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3 Pair of Linear Equations in Two Variables
3.1 Introduction
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs Rs 3, and a game of Hoopla costs Rs 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent Rs 20.
May be you will try it by considering different cases. If she has one ride, is it possible? Is it possible to have two rides? And so on. Or you may use the knowledge of Class IX, to represent such situations as linear equations in two variables.
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she played Hoopla by y. Now the situation can be represented by the two equations:
y = 1/2 (1)
3x + 4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of finding these, which we will study in this chapter.
3.2 Pair of Linear Equations in Two Variables
Recall, from Class IX, that the following are examples of linear equations in two variables:
2x + 3y = 5
x – 2y – 3 = 0
and x – 0y = 2, i.e., x = 2
You also know that an equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a^{2} + b^{2} ≠ 0). You have also studied that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal.
For example, let us substitute x = 1 and y = 1 in the left hand side (LHS) of the equation 2x + 3y = 5. Then
LHS = 2(1) + 3(1) = 2 + 3 = 5,
which is equal to the right hand side (RHS) of the equation.
Therefore, x = 1 and y = 1 is a solution of the equation 2x + 3y = 5.
Now let us substitute x = 1 and y = 7 in the equation 2x + 3y = 5. Then,
LHS = 2(1) + 3(7) = 2 + 21 = 23
which is not equal to the RHS.
Therefore, x = 1 and y = 7 is not a solution of the equation.
Geometrically, what does this mean? It means that the point (1, 1) lies on the line representing the equation 2x + 3y = 5, and the point (1, 7) does not lie on it. So, every solution of the equation is a point on the line representing it.
In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
Now, consider Equations (1) and (2) given above. These equations, taken together, represent the information we have about Akhila at the fair.
These two linear equations are in the same two variables x and y. Equations like these are called a pair of linear equations in two variables.
Let us see what such pairs look like algebraically.
The general form for a pair of linear equations in two variables x and y is
ax_{1} + b_{1}y + c_{1} = 0
and a_{2}x + b_{2}y + c_{2} = 0,
where a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2} are all real numbers and a_{1}^{2} + b_{1}^{2} ≠ 0, a_{2}^{2} + b_{2}^{2} ≠ 0.
Some examples of pair of linear equations in two variables are:
2x + 3y – 7 = 0 and 9x – 2y + 8 = 0
5x = y and –7x + 2y + 3 = 0
x + y = 7 and 17 = y
Do you know, what do they look like geometrically?
Recall, that you have studied in Class IX that the geometrical (i.e., graphical) representation of a linear equation in two variables is a straight line. Can you now suggest what a pair of linear equations in two variables will look like, geometrically? There will be two straight lines, both to be considered together.
You have also studied in Class IX that given two lines in a plane, only one of the following three possibilities can happen:
 The two lines will intersect at one point.
 The two lines will not intersect, i.e., they are parallel.
 The two lines will be coincident.
We show all these possibilities in Fig. 3.1:
In Fig. 3.1 (a), they intersect.
In Fig. 3.1 (b), they are parallel.
In Fig. 3.1 (c), they are coincident.
Fig. 3.1
Both ways of representing a pair of linear equations go handinhand — the algebraic and the geometric ways. Let us consider some examples.
Let us take the example given in Section 3.1. Akhila goes to a fair with Rs 20 and wants to have rides on the Giant Wheel and play Hoopla. Represent this situation algebraically and graphically (geometrically).
The pair of equations formed is :
y = (1/2)x
i.e., x – 2y = 0 (1)
3x + 4y = 20 (2)
Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in Table 3.1.
Table 3.1
x  0  2 
y = (1/2)x  0  1 
(i)
x  0  20/3  4 
y = (203x)/4  5  0  2 
(ii)
Recall from Class IX that there are infinitely many solutions of each linear equation. So each of you can choose any two values, which may not be the ones we have chosen. Can you guess why we have chosen x = 0 in the first equation and in the second equation? When one of the variables is zero, the equation reduces to a linear equation in one variable, which can be solved easily. For instance, putting x = 0 in Equation (2), we get 4y = 20, i.e., y = 5. Similarly, putting y = 0 in Equation (2), we get 3x = 20, i.e., x = 20/3. But as 20/3 is not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2 which gives x = 4, an integral value.
