Practice Abstract Algebra Exams

Abstract for algebra:

In algebra the algebraic expressions represent mathematical ideas and operations in a short way using symbols and variables.The algebra equation states that one expression names the same number as another expression. The symbol in algebra which can be assigned different numerical values is called a variables. Algebra uses known quantities to find the unknown quantities. Here it is about the practice for abstract algebra exams. I like to share this algebra 2 problems and solutions with you all through my article.


practice abstract algebra exams - Examples:


practice abstract algebra exams - Problem 1: Simple equations:

solving simple equation: 7x + 13 = 4x + 43.

Solution:

since 7x is larger than 4x,

we decide to shift the letter terms to left and the number terms to right side.

7x + 13 = 4x + 43

we want to remove 4x from the right side,

we can do so by subtracting 4x from both sides.

7x + 13 – 4x = 4x + 43 – 4x

3x + 13 = 43

we want to remove 13 from both sides,

we do so by subtracting 13 from both sides.

3x + 13 – 13= 43 – 13

3x = 30

Dividing both sides by 3.

3x/3 = 30 / 3

x = 10

check to substituting the x in as:

left side: 7(10) + 13 = 70 + 13 = 83

right side: 4 (10) + 43 = 40 + 43 = 83

practice abstract algebra exams - Problem 2: Algebraic expressions:

Simplify the algebraic expression: 17m² – 9m + 5m – 3m² – 7m + 11

solution:

By rearranging the given terms as,

we have,

=17m² – 3m² + 5m – 9m – 7m + 11

By equating, we get

= (17 – 3) m² + (5 – 9 – 7) m + 11

To simplify the equation as

=14m² + (– 4 – 7) m + 11

=14m² + (–11) m + 11

=14m² – 11m + 11.

This is the equation for the given algebraic expression.

practice abstract algebra exams - Problem 3: Algebraic identities:

Evaluate 102 × 104 without directly multiplying the two given numbers.

Solution:

Taking 102 as (100 + 2) and 104 as (100 + 4)

we get the multiplier as

= 102 × 104

Taking the multiplier as

= (100 + 2) × (100 + 4)

By multiplying and adding the identities we get,

= 100² + (2 + 4) 100 + 2 × 4

= 10000 + 6 × 100 + 8

= 10000 + 600 + 8

= 10608

This is the solution for the given algebraic identities.

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practice abstract algebra exams - Problems for practice exam:


1. solving simple equation:18 – 5x = 3x – 6.

Answer: x = 3

2. Evaluate the algebraic identities 103 × 105 without directly multiplying the two given numbers.

Answer: = 10815

3. Simplify the algebraic expression: 16m² – 9m + 5m – 3m² – 7m + 12.

Answer: =13m² – 11m + 12.

How to Use Algebra

Algebra:

Algebra word problems are very useful to answer real-life problems. Remember the famous words of Albert Einstein

Definition:

That branch of math which treats of the relations and properties of quantity by means of letters and other symbols.

A branch of mathematics in which signs, usually letters of the alphabet, represent numbers or members of a specified set and are used to represent quantities and to express general relationships that hold for all members of the set.


How to use Algebra:


An algebraic expression is every combination of variables and constants with mathematical operations (addition, subtraction, multiplication, division, roots and powers). Writing algebraic expressions is a method of determining and combining constants, variables, and math convention using the order of operations. The next shows how to put in writing algebraic expressions, with an example in parentheses.

Step 1:

Write down what the algebraic expression is to show. (A toy line has toys of 3 different sizes and both with a different price. Small costs $5, medium costs $10, and large costs $15. Write as an algebraic expression.)

Step 2:

Determine the constant(s). (5, 10, 15 for price of each size toy.)

Step 3:

Determine the variable(s). (x, y, z for the quantity of each size toy.)

Step 4:

Determine the operations that need to be done. (Multiply both prices by the quantity and find the sum total of all 3 sizes.)

Step 5:

Mingle the constants, variables, and operations using the conventions of the order of operations. This is the algebraic expression. (5x + 10y + 15z)

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Use Algebra expressions with detailed solutions:


Problem 1:

Answer the expression 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

-15x - 10 - x + 3 = -16x - 20 +13

-16x - 7 = -16x – 7

0 = 0

Problem 2:

Reduce the algebraic expression 2(a -3) + 4b - 2(a -b -3) + 5

Solution:

= 2(a -3) + 4b - 2(a -b -3) + 5

= 2a - 6 + 4b -2a + 2b + 6 + 5

= 6b + 5

Problem 3:

Reduce |x - 2| - 4|-6|

Solution:

= |x - 2| - 4|-6|

Alternate |x - 2| by -(x - 2) and |-6| by 6.

= |x - 2| - 4|-6| = -(x - 2) -4(6)

= -x -22

Problem 4:

Estimate f (2) - f (1), f(x) = 6x + 1

Solution:

f (2) - f (1) = (6*2 + 1) - (6*1 + 1)

= 6

Problem 5:

Answer the equation |-2x + 2| -3 = -3

Solution:

|-2x + 2| -3 = -3

|-2x + 2| = 0

x = 1