Pair of Linear Equations in Two Variables

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Class 10 maths chapter 3 NCERT Solutions Pair of Linear Equations in Two Variables is concerned with understanding and exploring the concept of graph plotting and straight-line formation using linear equations in two variables. Students preparing for class 10 maths chapter 3 will be able to clear all their concepts on linear equations at the root level. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

NCERT solutions for class 10 maths chapter 3 of the section, concentrate on the essential concepts such as the Algebraic methods for solving a pair of linear equations in two variables, such as the elimination method, cross-multiplication methods, and equations reducible to a pair of linear equations in two variables. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Pair of linear equations in two variables class 10 Chapter 3 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The Class 10 Maths NCERT Solutions Chapter 3 are available in pdf format below, and some of them are also included in the exercises.

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Exercise 3.1

Question 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." (Isnt this interesting?) Represent this situation algebraically and graphically.

Solution

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Consider Aftab's age as $x$ and his daughter's age as $y.$

Then , seven years ago,

Aftab's age $=x−7$

His daughter's age $=y−7$

According to the question,

$x−7=7(y−7)$

$x−7=7y−49$

$x−7y=−49+7$

$x−7y=−49+7$

$x−7y=−42$ …(i)

After three years,

Aftab's age $=x+3$

His daughter's age $=y+3$

According to the question,

$x+3=3(y+3)$

$x+3=3y+9$

$x−3y=9−3$

$x−3y=6$ …(ii)

Representing equation $(i)$ and $(ii)$ geometrically, we plot these equations by finding points on the lines representing these two equations

$x−7y=−42⟹x=7y−42$

$x=7y−42$ | $42$ | $35$ | $49$ |

$y$ | $12$ | $11$ | $13$ |

$x−3y=6⟹x=3y+6$

$x=3y+6$ | $42$ | $36$ | $48$ |

$y$ | $12$ | $10$ | $14$ |

From the graph we can see that two lines will intersect at a point.

$x−7y=−42$

$x−3y=6$

On subtracting the two equations, we get

$4y=48$

or $y=12$

Substituting value of $y$ in (2),

$x−36=6$

$∴x=42$

Question 2

The coach of a cricket team buys $3$ bats and $6$ balls for $c3900.$ Later, she buys another bat and $3$ more balls of the same kind for $c1300.$ Represent this situation algebraically and geometrically

Solution

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Reffer imagr.

let rate of bat = $xRs.$ and rate of ball = $yRs.$

$3x+6y=3900.....(i)$

and $x+3y=1300.....(ii)$

or $x=(1300−3y)$

now equation (i) can be written as

$3(1300−3y)+6y=3900$

$3900−9y+6y=3900$

$∴y=0$

Ans:- $x=1300$

Question 3

The cost of $2kg$ of apples and $1kg$ of grapes on a day was found to be $Rs160$. After a month, the cost of $4kg$ of apples and $2kg$ of grapes is $Rs300$. Represent the situation algebraically and geometrically.

Solution

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$x$ | $65$ | $55$ | $45$ | $35$ |

$y$ | $20$ | $40$ | $60$ | $80$ |

So the algebraic representation can be as follows:

$2x+y=160$

$4x+2y=300⇒2x+y=150$

$4x+2y=300⇒2x+y=150$

The situation can be represented graphically by plotting these two equations.

$2x+y=160⇒y=160−2x$

$2x+y=150⇒y=150−2x$

$x$ | $70$ | $60$ | $50$ | $40$ |

$y=160−2x$ | $20$ | $40$ | $60$ | $80$ |

$x$ | $65$ | $55$ | $45$ | $35$ |

$y=150−2x$ | $20$ | $40$ | $60$ | $80$ |

Exercise 3.2

Question 1

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 Rs $50,$ whereas $7$ pencils and $5$ pens together cost Rs $46.$ Find the cost of one pencil and that of one pen.

(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 Rs $50,$ whereas $7$ pencils and $5$ pens together cost Rs $46.$ Find the cost of one pencil and that of one pen.

