Abstract Algebra Homework

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Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras. The phrase abstract algebra was coined at the turn of the 20th century to distinguish this area from what was normally referred to as algebra, the study of the rules for manipulating formulae and algebraic expressions involving unknowns and real or complex numbers, often now called elementary algebra. The distinction is rarely made in more recent writings. Contemporary mathematics and mathematical physics make intensive use of abstract algebra.
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Sample Problem for abstract algebra homework:

Abstract algebra homework problem 1:

Find the product of (7U - 3V) and (4U + 5V), verify the result for U = 3, V = 2.

Solution:

The required product is

(7U - 3V)(4U + 5V)

= 7U (4U + 5V) - 3V (4U + 5V)

= 28U2 + 35UV -12VU -15V2

= 28U2 + 23UV -15V2

Verification L.H.S

= (7U - 3V) (4U + 5V)

= (7 * 3- 3 *2) (4 * 3+ 5 * 2)

= (21- 6) (12 + 10) = 330

Verification on R.H.S

= ...
... 28U2 + 23UV -15V2

= 28 * 32 + 23 * 3 2 -15 * 22

= 252 +138 - 60 = 330

Therefore, L.H.S. = R.H.S.

Hence, our answer is exactly correct.

Abstract algebra homework problem 2:

Find the value of greatest common factor of 2x5 and 12x2 and then factorize 2x5+12x2.

Solution:

Here, the largest number that divides the coefficients of 2x5 and 12x2 is 2.

The highest power of x that is a factor of x5 and x2 is x2.

The greatest common factor of 2x5 and 12x2 is therefore 2x2

Thus,

2x5 + 12x2 = 2x2 * x3 + 2x2 * 6

= 2x2 (x3 + 6)

So, given expression can be expressed as the product of two factors

One of which is the greatest common factor.

Abstract algebra homework problem 3:

Find the coefficients of x3, x2 and x terms in the product of 7x3 – 6x2 – 9x + 8 and 5x2 – 3x + 5 without doing actual multiplication.

Solution:

To get the coefficient of x3:

Coefficient of x3 term in A × Constant term in B = 7 × 5 = 35.

Coefficient of x2 term in A × Coefficient of x term in B = – 6 × – 3 = 18.

Coefficient of x term in A × Coefficient of x2 term in B = – 9 × 5 = –45.

Constant term in A × Coefficient of x3 term in B = 8 × 0 = 0.

So the coefficient of x3 in A × B is 35 + 18 + (– 45) + 0 = 8.

So the coefficient of x3= 8

To get the coefficient of x2:

Coefficient of x2 term in A × Constant term in B = – 6 × 5 = –30.

Coefficient of x term in A × Coefficient of x term in B = –9 × –3 = 27.

Constant term in A × Coefficient of x2 term in B = 8 × 5 = 40.

So the coefficient of x2 in A × B is (–30) + 27 + 40 = 37.

So the coefficient of x2 = 37

To get the coefficient of x term:

Coefficient of x term in A × Constant term in B = –9 ×5 = –45.

Constant term in A × Coefficient of x term in B = 8 × –3 = –24.

So the coefficient of x term in A × B is (–45) + (–24) = – 69.

So the coefficient of x = - 69.

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Practice Problem for abstract algebra homework:

Problem 1:

Solve the given equation x + 9 = 18 - 2x

Answer: x = 3

Problem 2:

Solve the given equation: 11s * 12s^3

Answer: 132s4

answers for basic algebra

Algebra is a branch of mathematics that substitutes letters instead of numbers. A basic algebraic equations represents a scale, what is done on one side of the scale with a number is also done to the other side of the scale. In basic algebra, use constant numbers. The basic Algebra includes real numbers, complex numbers, matrices, vectors and so on. While converting from Arithmetic to Algebra will look something like this: Arithmetic: 3 + 4 = 3 + 4 in basic algebra it would look like: x + y = y + x.

answers for basic algebra - Terms

A term in a basic algebraic expression involving letters and/or numbers (called factors), multiplied together.

3x2 - 7ab + 2evp has three terms.

First term: 4x2; have factors 4 and x2

Second term: -8ab; has factors -8, a and b

Third Term: 9evp; has factors 9, e and vp

The 4, -8 and 9 are called coefficients of the terms

Like Terms:

Like terms are terms with same variables raised to the same power.

9x3 and 3x3 are like terms.

-2x2 and 7y2 are not like terms, because the variable is unlike. Let us see some examples on answers for basic algebra.

answers for basic algebra - Examples

Example 1

Simplify 13x + 7y - 2x + 6a

The only like terms in this expression are 13x and -2x. We cannot do anything with the 7y or 6a.

So we group together the terms we can subtract, and leave the rest:

(13x - 2x) + 6a + 7y

= 6a + 11x + 7y

Answer : 6a + 11x + 7y

We usually present our variables in alphabetical order, but it is not essential.

Example 2

Simplify - [9(a- 2b) - 4b]

- [9(a - 2b) - 4b]

= - [9a - 14b - 4b]

= - [9a - 18b]

= - 9la + 18b

Answer : - 9la + 18b

Example 3

Solve the equation

x - 6 = 10

Solution:

First we add 6 to both sides, so, we will remove the -6 on the left.

x - 6 = 10

x - 6 + 6 = 10 + 6

x = 16

So, the answer x value needs to be 16 to make the equation true.

Check the obtained answer in the equation x – 6 = 10:

16 - 6 = 10.

Hence be proved.

Answer: 16

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