Plot the points A(0, 0), B(2, 1) and P(0, 5), Q(4, 2), corresponding to the solutions in Table 3.1. Now draw the lines AB and PQ, representing the equations x – 2y = 0 and 3x + 4y = 20, as shown in Fig. 3.2.
Fig. 3.2
In Fig. 3.2, observe that the two lines representing the two equations are intersecting at the point (4, 2). We shall discuss what this means in the next section.
Romila went to a stationery shop and purchased 2 pencils and 3 erasers for Rs 9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought 4 pencils and 6 erasers of the same kind for Rs 18. Represent this situation algebraically and graphically.
Let us denote the cost of 1 pencil by Rs x and one eraser by Rs y. Then the algebraic representation is given by the following equations:
2x + 3y = 9 (1)
4x + 6y = 18 (2)
To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.
These solutions are given below in Table 3.2.
x  0  4.5 
y = (9  2x)/3  3  0 
(i)
x  0  3 
y = (18  4x)/6  3  1 
(ii)
We plot these points in a graph paper and draw the lines. We find that both the lines coincide (see Fig. 3.3). This is so, because, both the equations are equivalent, i.e., one can be derived from the other.
Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.
Fig. 3.3
Two solutions of each of the equations :
x + 2y – 4 = 0 (1)
2x + 4y – 12 = 0 (2)
are given in Table 3.3
Table 3.3
x  0  4 
(4x)/2  2  0 
(i)
x  0  6 
(122x)/2  3  0 
(ii)
To represent the equations graphically, we plot the points R(0, 2) and S(4, 0), to get the line RS and the points P(0, 3) and Q(6, 0) to get the line PQ.
Fig. 3.4
We observe in Fig. 3.4, that the lines do not intersect anywhere, i.e., they are parallel.
So, we have seen several situations which can be represented by a pair of linear equations. We have seen their algebraic and geometric representations. In the next few sections, we will discuss how these representations can be used to look for solutions of the pair of linear equations.
3.3 Graphical Method of Solution of a Pair of Linear Equations
In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. Can we solve them in each case? And if so, how? We shall try and answer these questions from the geometrical point of view in this section.
Let us look at the earlier examples one by one.

In the situation of Example 1, find out how many rides on the Giant Wheel
Akhila had, and how many times she played Hoopla.
In Fig. 3.2, you noted that the equations representing the situation are geometrically shown by two lines intersecting at the point (4, 2). Therefore, the
point (4, 2) lies on the lines represented by both the equations x – 2y = 0 and 3x + 4y = 20. And this is the only common point.Let us verify algebraically that x = 4, y = 2 is a solution of the given pair of equations. Substituting the values of x and y in each equation, we get 4 – 2 × 2 = 0 and 3(4) + 4(2) = 20. So, we have verified that x = 4, y = 2 is a solution of both the equations. Since (4, 2) is the only common point on both the lines, there is one and only one solution for this pair of linear equations in two variables.
Thus, the number of rides Akhila had on Giant Wheel is 4 and the number of times she played Hoopla is 2.

In the situation of Example 2, can you find the cost of each pencil and each
eraser?
In Fig. 3.3, the situation is geometrically shown by a pair of coincident lines. The solutions of the equations are given by the common points.
Are there any common points on these lines? From the graph, we observe that every point on the line is a common solution to both the equations. So, the equations 2x + 3y = 9 and 4x + 6y = 18 have infinitely many solutions. This should not surprise us, because if we divide the equation 4x + 6y = 18 by 2 , we get 2x + 3y = 9, which is the same as Equation (1). That is, both the equations are equivalent. From the graph, we see that any point on the line gives us a possible cost of each pencil and eraser. For instance, each pencil and eraser can cost Rs 3 and Re 1 respectively. Or, each pencil can cost Rs 3.75 and eraser can cost Rs 0.50, and so on.

In the situation of Example 3, can the two rails cross each other?
