Algebra problems are not just based on the algebraic expressions but also depends on various types of equations in maths in which a quantity or a variable is unknown to us. Many of you would be familiar with the word problems, but are you aware of the fact and the problems that are related to the variables and constants? When you say 5 it means the number 5 but what if you say x = 5 or 5y? This is when algebra came into existence. Algebra is the branch of the Mathematics that not only deals with the numbers but also deals with the variables and alphabets. The versatility of algebra is much deep and conceptual. All the non-numeric characters in algebra represent the variable and numerics represent the constants. We will study in this article about the algebra problems with solutions and take a look at the algebra word problems as well.

Let us first revise on the basic algebra identities.

Given below are the algebra identities that you must learn to solve Algebra problems.

(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

In class 6, you will be introduced to the algebra concept. Here, you will learn about how the unknown values can be represented in terms of the variables. The given expression can be solved only when you know the value of the unknown variable.

Consider the following example:

Six less than a number is equal to two. Determine the number

Solution:

Let the given number be denoted as ‘x’.

According to condition, you have x – 6 = 2

By solving it algebraically for x, you get,

x = 2 + 6 = 8

Hence, your required number is 8.

There are different types of terminology that are used in the algebraic equations such as term, factor, and coefficient. Let us understand these terms with an example.

Consider 4x + 5y to be an algebraic expression, then 4x and 5y are called the terms. Since the variables used here are x and y, hence, x and y are called the factors of 4x + 5y. And the numerical factor that is attached to the variables is the coefficient. 4 and 5 here are the coefficients of x and y respectively in the given expression.

Any expression having one or more terms is known as a polynomial. Specifically, a one-term expression is known as a monomial, a two-term expression is known as a binomial, and a three-term expression is known as a trinomial.

The terms that have the same algebraic factors are the like terms. The terms that have different algebraic factors are called, unlike terms. Hence, the terms 4xy and – 3xy are called like terms, but the terms 4xy and – 3x are called, unlike terms.

Let us now learn about how to solve the algebra word problems by taking some examples.

A man starts his car from Delhi to Amritsar at 6.00 am. Assume the uniform speed of his car to be x km/h. At 12:00 noon, he learns that he is 50 km away from Amritsar. Determine the distance between Delhi and Amritsar.

Solution:

The time taken by the man to reach Amristar = 12:00 noon – 6.00 am = 6 hours

The uniform speed of the car = x km/h

Therefore, the total distance covered by the man = Time x speed = 6x km

Hence, the distance between Delhi and Amristar is (6x + 50) km.

The present age of Deepak is one-third of his mother’s present age. If his mother’s age was five times than his age 6 years ago, determine their present ages.

Solution:

Let the present age of Deepak’s mother be x years

Hence, the present age of Deepak = x/3 years

6 years ago, the mother’s age = (x – 6) years

Hence, Deepak’s age 6 years ago = ( \[\frac{x}{3}\] - 6 )years

According to the problem, 6 years ago, the mother’s age was 5 times than that of Deepak’s age.

Hence,(x - 6) = 5 x (\[\frac{x}{3}\] - 6)

Solving this equation, you get

x - \[\frac{5x}{3}\] = -30 + 6

= \[\frac{3x - 5x}{3}\] = -24

= \[\frac{-2x}{3}\] = -24

= 2x = 24 x 3

Hence, x = 72/2 = 36

Therefore, the present age of Deepak’s mother is 36 years.

The present age of Deepak = x/3 = 36/3 = 12 years.

FAQ (Frequently Asked Questions)

1. How To Solve Linear Equations?

The solution which is also called the root of an equation refers to the value of the variable which satisfies the equation. For the linear equations, there is at least one solution for the given equation. To solve these equations you use several algebraic methods which include expanding the expressions, grouping the terms, and factorizing. There are several methods to solve the linear equations:

The Cross multiplication method

The Replacement method or the substitution method

The Hit and trial method

2. What is The Method For Solving Linear Equations?

The method for solving a linear equation is as follows:

If the given equation contains any kind of fractions use the least common denominator to clear out the fractions. You can do this by multiplying both sides of the equation by the LCD.

Also, if there are variables present in the denominators of the fractions then identify the values of the variable that will give you a division by zero as you need to avoid these values in the solution.

Simplify both sides of the equation. This means that you need to clear out any parenthesis and combining the like terms.

Use the first two steps above to get all the terms with the variable in them on one side of the equation by combining it into a single term and all constants on the other side.

If the coefficient of the variable is not one then use the third or fourth step above to make the coefficient one.

The last step is to verify the answer. This is the final step and the often the most skipped step, yet it is probably the most important step in the whole process.