Algebra Problems
Algebra problems are not only based on algebraic expressions but also on various types of equations in Maths where a quantity or variable is unknown to us. Many of us are familiar with the word problem, but are we aware of the fact and problems related to variables and constants? When we say 5 it means a number but what if we say x=5 or 5y or something like that?
This is where algebra came into existence algebra is that branch of mathematics which not only deals with numbers but also variable and alphabets. The versatility of Algebra is very deep and very conceptual, all the non-numeric character represents variable and numeric as constants. Let us solve some problems based algebra with solutions which will cover the syllabus for class 6, 7, 8. Below are some of the examples of algebraic expressions.
For example.
| 1. | -5y+3=2(4y+12) |
| 2. | \(\frac{4}{x^{2}-2x}-\frac{2}{x-2}=-\frac{1}{2}\) |
| 3. | x√x=-x |
| 4. | |x-a|= a2-x2 |
| 5. | \(4^{x^{2}+1}-2^{x^{2}+2}=8\) |
| 6. | log2(2x-1)+x=log4(144) |
| 7. | \(\left\{\begin{matrix}x^{2}+y^{2}=17+2x & \\ (x-1)^{2}+(y-8)^{2}=34 & \end{matrix}\right.\) |
Algebra Word Problems deal with real-time situations and solutions which can be solved using algebra.
Basic Algebra Identities
- (a + b)2 = a2 + b2 + 2ab
- (a – b)2 = a2 + b2 – 2ab
- a2 – b2 = (a + b)(a – b)
- a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab
- a3 + b3 = (a + b)(a2 – ab + b2)
- a3 – b3 = (a – b)(a2 + ab + b2)
- (a + b)3 = a3 + 3ab(a + b) + b3
- (a – b)3 = a3 – 3ab(a – b) – b3
Also, see:
Algebra problems With Solutions
Example 1: Solve, (x-1)2 = [4√(x-4)]2
Solution: x2-2x+1 = 16(x-4)
x2-2x+1 = 16x-64
x2-18x+65 = 0
(x-13) (x-5) = 0
Hence, x = 13 and x = 5.
Algebra Problems for Class 6
In class 6, students will be introduced with an algebra concept. Here, you will learn how the unknown values are represented in terms of variables. The given expression can be solved only if we know the value of unknown variable. Let us see some examples.
Example: Solve, 4x + 5 when, x = 3.
Solution: Given, 4x + 5
Now putting the value of x=3, we get;
4 (3) + 5 = 12 + 5 = 17.
Example: Give expressions for the following cases:
(i) 12 added to 2x
(ii) 6 multiplied by y
(iii) 25 subtracted from z
(iv) 17 times of m
Solution:
(i) 12 + 2x
(ii) 6y
(iii) z-25
(iv)17m
Algebra Problems for Class 7
In class 7, students will deal with algebraic expressions like x+y, xy, 32x2-12y2, etc. There are different kinds of the terminology used in case algebraic equations such as;
- Term
- Factor
- Coefficient
Let us understand these terms with an example. Suppose 4x + 5y is an algebraic expression, then 4x and 5y are the terms. Since here the variables used are x and y, therefore, x and y are the factors of 4x + 5y. And the numerical factor attached to the variables are the coefficient such as 4 and 5 are the coefficient of x and y in the given expression.
Any expression with one or more terms is called a polynomial. Specifically, a one-term expression is called a monomial; a two-term expression is called a binomial, and a three-term expression is called a trinomial.
Terms which have the same algebraic factors are like terms. Terms which have different algebraic factors are unlike terms. Thus, terms 4xy and – 3xy are like terms; but terms 4xy and – 3x are not like terms.
Example: Add 3x + 5x
Solution: Since 3x and 5x have the same algebraic factors, hence, they are like terms and can be added by their coefficient.
3x + 5x = 8x
Example: Collect like terms and simplify the expression: 12x2 – 9x + 5x – 4x2 – 7x + 10.
Solution: 12x2 – 9x + 5x – 4x2 – 7x + 10
= (12 – 4)x2 – 9x + 5x – 7x + 10
= 8x2 – 11x + 10
Algebra Problems for Class 8
Here, students will deal with algebraic identities. See the examples.
Example: Solve (2x+y)2
Solution: Using the identity: (a+b)2 = a2 + b2 + 2 ab, we get;
(2x+y) = (2x)2 + y2 + 2.2x.y = 4x2 + y2 + 4xy
Example: Solve (99)2 using algebraic identity.
Solution: We can write, 99 = 100 -1
Therefore, (100 – 1 )2
= 1002 + 12 – 2 x 100 x 1 [By identity: (a -b)2 = a2 + b2 – 2ab
= 10000 + 1 – 200
= 9801
Algebra Word Problems
Question 1: There are 47 boys in the class. This is three more than four times the number of girls. How many girls are there in the class?
Solution: Let the number of girls be x
As per the given statement,
4 x + 3 = 47
4x = 47 – 3
x = 44/4
x = 11
Question 2: The sum of two consecutive numbers is 41. What are the numbers?
Solution: Let one of the numbers be x.
Then the other number will x+1
Now, as per the given questions,
x + x + 1 = 41
2x + 1 = 41
2x = 40
x = 20
So, the first number is 20 and second number is 20+1 = 21
Linear Algebra Problems
There are various methods For Solving the Linear Equations
- Cross multiplication method
- Replacement method or Substitution method
- Hit and trial method
There are Variety of different Algebra problem present and are solved depending upon their functionality and state. For example, a linear equation problem can’t be solved using a quadratic equation formula and vice verse for, e.g., x+x/2=7 then solve for x is an equation in one variable for x which can be satisfied by only one value of x. Whereas x2+5x+6 is a quadratic equation which is satisfied for two values of x the domain of algebra is huge and vast so for more information. Visit BYJU’S. where different techniques are explained different algebra problem.