NCERT Solutions: Pair of Linear Equations in Two Variables (Exercise 3.1)

# Pair of Linear Equations in Two Variables (Exercise 3.1) NCERT Solutions - Mathematics (Maths) Class 10

Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.
Ans:
(i)

Let number of boys = x
Number of girls = y
Given that total number of student is 10.
So, x + y = 10
Subtract y both side we get,
x = 10 – y
Putting y = 0 , 5, 10 we get,
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0

 x 10 5 y 0 5

Given that If the number of girls is 4 more than the number of boys.
So, y = x + 4
Putting x = -4, 0, 4, and we get,
y = - 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8

 x -4 0 4 y 0 4 8

Graphical representation:
Therefore, number of boys = 3 and number of girls = 7

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.
According to the question, the algebraic expression cab be represented as;
5x + 7y = 50
7x + 5y = 46
For, 5x + 7y = 50 or  x = (50 - 7y) / 5, the solutions are;

 x 3 0 y 5 7.14

For 7x + 5y = 46 or x = (46 - 5y) / 7, the solutions are;

 x 6.57 3 y 0 5

Hence, the graphical representation is as follows:

From the graph, it can be seen that the given lines cross each other at point (3, 5).So, the cost of a pencil is Rs. 3 and cost of a pen is Rs. 5.

Q2. On comparing the ratios a1 / a2 , b1 / b2 , c1 / c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Ans:
(i) Given expressions;
5x − 4y + 8 = 0
7x + 6y − 9 = 0

Comparing these equations with a1x + b1y + c1 = 0
And a2x + b2y + c2 = 0
We get,
a1 = 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9
(a1 / a2) = 5 / 7
(b1 / b2) = -4 / 6 = -2 / 3
(c1 / c2) = 8 / -9
Since, (a/ a2) ≠ (b1 / b2)
So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;
9x + 3y + 12 = 0
18x + 6y + 24 = 0

Comparing these equations with a1x + b1y + c1 = 0
And a2x + b2y + c2 = 0
We get,
a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2= 24
(a1/ a2) = 9 / 18 = 1 / 2
(b1 / b2) = 3 / 6 = 1 / 2
(c1 / c2) = 12 / 24 = 1 / 2
Since (a1/ a2) = (b1 / b2) = (c1 / c2)
So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;
6x – 3y + 10 = 0
2x – y + 9 = 0

Comparing these equations with
a1x + b1y + c1 = 0
And a2x + b2y + c2= 0
We get,
a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2= -1, c2= 9
(a1 / a2) = 6 / 2 = 3 / 1
(b1 / b2) = -3 / -1 = 3 / 1
(c1 / c2) = 10 / 9
Since (a1 / a2) = (b1 / b2) ≠ (c1 / c2)
So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

Q3. On comparing the ratio, (a1 / a2), (b1 / b2), (c1 / c2) find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) (3 / 2)x + (5 / 3)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) (4 / 3)x + 2y = 8 ; 2x + 3y = 12
Ans:
(i) Given: 3x + 2y = 5 or 3x + 2y -5 = 0
and 2x – 3y = 7 or 2x – 3y -7 = 0
Comparing these equations with a1x + b1y + c1 = 0
And a2x + b2y + c2 = 0
We get,
a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2= -3, c2= -7
(a1 / a2) = 3 / 2
(b1 / b2) = 2 / -3
(c1 / c2) = -5 / -7 = 5 / 7
Since, (a1 / a2) ≠ (b1 / b2)
So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) Given: 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a1 = 2, b1 = -3, c1 = -8
a2= 4, b2= -6, c2 = -9
(a1 / a2) = 2 / 4 = 1 / 2
(b1 / b2) = -3 / -6 = 1 / 2
(c1 / c2) = -8 / -9 = 8 / 9
Since , (a1/ a2) = (b1/ b2) ≠ (c1 / c2)
So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii) Given: (3 / 2)x + (5 / 3)y = 7 and 9x – 10y = 14
Therefore,
a1= 3 / 2, b1= 5 / 3, c1 = -7
a2= 9, b2= -10, c2= -14
(a1/ a2) = 3 / (2 × 9) = 1 / 6
(b1/ b2) = 5 / (3× - 10)= -1 / 6
(c1 / c2) = -7 / -14 = 1 / 2
Since, (a1/ a2) ≠ (b1/ b2)
So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) Given: 5x – 3y = 11 and – 10x + 6y = –22
Therefore,
a1= 5, b1 = -3, c1 = -11
a2= -10, b2 = 6, c2= 22
(a1/ a2) = 5 / (-10) = -5 / 10 = -1 / 2
(b1/ b2) = -3 / 6 = -1 / 2
(c1 / c2) = -11 / 22 = -1 / 2
Since (a1/ a2) = (b1/ b2) = (c1 / c2)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v) Given: (4 / 3)x + 2y = 8 and 2x + 3y = 12
a1= 4 / 3 , b1= 2 , c1 = -8
a2= 2, b2 = 3 , c2= -12
(a1/ a2) = 4 / (3 × 2)= 4 / 6 = 2 / 3
(b1/ b2) = 2 / 3
(c1 / c2) = -8 / -12 = 2 / 3
Since (a1/ a2) = (b1/ b2) = (c1 / c2)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ans: (i) Given, x + y = 5 and 2x + 2y = 10
(a1/ a2) = 1 / 2
(b1/ b2) = 1 / 2
(c1 / c2) = 1 / 2
Since (a1/ a2) = (b1/ b2) = (c1 / c2)
∴The equations are coincident and they have infinite number of possible solutions.
So, the equations are consistent.
For, x + y = 5 or x = 5 – y

