Example

Simplify  1    +      2   

             x – 3    (x – 3)²

        =    1    +      2                  (Step1: Identify common factors)

            y – 3    (y – 3)²              (y – 3)(y – 3)=> (y – 3)² = C.factor

                 

        =     (x – 3)        +      2     (Step2 : Combine to 1 fractional term)

            (y – 3)(y – 3)    (x – 3)²

Step3: Simplify

               = x – 3 + 2

                   (x – 3)²

               = x – 1

                 (x – 3)²

Practice

1. Simplify the following:

       1     +      2   

     x -  2       x  -  3

(b)    1     +        2       

     x² -  9       x  -  3

(c)    1     +        2       

     x -  2          (x  -  2)²

2. Solve for x

(a)  x  + x - 3.   = 6

       3       4

b.         3       = 6

         x – 4

<< End of Algebra Topics O level at Sec 2, NA at Sec3>>

AL1-L32 Algebra Practice : Notation, Algebraic Expression

(Part 1) TEACHING(15-20mins)                                         

1. Using a letter to represent a known no, notation, representation and interpretation

2. Simplify linear expression (excluding bracket)

3. Solving simple linear equation (whole number coefficient)

(Part2) Core PRACTICE (35-55mins)                 

State the algebraic expression for each of the following.

1.  Add 5 to K

2.  Y more than 8

3.  Subtract x from 12

4. 11 less than z

Find the value of each expression when p =6

5    10 p – 13

6.    12 – 4 p

                3

7.    3p + 1

         4      2

8.    7p + 9

            3

Simplify

9.    3y +4y – 2y

10.  8c + 5 – 2c -3

11.  5t + 8 – t

(II) Word Problems 

12. Find the value of 8𝑤− 3w if 𝑤= 4 (18/s/17)
                                           2

13.  What is the value of 21w – w +1 when w = 2? 

                                        10

14. *  What is the value of 10 + 3y  when y = 4?

                                                 2

For Question 15 and 16.
John has y sweet. Peter has 4 times as many sweets as John. Peter bought another 9 sweets.

15. Find the number of sweets Peter has more than John after he bought another 9 sweets in terms of y.

16. If John has 27 sweets. How many more sweet does John have than Peter?

Extra

1.  Mary had $4y. After buying some cloth at $7 per metre, she had $y left. How many metre of cloth did she buy?

2.   At a concert, there were X men and twice the number of women. During the interval, 4 men and 7 women left the concert.

(a)   How many people left during the interval?

(b)   How many people were there at the concert after the interval?

(c)   If X = 400, how many people were there at the convert after the interval?

3.*  Randy sold (4k + 1) tickets on Monday. He sold K more tickets on Tuesday than on Monday. How many tickets did he sell altogether? Give your answer in terms of k in the simplest form.

4.*  Three boys collected plastic bottles for recycling. Jona collected 2Y bottles which was half as many as what Zack collected. Zack collected 7 bottles more than Lionel. Hoe many bottle did they collect altogether? Give your answer in terms of Y in the simplest form.

5.*  Joyce had 4w apples. She ate 1 apple and gave w apples to her friend. Joyce’s sisters then ate half of the remaining apples. Hoe many apples had she left? Give your answer in terms of w in the simplest form.

6. Find the value of each expression when q = 2

a    11q – 13

b.   42 – 4q

                5

c.    9q + 27

         4

d.    14 p + 5
             7

7. State the algebraic expression.

a.  Add 2p to 10

b.  12 more than K

c. Subtract 8 from 3J

d. Y less than 4

8.  Simplify

a.  3y + 7 + y – 2

b.  4d - 5 – 2d - 3

TheMathbooklets Content Page 
Practice Content Page 

Algebra Answer Key

1.         k+5

2.         Y+8

3.         12 –x

4.         z-11

5.         47

6.         6

7.         5

8.         17

9.         5y

10.      6c + 2

11.      4t + 8

12.      26

13.      3.2

14.      60

15.      4y + 9

16.      117

AL2 Algebra : Equation

ALGEBRA

Mathematics where letters (example: X, V or U) and other general symbols are used to represent numbers and quantities in equations

The equal sign   =

The symbol shows what is on the left of the symbol is the same as what is on the right.

Example

                        4 = 1+ 1 + 1 + 1

Using the balance

The diagram (a simple balance) is balanced because it is horizontal (straight line parallel to the table) and we can say that the left side is equal to the right side

                  

Example

Four red balls are placed on the right size of the balance. The balance tilted towards the right because the right size is heavier. Now, the left and right sides are not balanced or equal anymore.

                     

Forming an equation

We have U, X, and Y. Which alphabets make both sides equal?

