NCERT Solutions Class 10th Maths Chapter – 3 Pair of Linear Equations in Two Variables Examples

Example 1. Check graphically whether the pair of equations
(1) x + 3y = 6
(2) 2x-3y= 12

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 3.2

and,
is consistent. If so, solve them graphically.

Example 2. Graphically, find whether the following pair of equations has no solution,
unique solution or infinitely many solutions:
(1) 5x-8y+1=0
(2) 3x- 24/5 y+3/5=0

Solution: Multiplying Equation (2) by5/3 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.

Example 3. 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:
(1) y = 2x-2
(2) y= 4x-4
and
Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations. They are given in Table 3.3.

Plot the points and draw the lines passing through them to represent the equations, as shown in Fig. 3.2.

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.

Example 4. Solve the following pair of equations by substitution method:
(1) 7x-15y=2
(2) x+2y= 3

Solution:
Step 1 – We pick either of the equations and write variable in terms of the other.
Let us consider the Equation
and write it as
(3) x+2y= 3
x = 3-2y

Step 2 – Substitute the value of x in Equation (1). We get
7(3-2y) – 15y = 2
2114y-15y
-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

Example 5. Solve the following question-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 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
(1) s-7-7 (-7), i.e., s-71 + 42=0
(2) s+3=3(t+3), i.e., s – 31 = 6
and
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.

Example 6. In a shop the cost of 2 pencils and 3 erasers is ₹9 and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.

Solution: The pair of linear equations formed were:
(1) 2x + 3y = 9
(2) 4x+6y= 18

We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
(3) x = 9-3y/2

Now we substitute this value of x in Equation (2), to get
4(9-3y)/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 because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.

Example 7. Two rails are represented by the equations
x+2y-4=0 and 2x+4y-12=0. Will the rails cross each other?

Solution: The pair of linear equations formed were:
(1) x + 2y – 4 = 0
(2) 2x+4y-12=0
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.

Example 8. 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 ₹ 9x and ₹ 7x and their
expenditures by 4y and ₹ 3y respectively. Then the equations formed in the situation
is given by:
(1) 9x-4y= 2000
(2) 7x-3y=2000
and

Step 1 – Multiply Equation (1) by 3 and Equation (2) by make the coefficients of y equal. Then we get the equations:
(3) 27x-12y= 6000
(4) 28x-12y= 8000

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 18,000 and 14,000, respectively.

Example 9. Use elimination method to find all possible solutions of the following pair of linear equations:
(1) 2x + 3y = 8
(2) 4x+6y= 7

Solution:
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:
(3) 4x+6y= 16
(4) 4x+6y= 7

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.

Example 10. 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 10x+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.
(1) (10x+y)+(10y+x) = 66
i.e., 11(x+y)= 66
i.e., x + y = 6

We are also given that the digits differ by 2, therefore,
either
(2) x-y=2
(3) y-x=2
or
Ifx – y = 2, then solving (1) and (2) by elimination, we
In this case, we get the number 42.
4 and y = 2.
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.