In Fig. 3.4, the situation is represented geometrically by two parallel lines. Since the lines do not intersect at all, the rails do not cross. This also means that the equations have no common solution.
A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.
We can now summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions as follows:
 the lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations).
 the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).
 the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
Let us now go back to the pairs of linear equations formed in Examples 1, 2, and 3, and note down what kind of pair they are geometrically.
 x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)
 x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of a_{1}/a_{2},b_{1}/b_{2} and c_{1}/c_{2} in all the three examples. Here, a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} denote the coefficents of equations given in the general form in Section 3.2.
Table 3.4
Sl No.  Pair of lines 
a_{1}/a_{2}  b_{1}/b_{2}  c_{1}/c_{2}  Compare the ratios 
Graphical representation  Algebraic interpretation 
1  x – 2y = 0 3x + 4y – 20 = 0 
1/3  2/4  0/20  a_{1}/a_{2}≠b_{1}/b_{2}  Intersecting lines  Exactly one solution (unique) 
2  2x + 3y – 9 = 0 4x + 6y – 18 = 0 
2/4  3/6  9/18  a_{1}/a_{2}=b_{1}/b_{2} =c_{1}/c_{2} 
Coincident lines  Infinitely many solutions 
3  x + 2y – 4 = 0 2x + 4y – 12 = 0 
1/2  2/4  4/12  a_{1}/a_{2}=b_{1}/b_{2} ≠c_{1}/c_{2} 
Parallel lines  No solution 
From the table above, you can observe that if the lines represented by the equation
a_{1}x + b_{1}y + c_{1} = 0
and a_{2}x + b_{2}y + c_{2} = 0
are
 intersecting, then a_{1}/a_{2} ≠ b_{1}/b_{2}
 coincident, then a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
 parallel, then a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
In fact, the converse is also true for any pair of lines. You can verify them by considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Check whether the pair of equations
x + 3y = 6 (1)
and 2x – 3y = 12 (2)
is consistent. If so, solve them graphically.
Let us draw the graphs of the Equations (1) and (2). For this, we find two solutions of each of the equations, which are given in Table 3.5
Table 3.5
x  0  6 
y = (6  x)/3  2  0 
x  0  3 
y = (2 x − 12)/3  – 4  2 
Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig. 3.5.
Fig. 3.5
Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
5x – 8y + 1 = 0 (1)
3x – (24/5)y + (3/5) = 0 (2)
Multiplying Equation (2) by , 5/3 we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought.
Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x – 2 (1)
and y = 4x – 4 (2)
Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations. They are given in Table 3.6.
Table 3.6
x  2  0 
y = 2x – 2  2  2 
x  0  1 
y = 4x – 4  – 4  0 
Plot the points and draw the lines passing through them to represent the equations, as shown in Fig. 3.6.
Fig. 3.5
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given problem.
3.4 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically. The graphical method is not convenient in cases when the point representing the solution of the linear equations has nonintegral coordinates like (√3, 2√7), (–1.75, 3.3), [(4/13), (1, 19)], etc. There is every possibility of making mistakes while reading such coordinates. Is there any alternative method of finding the solution? There are several algebraic methods, which we shall now discuss.
3.4.1 Substitution Method :
We shall explain the method of substitution by taking some examples.
Solve the following pair of equations by substitution method:
7x – 15y = 2 (1)
x + 2y = 3 (2)
Step 1 :
We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
x + 2y = 3
and write it as x = 3 – 2y (3)
Step 2 :
Substitute the value of x in Equation (1). We get
7(3 – 2y) – 15y = 2
i.e., 21 – 14y – 15y = 2
i.e., – 29y = –19
Therefore, y = 19/29
Step 3 :
Substituting this value of y in Equation (3), we get
x = 3 – 2(19/29) = (49/29)
Therefore, the solution is x = 49/29, y = 19/29.
Verification : Substituting x = 49/29 and y = 19/29, you can verify that both the Equations (1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:
Step 1 :
Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2 :
Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
Step 3 :
Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
Remark :
We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.
