subject: Introduction To College Level Algebra [print this page] 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.
The part of algebra called elementary algebra is often part of the curriculum in secondary education and introduces the concept of variables representing numbers.
Solving College Level Algebra:
Example in Solving college level algebra 1:
f(x) = x2 - 2x + 3, find f(-5).
Solution:
The given equation is f(x) = x2 2x + 3
Here we have to substitute (-5) in the equation in the place of x.
f(-5) = (-5)2 2(-5) + 3
By resolving the above equation we get
= 25 + 10 + 3
The final solution is = 38
Example in Solving college level algebra 2:Solve 5x2 + 2x - 3 = 0 for x.
Solution:
Given equation is 5x2+2x-3=0 for x.
Solve the given equation we get
. (5x - 3)(x + 1)=0
Here by equating the values to zero
5x-3=0, x+1=0
Here keep constant terms on one side and move the variables to other side.
5x=3 , x=-1
The final solution is X=3/5, x = -1.
Solving college algebra online:
Example Solving college level algebra 3: Solve 6x2 + 3x - 3 = 0 for x.
Sol: Given equation is 6x2+3x-3=0 for x.
Solve the given equation we get
. (6x - 3)(x + 1)=0
Here by equating the values to zero
6x-3=0, x+1=0
Here keep constant terms on one side and move the variables to other side.
6x=3 , x=-1
The final solution is X=1/2, x = -1.
Example Solving college algebra 4 : Factorize x2+ 7x + 10.
Sol: Here a = coefficient of x2 = 1
b = coefficient of x=7
c = constant term=10
We find a =10=10= 5,5+2=7=b. Hence
x2 +7x+10=1/2(2x+10) (x+2)=(x+5)(x+2).
Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows:
x2 +7x+10=x2 + (5+2)x+10
= x2 +5x+2x+10
= x(x+5) + (1)(x+2) = (x+2) (x+5).
Practice Problems in Solving College Level Algebra
1)Solve the value of a and b if ax3 + bx2 + 7x + 9 and x3 + ax2 2x + b 4 when divided by x +2 responds remainders 5 and 8 respectively.
Answer: a , b =
2) Solve the algebraic equations 2x + z = 5, x + 2y + z = 3, 3y - 2z = 2
Answer:
The final answer is x = 2, y = 0, z = 1
Introduction to algebra:
An unknown quantity is called a variable.
A statement or equality of two algebriac expressions involving a variable is called an equation.
An equation which contains only one varaible of degree 1 , is called a simple linear equations.
A word problem is a mathematical problem stated in words.
Rules for Solving An algebraic equations
1) Same number can be added to both sides of an equation.
2) Same number can be subtracted from both sides of an equation.
3) Both sides of an equations can be multiplied by the same non-zero number.
4) Both sides of an equation can be divided by the same non-zero number.
Addition and Substraction Pre-algebra Problems
Solve :
1) Two numbers add up to 12. If one number is 7 then find the other number.
Solution: Given, One number = 7
Let the other number be x
Addition of these two numbers = 12
x + 7 = 12
substract 7 on both sides
x + 7 -7 = 12 - 7 ( 7 - 7 = 0)
x = 5
The other number = x = 5
Solve word problems:
2) A value 13 is taken away from 96. what is the result.
Solution: Given, 13 is taken away from 96, means we have substract them
96 - 13 = 86
Solve problems:
3) Sixty-five less than a number is 25. Find the number.
Solution: Let the number b x
65 less than the number ( means substraction)
65 - x = 25
substract 25 on both sides
65 - 25 - x = 25 - 25
40 - x = 0
add x on both sides,
40 - x + x = 0 + x
40 = x
The number is 40
pre-algebra word problem:
4) Thirty-eight more than a number is 62. Find the number.
Solution: Let the number be x
Thirty- eight more than number
38 + x is 62
3 8 + x = 62
substract on both sides by 38
38 - 38 + x = 62 - 38
0 + x = 24
x = 24
The number is 24
Solve
5) A number is taken away from 350 and the result is 175. what is the number?
Solution: Let the number be m
This number 'm' is taken away from 350
350 - m and its result is 175
so, we will put is as,
350 - m = 175
substract 175 on both sides
350 - 175 - m = 175 - 175
125 - m = 0
add m on both sides
125 - m + m = 0 + m
125 = m
The number is 125.
Multiplication and Division Pre-algebra Problems
word problems:
1) A number multiplied by 8 is 72. Find the number
Solution: Let the number be x
Number multiplied by 8 = x * 8 is equal to 72
x * 8 = 72
divide throughout by 8
( x * 8 ) / 8 = 72 / 8 ( 8 * 1/8 = 1)
1 * x = 9 ( 8 x 9 = 72)
x = 9
The number is 9
Solve:
2) Three-fourth of a number is 12. Find the number.
Solution: Let the number be x
Three-fourth of a number = ( 3/4) * x
(3/4) * x = 12
3 x / 4 = 12
multiply throughout by 4
(3 x / 4) * 4 = 12 * 4
3x * 1 = 48 ( 4 * 1/4 = 1)
Divide throughout by 3
3x / 3 = 48/ 3
x * 1 = 16
x = 16
The number is 16
worked problems:
3) A number divided by 11 is 6. Find the number
Solution: Let the number be x
Number divided by 11 = x / 11
x / 11 = 6
multiply throughout by 11
( x / 11) * 11 = 6 * 11
x * 1 = 66
x = 66
The number is 66
Pre-algebra problem:
4) What number do we have to divide 1050 by to get 70?
Solution: Let the number be x
1050 divided by a number = 1050 / x
1050 / x = 70
multiply throughout by x
(1050 / x) * x = 70 * x
1050 * x / x = 70 * x
1050 * 1 = 70 * x
1050 = 70 * x
divide throughout by 70
1050 / 70 = (x * 70) / 70
105 / 7 = x * 1
15 = x
The number is 15
Some Complex Problems
Solve the following:
1) A number is 25 more than '(5)/(6)' th part. Find the number .
Solution: Let the number be x
'(5)/(6)' x +25 = x
The denominator is 6, we multiply all terms with 6
6( '(5)/(6)' x) +6(25) = 6x
x + 150 = 6x
x - x + 150 = 6 x - x
0 + 150 = 5 x
divide both sides by 5
'(150)/(5)' = '(5x)/(5)'
30 = x
The number is 30.
Word problem:
2) 17 less than four times a number is 11. Find the number.
