Class -10 chapter – 3, Linear Equation in One Variable
(A) Types of linear equation on the basis of number of variables
(1) A linear equation in one variable:-
It is that type of linear equation in which only one variable is used.
Example:
(i) 4y+5=10 (y is only variable)
(ii) 3x+2= 7 (x is only one variable)
(2) A linear equation in two variables:-
It is that type of linear equation in which two variables are used.
Example: 3x+4y =5 (xand y are two variables)
(3) A linear equation in three variables:-
It is that type of linear equation in which three variables are used.
Example: x+y+z = 0 (x, y and z are three variables)
(B) What is a pair of linear equations?
A word “pair” comes in “a pair of equations”. The word “pair” basically represents two things in a group.
In the above context, we can say that “a pair of equations” will represent two equations at a glance.
In the same way, “a pair of linear equations” will indicate such group in which there will be two linear equations.
(C) Characteristics of a pair of linear equations
We are going to assume two linear equations representing a pair of linear equations as
a1x+b1y +c1=0 —————equation (1)
a2x+b2y +c2=0 —————equation (2)
(i) a1 and a2 are coefficients of the variable x in equations (1) and (2) respectively.
(ii) b1 and b2 are coefficients of the variable y in equations (1) and (2) respectively.
(ii) c1 and c2 are constant terms in equations (1) and (2) respectively.
In the table below detailed properties of the assumed two linear equations have been described:-
| Different conditions | Graphical representation | Algebraic explanation | |
| (1) | If (a1/a2)≠(b1/b2) | The curves of the two equations will intersect each other at one point. | The pair of linear equations will have exactly one solution which sometimes considered as unique solution |
| (2) | If (a1/a2)= (b1/b2)= (c1/c2) | The curves of the two equations will coincide upon each other. | The pair of linear equations will have infinitely many solutions. |
| (3) | If (a1/a2)= (b1/b2) ≠ (c1/c2) | The curves of the two equations will be parallel to each other. | The pair of linear equations will not have any solution. |
N.B.(Note Book)
(i) The meaning of solving equations means to find the values of variables in the same equations.
(ii) Solutions of particular equations mean values of variables which will satisfy all the given equations to which variables are related.
(iii) The meaning of a pair of linear equation refers that there are two linear equations in the same variables.
(D) Methods for solving a pair of linear equations in two variables:-
A pair of linear equations means there are two linear equations. There are various methods for solving a pair of linear equations in two variables which are listed below:-
(1) Graphical method,
(2) Substitution method,
(3) Elimination method,
(4) Cross-multiplication method.
Now we are going to discuss the different methods of solving a pair of linear equations one by one.
(1) Graphical method:-
(i) The best quality of this method is that values of abscissa and ordinate of a point at which curves of the two given linear equations intersect each other give the values of variables.
(ii) The curve of a linear equation (in one or two or three variables) is always a straight line.
(iii) Steps for drawing curves of a pair of linear equations are as follows:
Step-1
(a) We suppose two linear equations as
a1x+b1y +c1=0 —————equation (1)
a2x+b2y +c2=0 —————equation (2)
(b) Make two tables for two equations in a pair of linear equations separately on the basis of which two straight lines will be drawn later. (Here for explanation we are taking equation (1) only into our consideration)
| x | ||
| y |
Step-2
(a) Now take 0 as the first value of x and obtain the corresponding value of y(let corresponding value of y is p)
| x | 0 | |
| y | p |
(b) Again take value of y as 0 and find from the same equation value of x on the basis of the value of y=0.(Let that obtained value of x is q).
| x | 0 | q |
| y | p | 0 |
(c) Plot the two points (0,p) and (q,0) on the graph paper.
(d) Using the above two points (0,p) and (q,0) draw a straight line which will represent a1x+b1y +c1=0 or equation (1).
Step-3
(a) Repeat the entire process of our 2nd step for the second equation of the given pair of linear equations i.e. a2x+b2y +c2=0.
(b) Draw another line for equation (2) on the basis of the two points so obtained in our 3rd step.
Step-4
The point of intersection of the two lines drawn on the graph paper for equations (1) and (2) respectively will be the required solution of the given pair of linear equations.
(2) Substitution method:-
(i) This method comes under algebraic method of solving a pair of linear equations.