Solve Q.1 of Exercise 3.1 by the method of substitution.
Let s and t be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is
s – 7 = 7 (t – 7), i.e., s – 7t + 42 = 0 (1)
and s + 3 = 3 (t + 3), i.e., s – 3t = 6 (2)
Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get
(3t + 6) – 7t + 42 = 0,
i.e., 4t = 48, which gives t = 12.
Putting this value of t in Equation (2), we get
s = 3 (12) + 6 = 42
So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.
Let us consider Example 2 in Section 3.3, i.e., the cost of 2 pencils and 3 erasers is Rs 9 and the cost of 4 pencils and 6 erasers is Rs 18. Find the cost of each pencil and each eraser.
The pair of linear equations formed were:
2x + 3y = 9 (1)
4x + 6y = 18 (2)
We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
x = (9  3y)/2 (3)
Now we substitute this value of x in Equation (2), to get
4(9 − 3 y ) / 2 + 6y = 18
i.e., 18 – 6y + 6y = 18
i.e., 18 = 18
This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen bcause both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. Observe that we have obtained the same solution graphically also. (Refer to Fig. 3.3, Section 3.2.) We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Let us consider the Example 3 of Section 3.2. Will the rails cross each other?
The pair of linear equations formed were:
x + 2y – 4 = 0 (1)
2x + 4y – 12 = 0 (2)
We express x in terms of y from Equation (1) to get
x = 4 – 2y
Now, we substitute this value of x in Equation (2) to get
2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0
i.e., –4= 0
which is a false statement.
Therefore, the equations do not have a common solution. So, the two rails will not cross each other.
3.4.2 Elimination Method
Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs 2000 per month, find their monthly incomes.
Let us denote the incomes of the two person by Rs 9x and Rs 7x and their expenditures by Rs 4y and Rs 3y respectively. Then the equations formed in the situation is given by :
9x – 4y = 2000 (1)
and 7x – 3y = 2000 (2)
Step 1 :
Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations:
27x – 12y = 6000 (3)
28x – 12y = 8000 (4)
Step 2 :
Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get
(28x – 27x) – (12y – 12y) = 8000 – 6000
i.e., x = 2000
Step 3 :
Substituting this value of x in (1), we get
9(2000) – 4y = 2000
i.e., y = 4000
So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are Rs 18,000 and Rs 14,000, respectively.
Verification 3 :
18000 : 14000 = 9 : 7. Also, the ratio of their expenditures = 18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3
Remarks :
 The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable. In the example above, we eliminated y. We could also have eliminated x. Try doing it that way.
 You could also have used the substitution, or graphical method, to solve this problem. Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method :
Step 1 :
First multiply both the equations by some suitable nonzero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2 :
Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3.
If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3 :
Solve the equation in one variable (x or y) so obtained to get its value.
Step 4 :
Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
Now to illustrate it, we shall solve few more examples.
Use elimination method to find all possible solutions of the following pair of linear equations :
2x + 3y = 8 (1)
4x + 6y = 7 (2)
Step 1 :
Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as :
4x + 6y = 16 (3)
4x + 6y = 7 (4)
Step 2 :
Subtracting Equation (4) from Equation (3),
(4x – 4x) + (6y – 6y) = 16 – 7
i.e., 0 = 9, which is a false statement.
Therefore, the pair of equations has no solution.
The sum of a twodigit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5).
According to the given condition.
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66
i.e., x+y= 6 (1)
We are also given that the digits differ by 2, therefore,
either x–y= 2 (2)
or y–x= 2 (3)
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.
Verification :
Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.
3.4.3 Cross  Multiplication Method
So far, you have learnt how to solve a pair of linear equations in two variables by graphical, substitution and elimination methods. Here, we introduce one more algebraic method to solve a pair of linear equations which for many reasons is a very useful method of solving these equations. Before we proceed further, let us consider the following situation.
The cost of 5 oranges and 3 apples is Rs 35 and the cost of 2 oranges and 4 apples is Rs 28. Let us find the cost of an orange and an apple.
Let us denote the cost of an orange by Rs x and the cost of an apple by Rs y. Then, the equations formed are :
5x + 3y = 35, i.e., 5x + 3y – 35 = 0 (1)
2x + 4y = 28, i.e., 2x + 4y – 28 = 0 (2)
Let us use the elimination method to solve these equations.
Multiply Equation (1) by 4 and Equation (2) by 3. We get
(4)(5)x + (4)(3)y + (4)(–35) = 0 (3)
(3)(2)x + (3)(4)y + (3)(–28) = 0 (4)
Subtracting Equation (4) from Equation (3), we get
[(5)(4) – (3)(2)]x + [(4)(3) – (3)(4)]y + [4(–35) – (3)(–28)] = 0
Therefore, x = –[(4)(–35) − (3)(−28)]/[(5)(4) − (3)(2)]
i.e., x= [(3)(– 28) − (4) (−35)] / (5)(4) − (2)(3) (5)
If Equations (1) and (2) are written as a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, then we have
a_{1} = 5, b_{1} = 3, c_{1} = –35, a_{2} = 2, b_{2} = 4, c_{2} = –28.
Then Equation (5) can be written as x = (b_{1}c_{2} − b_{2} c_{1})/(a_{1}b_{2} − a_{2}b_{1})
Similarly, you can get y= (c_{1}a_{2} − c_{2}a_{1})/(a_{1}b_{2} − a_{2}b_{1})
By simplyfing Equation (5), we get
x= (−84 + 140)/(20 − 6) = 4
Similarly, y = [(−35)(2) − (5)(−28)]/[20 − 6] = (−70 + 140)/14 = 5
Therefore, x = 4, y = 5 is the solution of the given pair of equations.
Then, the cost of an orange is Rs 4 and that of an apple is Rs 5.
Verification :
Cost of 5 oranges + Cost of 3 apples = Rs 20 + Rs 15 = Rs 35. Cost of 2 oranges + Cost of 4 apples = Rs 8 + Rs 20 = Rs 28.
Let us now see how this method works for any pair of linear equations in two variables of the form
a_{1}x + b_{1}y + c_{1} = 0 (1)
and a_{2}x + b_{2}y + c_{2} = 0 (2)
To obtain the values of x and y as shown above, we follow the following steps:
Step 1 :
Multiply Equation (1) by b_{2} and Equation (2) by b_{1}, to get
b_{2}a_{1}x + b_{2}b_{1}y + b_{2}c_{1} = 0 (3)
b_{1}a_{2}x + b_{1}b_{2}y + b_{1}c_{2} = 0 (4)
Step 2 :
Subtracting Equation (4) from (3), we get:
(b_{2}a_{1} – b_{1}a_{2}) x + (b_{2}b_{1} – b_{1}b_{2}) y + (b_{2}c_{1}– b_{1}c_{2}) = 0
i.e., (b_{2}a_{1} – b_{1}a_{2}) x = b_{1}c_{2} – b_{2}c_{1}
So, x = (b_{1}c_{2} − b_{2}c_{1})/(a_{1}b_{2} − a_{2}b_{1}), provided a_{1}b_{2} – a_{2}b_{1} ≠ 0 (5)
Step 3 :
Substituting this value of x in (1) or (2), we gety = (c_{1}a_{2} − c_{2}a_{1}) / (a_{1}b_{2} − a_{2}b_{1}) (6)
Now, two cases arise :
Case 1 :
a_{1}b_{2} – a_{2}b_{1} ≠ 0. In this case a_{1}/a_{2} ≠ b_{1}/b_{2}. Then the pair of linear equations has
Case 2 :
a_{1}b_{2} – a_{2}b_{1} = 0. If we write a_{1}/a_{2} = b_{1}/b_{2} = k , then a_{1} = k a_{2}, b_{1} = k b_{2}.
Substituting the values of a1 and b1 in the Equation (1), we get
k (a_{2}x + b_{2}y) + c_{1} = 0. (7)
It can be observed that the Equations (7) and (2) can both be satisfied only if c_{1} = k c_{2}, i.e., c_{1} / c_{2} = k
If c_{1} = k c_{2}, any solution of Equation (2) will satisfy the Equation (1), and vice versa. So, if a_{1}/a_{2} = b_{1}/b_{2} = c_{1} / c_{2} = k, then there are infinitely many solutions to the pair of linear equations given by (1) and (2).
