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Whenever any equation is represented by ax + by + c where x and y are two variables, it is known as a linear equation consisting of two variables. The values of a, b, and c here is not equal to zero and are real numbers. On the other hand, when two such linear equations have the same set of variables, they are known as the pair of linear equations in two variables. Usually, when we solve a pair of linear equations, we get a point or set of points on the coordinate plane as a solution. The line on which the point lies is represented by the equation. The third chapter of class 10 mathematics is about how to solve the following pair of linear equations. The chapter is divided into seven parts with six exercises. The NCERT solutions will help you to understand the sums of this chapter for your examinations.

The first part that you will learn in this chapter is the introduction, followed by the graphical representation of a pair of linear equations. The next parts will be on solving these linear equations by applying various methods. The chapter discusses in detail, how to solve a pair of linear equations in two variables. In junior classes, we have learned how to solve one variable and one equation system. That was quite easy, but now we have two variables and two-equation methods. Of course, we need two different equations with the same variables to solve the values for both the variables.

Polynomial: An algebraic expression with variables having non-negative integers as power and equality on both sides is known as a polynomial.

Degree: The largest value of power given to the variables is known as the degree of the polynomial. A degree 1 polynomial is linear. While degree 2 is quadratic in nature and degree 3 polynomial is cubic. For example, 5x + 7y = 6 is a linear polynomial. Further, we will see how to solve the following pair of linear equations.

A pair of linear equations can be represented both algebraically as well as graphically. Let's look at the equations. The first equation is in the form of a1x + b1y + c1 = 0 and the second equation is given by the form a2x + b2y + c2 = 0 where a1, a2, b1, b2, c1 and c2 are real numbers. Again, a12 + b12 and a22 + b12 is not equal to zero.

Here three important conditions ariseâ€”

If a1/a2 is not equal to b1/b2, then the pair is consistent.

If a1/a2 is equivalent to b1/b2 but is not similar to c1/c2, then the pair is said to be inconsistent.

If a1/a2 is equal to b1/b2 and also equal to c1/c2, then the pair is consistent and dependent.

The graphical representation of these equations is also simple. Two lines in a plane can intersect each other, run parallel to each other, or coincide with each other. Consider a pair of linear equations and solve the following pair of linear equations by the elimination method and the substitution method. After this, you will get different values for x and y by putting them in the equations. It can be easily plotted on graph paper.

(image to be added soon)

We can solve these equations graphically, but this is not at all feasible. Graphical methods are the most time consuming and tedious. In cases of non-integral coordinates, plotting it on a graph paper becomes very difficult. Hence we have algebraic ways to solve such problems. The plans include the elimination method, cross multiplication, and method of substitution.

**Find out whether the equations given below are consistent or inconsistent. 2x + 3y = 5 and 3x â€“ 2y = 9**

Answer: a1/a2 = 2/3, b1/b2 = 3/-2 and c1/c2 = 5/9.

Since a1/a2 is not equal to b1/b2, the pair is consistent. These lines intersect each other at some point and have only one single solution.

**Â Â Â Â Â 2. Determine whether the pair of linear equations given below has a unique solution, an infinite number of solutions, or no solution. 3x + y = 10 and 2x + 3y = 7**

Answer: a1/a2 = 3/2, b1/b2 = 1/3 and c1/c2 = 10/7.

Since a1/a2 is not equal to b1/b2, the pair is consistent. These lines intersect each other at some point and have only one single solution. Hence, they have a Unique Solution as they cross at an end.

FAQ (Frequently Asked Questions)

1. How to solve the following pair of linear equations by the elimination method and the substitution method?

**Method of Elimination: **Multiply the two equations with specific integral constants if the coefficient of any one of the variables is not equal in the two equations. It should make any variable's coefficients in both the equations equivalent. To get a single variable equation either add or subtract the equations. Get the value of another variable by substituting the value that you got by solving the above equations in the previous step.

**Method of Substitution: **Calculate the value of one variable in terms of the other variable. Say x in terms of y or vice versa. To get a single variable equation, substitute this value which you found out to the next equation and get the solution. Substitute the value calculated in the previous step in the first equation to get the value of the other variable.

2. Solve the following pair of linear equations by substitution method, elimination method, or cross-multiplication method. 1/2x + 1/3y = 2 and 1/3x + 1/2y.

Let us assume some variables. Consider 1/x = a and 1/y = b, then the equation will change given as follows.

a/2 + b/3 = 2

â‡’ 3a+2b-12 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(Equation Number 1)

a/3 + b/2 = 13/6

â‡’ 2a+3b-13 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(Equation Number 2)

Now, we will be applying the cross-multiplication method to get values for both the variables,Â

a/(-26-(-36) ) = b/(-24-(-39)) = 1/(9-4)

a/10 = b/15 = 1/5

a/10 = 1/5 and b/15 = 1/5

So, a = 2 and b = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3