 x 0 5 y 5 0

For 2x + 2y = 10 or x = (10 - 2y) / 2

 x 0 5 y 5 0

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other. Therefore, the equations have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16
(a1/ a2) = 1 / 3
(b1/ b2) = -1 / -3 = 1 / 3
(c1 / c2) = 8 / 16 = 1 / 2
Since, (a1/ a2) = (b1/ b2) ≠ (c1 / c2)
The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0
(a1/ a2) = 2 / 4 = 1 / 2
(b1/ b2) = 1 / -2
(c1 / c2) = -6 / -4 = 3 / 2
Since, (a1/ a2) ≠ (b1/ b2)
The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.
Now, for 2x + y – 6 = 0 or y = 6 – 2x

 x 0 3 y 6 0

And for 4x – 2y – 4 = 0 or y = (4x - 4) / 2

 x 0 1 y -2 0

So, the equations are represented in graphs as follows:

From the graph, it can be seen that these lines are intersecting each other at only one point (2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
(a1/ a2) = 2 / 4 = 1 / 2
(b1/ b2) = -2 / -4 = 1 / 2
(c1 / c2) = 2 / 5
Since, a1/ a2= b1/ b2≠ c1 / c2
Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

Q5. 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.
Ans:
Let us consider.
The width of the garden is x and length is y.
Now, according to the question, we can express the given condition as;
x – y = 4
and
y + x = 36
Now, taking x – y = 4 or y = x + 4

 x 4 0 y 0 -4

For y + x = 36, y = 36 – x

 x 12 20 y 24 16

The graphical representation of both the equation is as follows:
From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

Q6. Given the linear equation 2x + 3y – 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
Ans: (i)
Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;
(a1/ a2) ≠ (b1/ b2)
Thus, another equation could be 2x – 7y + 9 = 0, such that;
(a1/ a2) = 2 / 2 = 1 and (b1/ b2) = 3 / -7
Clearly, you can see another equation satisfies the condition.
(ii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;
(a1/ a2) = (b1/ b2) ≠ (c1 / c2)
Thus, another equation could be 6x + 9y + 9 = 0, such that;
(a1/ a2) = 2 / 6 = 1 / 3
(b1/ b2) = 3 / 9= 1 / 3
(c1 / c2) = -8 / 9

Clearly, you can see another equation satisfies the condition.
(iii) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;
(a1/ a2) = (b1/ b2) = (c1 / c2)
Thus, another equation could be 4x + 6y – 16 = 0, such that;
(a1/ a2) = 2 / 4 = 1 / 2 ,(b1/ b2) = 3 / 6 = 1 / 2, (c1 / c2) = -8 / -16 = 1 / 2
Clearly, you can see another equation satisfies the condition.

Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Ans:
Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.
For, x – y + 1 = 0 or x = 1 + y

 x 0 -1 y 1 0

For, 3x + 2y – 12 = 0 or x = (12 - 2y) / 3

 x 0 4 y 6 0

Hence, the graphical representation of these equations is as follows:

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

The document Pair of Linear Equations in Two Variables (Exercise 3.1) NCERT Solutions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Pair of Linear Equations in Two Variables (Exercise 3.1) NCERT Solutions - Mathematics (Maths) Class 10

 1. What is a pair of linear equations in two variables?
Ans. A pair of linear equations in two variables is a set of two equations that involve two variables and are linear in nature. The general form of a pair of linear equations in two variables is ax + by = c and dx + ey = f. Here, x and y are the variables, and a, b, c, d, e, and f are constants.
 2. What is the importance of solving pair of linear equations in two variables?
Ans. Solving a pair of linear equations in two variables helps in finding the values of the variables that satisfy both the equations simultaneously. This is useful in various real-life situations, such as in finding the cost and revenue of a business, determining the distance between two points, and analyzing the motion of objects.
 3. How many solutions can a pair of linear equations in two variables have?
Ans. A pair of linear equations in two variables can have three types of solutions - unique solution, infinite solutions, and no solution. If the equations have different slopes, they will intersect at one point, and there will be a unique solution. If the equations have the same slope and intersect at all points, there will be infinite solutions. If the equations have the same slope and do not intersect, there will be no solution.
 4. What is the graphical method of solving a pair of linear equations in two variables?
Ans. The graphical method of solving a pair of linear equations in two variables involves plotting the equations on a graph and finding the point of intersection. The point of intersection represents the values of the variables that satisfy both the equations. If the lines are parallel, there is no point of intersection, and hence no solution. If the lines are coincident, all points on the line are solutions, and hence there are infinite solutions.
 5. What is the substitution method of solving a pair of linear equations in two variables?
Ans. The substitution method of solving a pair of linear equations in two variables involves solving one equation for one variable in terms of the other and substituting it in the other equation. This reduces the number of variables to one, and we can solve for the value of that variable. This value can then be substituted in either of the equations to find the value of the other variable. This method is useful when one of the variables has a coefficient of one.

## Mathematics (Maths) Class 10

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