Solution

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and number of girls $=y.$

According to the question,

$x+y=10$

or, $y=10−x$ ………(i)

$y=x+4$ ………..(ii)

or, $y=10−x$ ………(i)

$y=x+4$ ………..(ii)

$x$ | $1$ | $2$ | $3$ | $4$ |

$y$ | $9$ | $8$ | $7$ | $6$ |

$x$ | $1$ | $2$ | $3$ | $4$ |

$y$ | $5$ | $6$ | $7$ | $8$ |

(ii) Consider price of a pencil is $x$

and price of a pen is $y.$

According to the question,

$5x+7y=50$.....(i)

$7x+5y=46$ .….(ii)

$7x+5y=46$ .….(ii)

$x$ | $1$ | $2$ | $3$ | $4$ |

$y$ | $7.8$ | $6.4$ | $5$ | $4.2$ |

$x$ | $1$ | $2$ | $3$ | $4$ |

$y$ | $6.4$ | $5.7$ | $5$ | $4.2$ |

Graphical solution :

Price of a pencil is Rs $3$

Price of a pen is Rs $5.$

Question 2

On comparing the ratios $a_{2}a_{1} ,b_{2}b_{1} $ and $c_{2}c_{1} $, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

$5x−4y+8=0;7x+6y−9=0$

$5x−4y+8=0;7x+6y−9=0$

Solution

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The given linear equation are

$⇒5x−4y+8=0....eq1$

$⇒a_{1}=5,b_{1}=−4,c_{1}=8$

$⇒7x+6y−9=0...eq2$

$⇒a_{2}=7,b_{2}=6,c_{2}=−9$

$⇒a_{2}a_{1} =75 $

$⇒b_{2}b_{1} =6−4 $

$⇒c_{2}c_{1} =98 $

comparing

$⇒a_{2}a_{1} ,b_{2}b_{1} ,c_{2}c_{1} $

$⇒a_{2}a_{1} =b_{2}b_{1} $

Hence,the line represented by eq1 and eq2 intersect at a point

$⇒5x−4y+8=0....eq1$

$⇒a_{1}=5,b_{1}=−4,c_{1}=8$

$⇒7x+6y−9=0...eq2$

$⇒a_{2}=7,b_{2}=6,c_{2}=−9$

$⇒a_{2}a_{1} =75 $

$⇒b_{2}b_{1} =6−4 $

$⇒c_{2}c_{1} =98 $

comparing

$⇒a_{2}a_{1} ,b_{2}b_{1} ,c_{2}c_{1} $

$⇒a_{2}a_{1} =b_{2}b_{1} $

Hence,the line represented by eq1 and eq2 intersect at a point

Question 3

On comparing the ratios $a_{2}a_{1} ,b_{2}b_{1} $ and $c_{2}c_{1} $, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) $3x+2y=5;2x−3y=7$

(i) $3x+2y=5;2x−3y=7$

(ii) $2x−3y=8;4x−6y=9$

(iii) $23 x+35 y=7;9x−10y=14$

(iv) $5x−3y=11;−10x+6y=−22$

(v) $34 x+2y=8;2x+3y=12$

Solution

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(i) $a_{2}a_{1} =23 ,b_{2}b_{1} =−32 ,c_{2}c_{1} =75 $

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

the lines intersect and have an unique consistent solution.

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

(ii) $a_{2}a_{1} =42 =21 ,b_{2}b_{1} =−6−3 =21 ,c_{2}c_{1} =98 $

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair

(iii) $a_{2}a_{1} =923 =61 ,b_{2}b_{1} =−1035 =−61 ,c_{2}c_{1} =147 =21 $

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

the lines intersect and have an unique consistent solution.

