Algebraic Expression

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Review – Types of Equations ,  PEMDAS ,  Identity

An expression which presents constants, variables, mathematical operations (addition, subtraction, multiplication and division) is known as Algebraic Expression.

45x,  4x+9,  (5/6)x2 – 9 ,    a2  + b 2   all are algebraic expressions.

Let us understand with the following equation

6x2 +4x2 + 6y+ 11

Where;

x and y are variables

6x2  ,  4x2  –   are called as  like terms  because both are having common variable x2

11  –  It is constant term as there is no variable with it.

6x2  –    The number written with variable is called as coefficient of a term

Equations

When the two algebraic expressions  are equal  , we solve it to find the values of variables.

Types of Equation

  1. Linear Equation    – With one variable

e.g.   10a + 4a – 6  = 15

Solve the following Equations

Ex.1                  5x + 3  = 7 – (3x – 3)

Answer;           5x  + 3  = 7 – 3x + 3

Transpose like terms on one side

5x  + 3x = 10 – 3

8x = 7

x = 7/8

Let us check the equation by substituting the value of  x = 7/8

5 x 7/8  + 3 =7 – 3×7/8  + 3

35/8   + 3   = 10 – 21/8

35 + 24 /8   = 80 -21 / 8

59/8   =   59/8

So, Left Hand Side  = Right Hand Side , that means our answer is correct.

Ex.2              7 +  15 x  = 30x – 5

Answer ,

As this is an equation we can rewrite this expression  as follows

30x – 5 =  7 + 15x

Transpose like terms toghether

30x – 15x =  7 + 5

15x  = 12

x = 12/15

Reduce the fraction to its lower term

x = 4/5

Now let us check  by substituting the value of x in equation

7+ 15 x 4/5  = 30 x 4/5 – 5

7 + 12 =24 -5

19  = 19

So, LHS  = RHS

Exercises

  1. 15 + x = 6                        12a +6 =3a-9                  3.      5b -90-6 =  24b +1-8b
  2. -5 – 6x  =  -4  + 7x         5.    3a +  10  = -16               6.        2(x-1)  + 6  = 12x

2. Linear Equation with two variables   – Follow the link for more explanation

like ; 5b – 3a  = 6   , where a, b are two varibles

 3.  Quadratic Equations with one variable

e.g.    50a2 + 20a – 6 = 0

PEMDAS

To solve any equation we need to apply PEMDAS rule.

P  –   Parenthesis

E  –  Exponents

M –  Multiplication

D –   Division

A  –  Addition

S  – Subtraction

Ex.3    Solve the equation   6(3a+9) –(-5 + 7a)  = 2(12a – 5)  + (6a – 7)

Answer

6 x 3a  + 6 x9  + 5  – 7a = 2 x 12a  – 2 x 5  + 6a – 7

18a  + 54   + 5  -7a  =   24a  – 10  + 6a  – 7

Transpose like terms together

18a – 7a  – 24 a – 6a  = -10  – 7   – 54  – 5

– 19 a  =  – 76

a  = -76 /-19

a = 4

Identity

An identity is an equation between two algebraic expressions that is true for all variables.

Following are the standard Identities

  1. x2 – y 2  =  (x-y)  (x+ y )

  2. ( x – y ) 2  =   x 2  – 2xy  + y2

  3. ( x  + y) 2   = x 2  + 2xy + y2

  4. ( x – y) 3 = x3  – 3x2y + 3xy2 – y3

  5. ( x + y ) 3 = x3  + 3x2y + 3xy2 + y3

  6. X3 – y 3  =  (x- y) (x2  +xy + y2 )

  7. X3 + y 3  =  (x+ y) (x2  – xy + y2 )

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