If c_{1} ≠ k c_{2}, then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.
We can summarise the discussion above for the pair of linear equations given by (1) and (2) as follows:
 When a_{1}/a_{2} ≠ b_{1}/b_{2}, we get a unique solution.
 When a_{1}/a_{2} = b_{1}/b_{2} = c_{1} / c_{2}, there are infinitely many solutions.
 When a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1} / c_{2}, there is no solution.
Note that you can write the solution given by Equations (5) and (6) in the following form :
x/(b_{1}c_{2} − b_{2}c_{1}) = y/(c_{1}a_{2} − c_{2}a_{1}) = 1/(a_{1}b_{2} − a_{2}b_{1}) (8)
In remembering the above result, the following diagram may be helpful to you :
The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first.
For solving a pair of linear equations by this method, we will follow the following steps :
Step 1 :
Write the given equations in the form (1) and (2).
Step 2 :
Taking the help of the diagram above, write Equations as given in (8).
Step 3 :
Find x and y, provided a1b2 – a2b1 ≠ 0
Step 2 above gives you an indication of why this method is called the crossmultiplication method.
From a bus stand in Bangalore , if we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur, the total cost is Rs 46; but if we buy 3 tickets to Malleswaram and 5 tickets to Yeshwanthpur the total cost is Rs 74. Find the fares from the bus stand to Malleswaram, and to Yeshwanthpur.
Let Rs x be the fare from the bus stand in Bangalore to Malleswaram, and Rs y to Yeshwanthpur. From the given information, we have
2x + 3y = 46, i.e., 2x + 3y – 46 = 0 (1)
3x + 5y = 74, i.e., 3x + 5y – 74 = 0 (2)
To solve the equations by the crossmultiplication method, we draw the diagram as given below.
Then x/[(3)(−74) − (5)(−46)] = y/[(−46)(3) − ( −74)(2)] = 1/[(2)(5) − (3)(3)]
i.e., x/[−222 + 230] = y/[−138 + 148] = 1/[10 − 9]
i.e., x/8 = y/10 = 1/1
i.e., x/8 = 1/1 and y/10 = 1/1
i.e., x = 8 and y = 10
Hence, the fare from the bus stand in Bangalore to Malleswaram is Rs 8 and the fare to Yeshwanthpur is Rs 10.
Verification :
You can check from the problem that the solution we have got is correct.
For which values of p does the pair of equations given below has unique solution?
4x + py + 8 = 0
2x + 2y + 2 = 0
Here a^{1} = 4, a^{2} = 2, b^{1} = p, b^{2} = 2.
Now for the given pair to have a unique solution : a^{1}/a^{2} = b^{1}/b^{2}
i.e., 4/2 ≠ p/2
i.e., p≠ 4
Therefore, for all values of p, except 4, the given pair of equations will have a unique solution.
For what values of k will the following pair of linear equations have infinitely many solutions?
kx + 3y – (k – 3) = 0
12x + ky – k = 0
Here, a^{1}/a^{2} = k/12, b^{1}/b^{2} = 3/k, c^{1}/c^{2} = (k  3)/k
For a pair of linear equations to have infinitely many solutions : a^{1}/a^{2} = b^{1}/b^{2} = c^{1}/c^{2}
So, we need k/12 = 3/k = (k  3)/k
or, k/12 = 3/k
which gives k^{2} = 36, i.e., k = ± 6.
Also, 3/k = (k  3)/k
gives 3k = k2 – 3k, i.e., 6k = k^{2}, which means k = 0 or k = 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the pair of linear equations has infinitely many solutions.
3.5 Equations Reducible to a Pair of Linear Equations in Two Variables
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. We now explain this process through some examples.