(iv) $a_{2}a_{1} =−105 =−21 ,b_{2}b_{1} =6−3 =−21 ,c_{2}c_{1} =−2211 =−21 $

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

(v) $a_{2}a_{1} =234 =32 ,b_{2}b_{1} =32 ,c_{2}c_{1} =128 =32 $

$∵a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Question 4

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) $2x−2y−2=0,4x−4y−5=0$

(ii) $x−y=8,3x−3y=16$

(iii) $2x+y−6=0,4x−2y−4=0$

(iv) $2x−2y−2=0,4x−4y−5=0$

Solution

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$2x+2y=10$ ...(ii)

$⇒x+y=5$

$⇒y=5−x$

x | 0 | 3 |

y | 5 | 2 |

$2x+2y=10$

$⇒2y=(10−2x)$

$⇒y=210−2x =5−x$ ...(iii)

x | 5 | 2 |

y | 0 | 3 |

So, the equation is consistent and has infinitely many solution

$3x−3y=16$ ....(ii)

$⇒x−y=8$

$⇒−x+y=−8$

$⇒y=−8+x$

$⇒y=x−8$

$⇒y=x−8$

x | 8 | 0 |

y | 0 | -8 |

$⇒3x=16+3y$

$⇒3x−16=3y$

$⇒y=33x−16 ⇒y=x−316 $

$x$ | $316 $ | $0$ |

$y$ | $0$ | $3−16 $ |

$4x−2y−4=0$

$2x+y=6$ ....(i)

$4x−2y=4$ ...(ii)

For equation (i), $2x+y=6⇒y=6−2x$

$x$ | $0$ | $3$ |

$y$ | $6$ | $0$ |

For equation (ii), $4x−2y=4⇒24x−4 =y$

$x$ | $1$ | $0$ |

$y$ | $0$ | $−2$ |

$x=2,y=2$ is the solution of the given pairs of equation . So. solution is consistent.

$4x−4y=5$ ...(ii)

$2x−2y=2⇒2x−2=2y$

$y=x−1$

$x$ | $0$ | $1$ |

$y$ | $−1$ | $0$ |

$⇒44x−5 =y$

Plot point $(0,−45 )$ and $(45 ,0)$ and join them to get the equation $4x−4y=5$ on a graph.

$x$ | $0$ | $45 $ |

$y$ | $−45 $ | $0$ |

The two lines never intersect, so, the solution is inconsistent.

Question 5

Half the perimeter of a rectangular garden, whose length is $4m$ more than its width, is $36m$. Find the dimensions of the garden.

Solution

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Let the width of the garden $=x$ meter

Then length=$(x+4)$ meter

Half perimeter $=36m$

So perimeter of garden $=(2×36)=72$ meters

According to the question

$⇒2(l+b)=72$

$⇒2(x+x+4)=72$

$⇒2x+2x+4=74⇒4x=64⇒x=16$ meters

Hence,the width of the garden $=16$ meters

The length of the garden $=(16+4)=20$ meters

Then length=$(x+4)$ meter

Half perimeter $=36m$

So perimeter of garden $=(2×36)=72$ meters

According to the question

$⇒2(l+b)=72$

$⇒2(x+x+4)=72$

$⇒2x+2x+4=74⇒4x=64⇒x=16$ meters

Hence,the width of the garden $=16$ meters

The length of the garden $=(16+4)=20$ meters

Question 6

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

(i) intersecting lines (ii) parallel lines (iii) coincident lines

Solution

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a) Intersecting lines

Solution: For intersecting line, the linear equations should meet following condition:

$a_{2}a_{1} =b_{2}b_{1} $

For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:

$4x+9y−8=0$

(b) Parallel lines

Solution: For parallel lines, the linear equations should meet following condition:

$a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $

For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:

$4x+6y–24=0$

(c) Coincident lines

Solution: For getting coincident lines, the equations should meet following condition;

$a_{2}a_{1} =b_{2}b_{1} =c_{2}c_{1} $

For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:

$4x+6y–16=0$

Question 7

Draw the graphs of $x−y+1=0$ and $3x+2y−12=0$. Determine the coordinates of the vertices of the triangle formed by these lines and $x$-axis and shade the triangular area.