                

                        X is heavier than the 4 red balls

                 

Both sides are balanced. V (the left side) = 4 red balls (right side)

                        V = 4 red balls. V balances both sides.

                   

                        U is lighter than the 4 red balls

We have U, X, and Y. Which alphabets make both sides equal?

What is X?                

                  

Since X (the left side) is heavier than the 4 red balls (the right side), we can add more red balls to try to balance both side. 

                   

After adding 1 red ball, both sides are still not balanced, we can add another red ball.

                

After adding 1 red ball, both sides are balanced.

              X = 6 red balls

What is U? 

                     

Since U (the left side) is lighter than the 4 red balls (the right side), we must by reduce or ‘lighten’ the right side. Remove 1 red ball from the right.

                 

                 Both sides are balanced or equal.

                                U = 3 red balls.

Forming An Equation

                        2U = 6 is an equation

An equation has 

            (1) Variables (those letters U, V, X) 

            (2) equal sign =

            (3) number (2, 6, 14m, 9 kg) 

           (4) operators (+ , - , x , ¸) 

We solve the equation to find the value of the variable that make the equality true.

This is also an equation:

                        3X + 2 = 8

An equation can also have 

       

Example

Form an equation from the diagram and find the value of U

                

In an equation

                                              U + 1 = 4

The variable is on the left side and equal to values(number) on the right side.

Moving the number (+1) to the right side by minus 1.

             

            We need to also minus 1 from the right side

Balancing the equation,

We minus 1 from the LEFT SIDE AND the RIGHT SIDE

           

                              U + 1 – 1 = 4 - 1

                              U = 3

Example

Find the value of X

                        X + 2 = 6

            X + 2 = 6        (Remove number from left side)

            X + 2 – 2 = 6 – 2 (by -2 from both side)

            X = 6 – 2

               = 4

Example

            Solve 12 = U + 5

METHOD:

Example

Solve 12 = U + 5

METHOD:

    L = R                     Step 1: WRITE L=R putting variable to left side

          12 = U + 5

    U + 5 = 12

    U + 5 – 5 = 12 – 5       Step 2: Move all numbers to right

                  U = 12 - 5

                  U = 7

Example

Solve A – 7 = 14

Step 1:                   L = R

            A – 7 = 14

Step 2 :      A – 7 + 7 = 14 + 7

Step 3:                   A = 21 

Solve for Non-one-value variable

Solve 3U - 2 = 7 

We need to find the value of U and not 3U. How do we find U? 

There are two ways to change to U:

1. Divide by itself  - >  / 3 , 3U / 3 = U
2. Multiply by x 1/number, 3U x 1/3 = U - > multiply by 1/number

Practice: Fill in the blank

 (1)    4U /    ____ = U    [U = 4]

        4U x   _____ = U [U = ¼]

(2)    2U / ______ = U        [U = 2]

        2U x ______ = U        [U = ½]

Fractional U, multiply by the inverse

Example: 

   ¾ U x ______ = U

Step: Inverse of ¾ = 4/3

         ¾ U x 4/3 = U

METHOD To Solve An Equation

Example

Solve 3U - 2 = 7      

                  L = R            Step1 : WRITE L = R and putting variables to left

          3U - 2 = 7

    3U - 2 + 2 = 7 + 2      Step 2 : Move all numbers to right

              3U =  9            Step3 : 1-UNIT Variable, solve

    3        3
      U = 3

Example

Solve ¼ A = 2

Method

        L = R              Step1: WRITE L = R and equation putting variables to left

          ¼ A = 2          Step 2 : Move all numbers to right

          ¼ A = 2           Step3 : 1-UNIT Variable

    ¼ A x 4 = 2 x 4

  A = 8

Example

Solve 2A + 3 = 11

Method

Step1:                  L = R

        2A + 3 = 11

Step2:     2A + 3 -3 = 11 -3

              2A = 8

Step3:          2A / 2 = 8 /2

      A = 4

Solving an Equation

Example

           Solve 3U - 2 = 7

                            L = R              Step1: WRITE L=R , putting variable to left side     

                     3U - 2 = 7

               3U - 2 + 2 = 7 + 2        Step2: Move all numbers to right

                          3U =  9             Step3: 1-UNIT Variable and solve

                          3        3            

                           U = 3

Finding The Value Of An Equation

When the variable is given a value, replace the variables with the value.

Example

Find the value of the equation when A = 4

(1) 4 + A 

= 4 + 4 (replace A with A = 4) 

= 8

(2) ½ A + 3

= ½ x 4 + 3 (replace A with A = 4)

= 2 + 3

= 5

(3) 5A / 2 + 8 

= 5/2 x 4 + 8 (replace A with A = 4)

= 5 x 2 + 8

=10 + 8 = 18

        ~~~~ END ~~~~ :)