subject: Different Methods In Algebra [print this page] In introduction to different methods in algebra, Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics. Let us see some different methods in algebra.
(Source: Wikipedia)
Example for Different Algebra Problems:
1.) Problem:
Solve the following equation: 2 (x - 1) = 8
Solution:
2 (x - 1) = 8
2x 2 = 8
2x = 8 + 2
2x = 10
X = 10 / 2
X = 5
2.) Free problems on cross multiplication:
Problem: Solve the following equation:
5y 4 / 2y = 8 / 10
Solution:
(5y 4) * 10 = 16y by cross multiplication
50y 40 = 16y
50y 16y = 40
34y = 40
Y = 40 / 34
Y = 20 / 17
3.) Fraction method:
Problem:
Add the fractions 12 / 24 + 10 / 24 and reduce your answer.
Solution:
The denominators are the equal, so we can skip. Add the numerators for the numerator in the result. 12 + 10 = 22. The addition of the two fractions are,
The denominators are different values, so we need build every fraction to a form wherever the two has the similar denominator. Hence 5 divides into 10 uniformly, we have to build the fraction 2 / 5 to a same fraction with a
2 / 5 = 2 / 5 * 1 = 2 / 5 * 2 / 2 = 4 / 10
Denominator is 10.
2 / 10 2 / 5 can now be written as 2 / 10 4 / 10
Merge the numerators for the numerator in the solution:
2 - 4 = -2.
2 / 10 4 / 10 = (2 4) / 10 = - 2 / 10
Therefore the solution is:
Now we have to reduce the fraction.
- 2 / 10 = 2 * 1 / 2 *5 = -1 / 5
5.) Quadratic equations method:
Problem: Find out the roots of the equation
X2 -1 = 0.
Solution: This given equation is equivalent to
X2 = 1
Because 1 has two square-roots {1, - 1}, the answer for this equation is
X= 1 or x = -1
6.) Factorization and Roots of Polynomials method:
Problem:
Write a polynomial with its roots x = 1, x = 2, and x = 3 / 4.
Solution:
Because f (1) = 0, the polynomial should have a linear factor (x - 1). Likewise, it has factors of the form (x - 2) and (x 3 / 4). So a feasible choice is
F (x) = (x 1) (x 2) (x 3 / 4 ).
Here other different two more choices:
F (x) = (x 1) (x 2) (4x 3),
Or
F ( x) = ( x 1 ) 2 ( x 2 ) ( x 3 / 4 ) ( x2 5x + 7 ).