(ii) Let us consider a pair of linear equations as;
a1x+b1y +c1=0 —————equation (1)
a2x+b2y +c2=0 —————equation (2)
(iii) Also we assume x and y as the first variable and second variables respectively in both the above assumed equations (1) and (2).
(iv) This method includes following steps which make easier for us to use this method.
Step-1
We will find the value of one variable in form of the second variable from either equation.
[Here for explanation we are taking equation (1) only into our consideration]
. a1x+b1y+c1=0
or, a1x = – c1– b1y
or, x = (- c1– b1y)×1/a1
Step-2
(a) Now, we are going to substitute the value of first variable so obtained in the 1st step in another equation.
(b) In this way, we will be able to calculate the numerical value of second variable.
. a2x+b2y+c2=0
or, a2 × [(- c1– b1y)×1/a1]+ b2y+c2=0
or, y = (a2c1-a1c2)/(a1b2-a2b1)
Step-3
In this step, we will substitute the numerical value of the second variable in either of the two equations in order to get the numerical value of the first variable.
(3) Elimination method:-
(i) Suppose a pair of linear equations is as follows:
a1x+b1y +c1=0 —————equation (1)
a2x+b2y +c2=0 —————equation (2)
(ii) Following steps make this method easy to use.
Step-1
(a) Multiply both the equations by appropriate non-zero constants to make the coefficients of a specific variable in both equations equal.
(b) Let we want to make coefficients of y equal. Then we will multiply the 1st equation by b2 and the 2nd equation by b1.
a1b2x+ b1b2y+ c1b2=0 —————equation (3)
a2b1x+ b1b2y+ c2b1=0 —————equation (4)
Step-2
(a) In this step, we will add or subtract one equation from other (as per situation) in such a way that the one variable gets eliminated.
(b) In this way, we will get a linear equation in one variable the numerical value of which will be calculated.
(c) here, we are going to subtract LHS and RHS of equation (4) from LHS and RHS of equation (3) respectively.
Therefore,
LHS of equation (3) – LHS of equation (4) = RHS of equation (3) – RHS of equation (4)
or, (a1b2x+ b1b2y+ c1b2) – (a2b1x+ b1b2y+ c2b1) = 0 – 0
or, a1b2x+ b1b2y+ c1b2 – a2b1x– b1b2y–c2b1 = 0
or, a1b2x+ c1b2 – a2b1x–c2b1 = 0 {Here (b1b2y) and (– b1b2y) will be vanished}
or, a1b2x– a2b1x = c2b1 – c1b2
or, x(a1b2– a2b1) = c2b1 – c1b2
or, x = (c2b1 – c1b2)/(a1b2– a2b1)
Step-3
In this step, we will substitute the value of the variable (x) in either of the original equations for getting the value of the eliminated variable (y).
Therefore, we are going to use equation (1),
. a1x+b1y+c1=0
or, a1×{(c2b1 – c1b2)/ (a1b2– a2b1)} +b1y+c1=0
or, y = (a2c1-a1c2)/(a1b2-a2b1)
N.B.(Note Book)
(i) The main concept in this method is to remove a specific variable from both the equations by making its coefficients equal in both equations.
(ii) Then find the numerical value of another variable.
(iii) This numerical value will be substituted in either equation to calculate the value of the variable which was eliminated.
(4) Cross-Multiplication method:-
(i) This method comes under algebraic method of solving a pair of linear equations.
(ii) Let us consider a pair of linear equations as;
a1x+b1y +c1=0 —————equation (1)
a2x+b2y +c2=0 —————equation (2)
According to this method; we make three expressions the value of which are equal to each other.
{x/(b1×c2 – b2×c1)} = {y/(c1×a2 – c2×a1)} = {1/(a1× b2 – a2× b1)}
1st expression 2nd expression 3rd expression
Now, on considering 1st and 3rd expressions,
. {x/(b1×c2 – b2×c1)} = {1/(a1× b2 – a2× b1)}
or, x = {1/(a1× b2 – a2× b1)}×(b1×c2 – b2×c1)
or, x = (b1×c2 – b2×c1)/(a1× b2 – a2× b1)
In the similar manner, on taking 2nd and 3rd expressions into our consideration,
. {y/(c1×a2 – c2×a1)} = {1/(a1× b2 – a2× b1)}
or, y = {1/(a1× b2 – a2× b1)}×(c1×a2 – c2×a1)
or, y = (c1×a2 – c2×a1)/(a1× b2 – a2× b1)
