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38 MATHEMATICS
3.1 Introduction
You must have come across situations like the one given below :
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.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 39
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 =
x (1)
3 x + 4 y = 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: 2 x + 3 y = 5 x – 2 y – 3 = 0 and x – 0 y = 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 2 x + 3 y = 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 2 x + 3 y = 5. Now let us substitute x = 1 and y = 7 in the equation 2 x + 3 y = 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 2 x + 3 y = 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.
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 41
Fig. 3. Both ways of representing a pair of linear equations go hand-in-hand — the algebraic and the geometric ways. Let us consider some examples.
Example 1 : 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).
Solution : The pair of equations formed is :
y =
x
i.e., x – 2 y = 0 (1)
3 x + 4 y = 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.
x 0 2 x 0
y = 2
x 0 1 y =
x 5 0 2
(i) (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
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equation in one variable, which can be solved easily. For instance, putting x = 0 in Equation (2), we get 4 y = 20, i.e., y = 5. Similarly, putting y = 0 in Equation (2), we get 3 x = 20, i.e., x =
. But as
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 – 2 y = 0 and 3 x + 4 y = 20, as
shown in 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.
Example 2 : Romila went to a stationery shop and purchased 2 pencils and 3 erasers
for ` 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 ` 18. Represent this
situation algebraically and graphically.
Solution : Let us denote the cost of 1 pencil by _x_ and one eraser by y. Then the
algebraic representation is given by the following equations:
2 x + 3 y = 9 (1)
4 x + 6 y = 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.
Fig. 3.
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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.
EXERCISE 3.
1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. 2. The coach of a cricket team buys 3 bats and 6 balls for 3900. Later, she buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically and geometrically. 3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation algebraically and geometrically.
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
Fig. 3.
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 45
point (4, 2) lies on the lines represented by both the equations x – 2 y = 0 and 3 x + 4 y = 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 2 x + 3 y = 9 and 4 x + 6 y = 18 have infinitely many solutions. This should not surprise us, because if we divide the equation 4 x + 6 y = 18 by 2 , we get 2 x + 3 y = 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 3 and 1 respectively. Or, each pencil can cost 3.75 and eraser can cost 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:
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 47
Fig. 3.
are (i) intersecting, then 1 1 2 2
a b a b
(ii) coincident, then 1 1 1 2 2 2
a b c a b c
(iii) parallel, then 1 1 1 2 2 2
a b c a b c
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.
Example 4 : Check graphically whether the pair of equations x + 3 y = 6 (1)
and 2 x – 3 y = 12 (2)
is consistent. If so, solve them graphically. Solution : 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.
Table 3.
x 0 6 x 0 3
y =
x 2 0 y =
x
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. We observe that there is a point B (6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, i.e., the given pair of equations is consistent.
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Example 5 : Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
5 x – 8 y + 1 = 0 (1)
3 x –
y (^) +
Solution : Multiplying Equation (2) by
we get 5 x – 8 y + 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.
Example 6 : 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. Solution : Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are : y = 2 x – 2 (1) and y = 4 x – 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.
x 2 0
y = 2 x – 2 2 – 2
x 0 1
y = 4 x – 4 – 4 0
Fig. 3.
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(i) x + y = 5, 2 x + 2 y = 10 (ii) x – y = 8, 3 x – 3 y = 16 (iii) 2 x + y – 6 = 0, 4 x – 2 y – 4 = 0 (iv) 2 x – 2 y – 2 = 0, 4 x – 4 y – 5 = 0
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. 6. Given the linear equation 2 x + 3 y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines 7. Draw the graphs of the equations x – y + 1 = 0 and 3 x + 2 y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x -axis, and shade the triangular region.
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 non-integral coordinates like (^) 3 , 2^7 ,
(^4) , 1 13 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.
Example 7 : Solve the following pair of equations by substitution method: 7 x – 15 y = 2 (1) x + 2 y = 3 (2) Solution : Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
x + 2 y = 3 and write it as x = 3 – 2 y (3)
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 51
Step 2 : Substitute the value of x in Equation (1). We get 7(3 – 2 y ) – 15 y = 2 i.e., 21 – 14 y – 15 y = 2 i.e., – 29 y = –
Therefore, y =
Step 3 : Substituting this value of y in Equation (3), we get
x = 3 –
Therefore, the solution is x =
, y =
Verification : Substituting x =
and y =
, 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.
Example 8 : Solve Q.1 of Exercise 3.1 by the method of substitution. Solution : 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 – 7 t + 42 = 0 (1)
and s + 3 = 3 ( t + 3), i.e., s – 3 t = 6 (2)
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 53
Solution : The pair of linear equations formed were: x + 2 y – 4 = 0 (1) 2 x + 4 y – 12 = 0 (2) We express x in terms of y from Equation (1) to get x = 4 – 2 y Now, we substitute this value of x in Equation (2) to get 2(4 – 2 y ) + 4 y – 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.
EXERCISE 3.
1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 (ii) s – t = 3
x – y = 4 3 2 6
s (^) t
(iii) 3 x – y = 3 (iv) 0.2 x + 0.3 y = 1. 9 x – 3 y = 9 0.4 x + 0.5 y = 2.
(v) (^2) x 3 y 0 (vi)
3 5 2 2 3
x y
3 x 8 y 0
13 3 2 6
x (^) y
2. Solve 2 x + 3 y = 11 and 2 x – 4 y = – 24 and hence find the value of ‘ m ’ for which y = mx + 3. 3. Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) The coach of a cricket team buys 7 bats and 6 balls for 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball.
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(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes 9 11
, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5 6
. Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
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.
Example 11 : 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 ` 2000 per month, find their monthly incomes.
Solution : Let us denote the incomes of the two person by 9 _x_ and 7 x and their expenditures by 4 _y_ and 3 y respectively. Then the equations formed in the situation is given by : 9 x – 4 y = 2000 (1) and 7 x – 3 y = 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: 27 x – 12 y = 6000 (3) 28 x – 12 y = 8000 (4) Step 2 : Subtract Equation (3) from Equation (4) to eliminate y , because the coefficients of y are the same. So, we get (28 x – 27 x ) – (12 y – 12 y ) = 8000 – 6000 i.e., x = 2000 Step 3 : Substituting this value of x in (1), we get 9(2000) – 4 y = 2000
i.e., y = 4000
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Step 2 : Subtracting Equation (4) from Equation (3), (4 x – 4 x ) + (6 y – 6 y ) = 16 – 7 i.e., 0 = 9, which is a false statement. Therefore, the pair of equations has no solution.
Example 13 : The sum of a two-digit 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?
Solution : 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 10 y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5). According to the given condition. (10 x + y ) + (10 y + 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. Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.
EXERCISE 3.
1. Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2 x – 3 y = 4 (ii) 3 x + 4 y = 10 and 2 x – 2 y = 2
(iii) 3 x – 5 y – 4 = 0 and 9 x = 2 y + 7 (iv)
(^2) 1 and 3 2 3 3
x (^) y (^) x y
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 57
2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces
to 1. It becomes 1 2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (iv) Meena went to a bank to withdraw 2000. She asked the cashier to give her 50 and 100 notes only. Meena got 25 notes in all. Find how many notes of 50 and 100 she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
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 35 and the cost of 2 oranges and 4 apples is 28. Let us find the cost of an orange and an apple.
Let us denote the cost of an orange by _x_ and the cost of an apple by y. Then, the equations formed are :
5 x + 3 y = 35, i.e., 5 x + 3 y – 35 = 0 (1) 2 x + 4 y = 28, i.e., 2 x + 4 y – 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