Solution

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Given,

$x−y+1=0$....(1)

$3x+2y−12=0$....(2)

put $x=0$ in (1)

$0−y+1=0⇒y=1$

$(0,1)$.....(a)

put $y=0$ in (2)

$3x+2(0)−12=0→x=4$

$(4,0)$.....(b)

put $x=0$ in (2)

$3(0)+2y−12=0→y=6$

$(0,6)$......(c)

put $y=0$ in (1)

$x−0+1=0⇒x=−1$

$(−1,0)$...(d)

From figure, the vertices of triangle are $(−1,0),(2,3),(4,0)$

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Practice more questions

Linear equations are made up of both constants and variables. A linear equation with two variables x and y is an equation of the form Ax + By + C = 0, where A, B, and C are real values.

A pair of linear equations in two variables is two linear equations in the same two variables. Linear equations with two variables are written in standard form:

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}, b_{2}, c_{1}, c_{2} are real numbers, such that a_{1}^{2} + b_{1}^{2} = 0, a_{2}^{2} + b_{2}^{2} = 0.

The necessary answer is the coordinates of the point (x, y) that satisfy the system of two linear equations in two variables. This is the point at which the two lines representing the two equations cross. There are two approaches to solving a pair of linear equations in two variables:

- Graphical Method
- Algebraic Method

**4. Graphical Method Of Solution Of A Pair Of Linear Equations**

While solving for the above 2 linear equations, one of the three cases might arise:

- If , the system is said to be consistent since it has a unique solution and a pair of straight lines representing the above equations meet at just one point.
- If , the system is dependent and has infinitely many solutions and the pair of lines representing the equations coincide.
- If , the system is inconsistent and has no solutions and the pair of lines representing the equations are parallel or they do not intersect at any point.

To solve the above equations, use one of the three Algebraic approaches listed below:

**Substitution Method**

Step 1 - From one equation, get the value of one variable, say y in terms of x or x in terms of y.

Step 2 – Substitute this number into the second equation to obtain the equation in one variable and the solution.

Step 3 – Now, substitute the value/solution acquired in step 2 into the equation obtained in step 1.

**Elimination Method**

Step 1 – If the coefficients of any one variable in both equations are not the same, multiply both equations by suitable non-zero constants to make the coefficients of any one variable numerically equal.

Step 2 – Add or subtract the acquired equations to generate an equation in one variable and solve it.

Step 3 – Substitute the value of the variable obtained in step 2 into either of the original equations to obtain the value of the other variable.

**Cross-Multiplication Method**

We write the equations in the following way:

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Q1. Explain the Graphical Method of Solutions of a Pair of Linear Equations.

Answer: Two lines are represented by the graph of a pair of linear equations with two variables.

- A system is considered to have a unique solution if the graphs of its two equations intersect at a single point.
- A system is considered to have no solution if the graphs of two equations are parallel lines.

- When the graphs of two equations in a system are two coincident lines, the system is said to have an unlimited number of solutions, indicating that it is consistent and dependent.

Q2. How many exercises are there in Chapter 3 of Class 10 Maths?

Answer: Pair of Linear Equations in Two Variables is NCERT Class 10 Maths Chapter 3. This chapter has 7 exercises with 29 well-researched questions, including both graph-based and algebraic sums. Toppr has answers for every exercise. Subject matter specialists developed the NCERT Solutions for Class 10 Maths Chapter 3. You should download them in PDF format so that you can study them whenever and wherever you like.

Q3. Where can I get the Class 10 Maths Chapter 3 solutions?

One of the scoring chapters is Class 10 Maths Chapter 3, Pair of Linear Equation in Two Variables. If you want to receive the solutions for this chapter, go to Toppr's official website. We provide NCERT Solutions for Class 10 Maths in PDF format so that you may utilize them offline. The solutions can also be downloaded via the Toppr App. All you have to do is download the app from the App/Play store, sign in, and get the study material.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability