How to solve a Algebraic Equation

Topic: Algebraic Equation

An algebraic equation over the rationals can always be converted to an equivalent one in which the coefficients are integers (where equivalence refers to the fact that the two equations will have the same solutions). For example, multiplying through by 42 = 2·3·7, the algebraic equation above becomes the algebraic expression.

For example:

Question:

Find x for the following sum given below

1.) 5 + 2x = 10 - 3x

Answer:

5 + 2x = 10 - 3x

  + 3x    + 3x
______   ______

5 + 5x = 10

-5       -5
______    _______

 5x/5  =  5/5

     x = 1 

Question:

Find X for the following sum given below

2.) 3(2x -5) = 10 (2x - 1)

Answer:

3(2x -5) = 10 (2x - 1)

 6x - 15 = 20x -10

     +15 = +15
   _____   ______

      6x = 20x + 5

    -20x = -20x
   _____   _______

-14x / -14 = 5 / 14


       x = -5 / 14

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Problem on Algebraic Surds

Surd with an algebraic expression can be simplified as shown in the example below.
Finding cube root and square root are basic of surd.


Topic : Solving a surd

In the problem shown below explains how to find square root of any algebraic expressions and how to verify the solutions. For more problems on surd you can see indice math.

Problem : Solve algebraically and check your potential solutions: √(t +12) − t = 0

(Alternately, you may type √x as square root(x) and show raising
to the nth power as ^n)

Solution :

√(t +12) − t = 0

square root(t + 12) - t = 0

Squaring both the sides we get..

t + 12 = t^2

=> 0 = t^2 – t – 12

=> t^2 – t - 12 = 0

=> t^2 – 4t + 3t – 12 = 0

=> t(t – 4) + 3(t – 4) = 0

=> (t – 4)(t + 3) = 0

=> t – 4 = 0 and t + 3 = 0

=> t = 4 and t = -3

Now verifying the solutions

when t = 4

L.H.S = root(t + 12) – t

= root(4+12) – 4

= root(16) – 4

= 4 – 4

= 0

= R.H.S

Hence t = 4 is a solution of the given equation.

when t = -3

L.H.S = root(t + 12) – t

= root(-3+12) – (-3)

= root(9) + 3

= -3 + 3

= 0

= R.H.S

Hence t = -3 is a solution of the given equation.



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