Solve the pair of equations:
2/x + 3/y = 13
5/x−4/y = −2
Let us write the given pair of equations as
2(1/x) + 3(1/y) = 13 (1)
5(1/x)−4(1/y) = −2 (2)
These equations are not in the form ax + by + c = 0. However, if we substitute
1/x = p and 1/y = q in Equations (1) and (2), we get
2p + 3q = 13 (3)
5p – 4q = – 2 (4)
So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3.
You know that p = 1/x and q = 1/y
Substitute the values of p and q to get 1/x = 2, i.e., x = 1/2 and 1/y = 3, i.e., y = 1/3
By substituting x = 1/2 and y = 1/3 in the given equations, we find that both the equations are satisfied.
Solve the following pair of equations by reducing them to a pair of linear equations :
5/(x −1)+1/(y − 2) = 2
6/(x −1)−3/(y − 2) = 1
Let us put 1/(x −1) = p and 1/(y − 2) = q. Then the given equations
5[1/(x −1)] + 1/(y − 2) = 2 (1)
6[1/(x −1)] − 3[1/(y − 2)] = 1 (2)
can be written as : 5p + q = 2 (3)
6p – 3q = 1 (4)
Equations (3) and (4) form a pair of linear equations in the general form. Now, you can use any method to solve these equations. We get p = 1/3 and q = 1/3
Now, substituting [1/(x −1)] for p, we have
[1/(x −1)] = 1/3,
i.e., x – 1 = 3, i.e., x = 4.
Similarly, substituting [1/(y − 2)] for q, we get
[1/(y − 2)] = 1/3
i.e., 3 = y – 2, i.e., y = 5
Substitute x = 4 and y = 5 in (1) and (2) to check whether they are satisfied.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Then the speed of the boat downstream = (x + y) km/h, and the speed of the boat upstream = (x – y) km/h
Also, time = distance/speed
In the first case, when the boat goes 30 km upstream, let the time taken, in hour, be t^{1}. Then
t^{1} = 30/(x− y)
Let t^{2} be the time, in hours, taken by the boat to go 44 km downstream. Then t^{2} = 44/(x+y). The total time taken, t^{1} + t^{2}, is 10 hours. Therefore, we get the equation
30/(x− y) + 44/(x+y)= 10 (1)
In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. We get the equation
40/(x− y) + 55/(x+y)= 13 (2)
Put 1/(x− y) = u and 1/(x+y) = v
On substituting these values in Equations (1) and (2), we get the pair of linear equations:
30u + 44v = 10 or 30u + 44v – 10 = 0 (4)
40u + 55v = 13 or 40u + 55v – 13 = 0 (5)
Using Crossmultiplication method, we get
u/[44( −13) − 55( −10)] = v/[40( −10) − 30(−13)] = 1/[30(55) − 44(40)]
i.e.,
i.e., u/−22 = v/−10 = 1/−110
i.e., u=1/5 , v= 1/11
Now put these values of u and v in Equations (3), we get
1/(x− y) = 1/5 and 1/(x+y) = 1/11 (6)
Adding these equations, we get
2x = 16
i.e., x= 8
Subtracting the equations in (6), we get
2y = 6
i.e., y= 3
Hence, the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.
Verify that the solution satisfies the conditions of the problem.
3.2 Summary
In this chapter, you have studied the following points:
 Two linear equations in the same two variables are called a pair of linear equations in two
variables. The most general form of a pair of linear equations is
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where a_{1}, a_{2}, b_{1}, b2, c_{1}, c_{2} are real numbers, such that a_{1}^{2} + b_{1}^{2} ≠ 0, a_{2} + b_{2} ≠ 0.
 A pair of linear equations in two variables can be represented, and solved, by the:
 graphical method
 algebraic method
 Graphical Method :
 If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.
 If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent).
 If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.
The graph of a pair of linear equations in two variables is represented by two lines.
 Algebraic Methods : We have discussed the following methods for finding the solution(s)
of a pair of linear equations :
 a_{1}/a_{2} ≠ b_{1}/b_{2}: In this case, the pair of linear equations is consistent.
 a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}: In this case, the pair of linear equations is inconsistent.
 a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2} : In this case, the pair of linear equation is dependent and consistent.
 There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.