NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation | SureSolv

NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation

Graphical representation of a pair of linear equations in two variables

NCERT-solutions-class-10-maths-chapter-3-part-1.jpg

In this 7 part series, we will cover the topic Pair of Linear Equations in two variables. This series will complement Chapter 3 of NCERT Class 10 Maths.

Each part will be made up of three components:

  • First: brief but comprehensive concepts explained in an easy to understand way,
  • Second: example problems that will be used for establishing the concepts more clearly and,
  • Third: the solutions to the exercises in the NCERT Chapter.

In this first of the seven part series we'll cover:

  1. What is a linear equation, especially corresponding to a word problem description,
  2. Why the name Linear equationGraphical representation,
  3. How to plot a linear equation on a two axes coordinate system,
  4. Conditions for unique solution of two linear equations in two variables,
  5. Solutions of Problems in Exercise 3.1 of Chapter 3, and
  6. Graphical solution of two linear equations in two variablesvery briefly.

What is a linear equation

All that we say here are for solving math problems, isn't it?

So to know what a linear equation is, and finally how to solve two linear equations in two variables, we'll start with a word problem.

Example problem 1.

Tukai purchased 2 marker pens and 5 jotter pens for Rs.50 and Bubai purchased 2 marker pens and 1 jotter pen from the same shop at the same prices for Rs.26. What is the price of a jotter pen?

Solution to example problem 1.

Let $x$ be the price of a marker pen and $y$ be the price of a jotter pen.


Note: The symbols $x$ and $y$ are variables that represent quantities of price of a marker pen and price of a jotter pen respectively. It is assumed that the prices are in Rupees. The values of $x$ and $y$ are unknown at the beginning. Your job is to find the value of unknown variable $y$ here. You will shortly know why we call these as variables.


Price paid by Tukai for 2 marker pens is,

$2x$, and

Price paid by Tukai for 5 jotter pens is,

$5y$.

Total price paid by Tukai to purchase 2 marker pens and 5 jotter pens is,

$2x+5y$.

This is what we call—an algebraic expression in two variables.

$x$ and $y$ are the two variables.

As this total price is Rs.50,

$2x+5y=50$.

The algebraic expression $2x+5y$ and the constant $50$ are joined by the equality symbol in your first equation in two variables.


Terminologies related to an equation

$2x+5y$ is called as the Left Hand Side (or LHS in short) of the equation and $50$ on the right of the equality symbol is called the Right Hand Side (or RHS in short).

$2x$ and $5y$ are the two terms of the algebraic expression.

The multiplier $2$ of variable $x$ and $5$ of variable $y$ are called the coefficients of variables $x$ and $y$ in the two terms.

Sometimes you may transform the RHS to 0 by shifting the RHS term (or expression) to the LHS by subtracting it in the LHS. Your equation you can rewrite in this way as,

$2x+5y-50=0$.

Basically you have added $(-50)$ to both sides of the equation. This third term of the equation $(-50)$ is called as the constant of the three-term equation.

A linear equation in two variables may be written in general as,

$ax+by+c=0$.

In your problem, $a=2$, $b=5$ and $c=-50$. But don't bother about this general form now. We won't need it for some time to come.


Coming back to the problem, by using the same method you can write your second equation of total price of 2 marker pens and 1 jotter pen purchased by Bubai as,

$2x+y=26$.

Let us reproduce the first equation for ease of looking at the two together.

$2x+5y=50$.

We will now know why the name linear equation, and also plot the two equations graphically onto a two axes coordinate system.

Why the name linear equation—Graphical representation

The primary requirement for any equation in two variables to be a linear equation is—the power of the two variables $x$ and $y$ must be unity 1.

In that case, when you plot the equation, say, $2x+5y=50$ on a graph paper with origin ($x=0$, $y=0$) as $O$ and the two mutually perpendicular axes as $x$ and $y$ both extending on negative and positive directions indefinitely, you will find the plot becomes a straight line.

The straight line for the linear equations $2x+5y=50$ is shown below.

ncert-class-10-maths-chapter-3-linear-eqns-1-1.jpg

Let's see how we obtained this straight line. The method is really simple in this case.

How to plot a straight line from a linear equation onto a two axes x-y coordinate system

The method to plot a straight line from the linear equation $2x+5y=50$ is:

  1. Step 1: Put $y=0$ in the equation to get $x=25$. This pair of (x, y) values represents the point (25, 0) on x-axis. The straight line corresponding to the equation must pass through this point as pair of values of $x$ and $y$ satisfies the equation. We call this pair of values as a solution to the linear equation.
  2. Step 2. Put $x=0$ in the equation to get $y=10$. This pair of (x, y) values represents the point (0, 10) on y-axis. In the same way, the straight line corresponding to the linear equation must also pass through this point.
  3. Step 3. Join the two points (25, 0) and (0, 10) and extend on both directions. You will get your straight line corresponding to the linear equation $2x+5y=50$.

Note: Each point on a straight line having an $\text{(x, y)}$ value pair is a solution to the linear equation corresponding to the straight line. As a straight line consists of infinitely many points, possible value pairs of $\text{(x, y)}$ will also be infinitely many. In other words, values of $x$ and $y$ are not only unknown, these can take any value. That's why these are called variables—values of variables are unknown and vary. A solution to a pair of linear equations produces a unique value pair for $x$ and $y$.


If you follow the same steps for the second linear equation $2x+y=26$, you will get the straight line corresponding to the second equation on to the same coordinate system with same origin.

With $y=0$ in $2x+y=26$ you will get $x=13$ and the first intersection point of x-axis and the straight line as the point (13, 0).

With $x=0$ in $2x+y=26$ you will get $y=26$ and the second intersection point of y-axis and the straight line as the point (0, 26).

Joining these two points you will get the second straight line plotted on the same coordinate system.

ncert-class-10-maths-chapter-3-linear-eqns-1-2.jpg

Well, you have drawn the two straight lines corresponding to the two linear algebraic equations on the two-axes (x, y) coordinate system.

Question is—what will you do with these graphical representation of linear equations!

Answer is simple—you are going to solve the two equations from this graphical representation.

That is also simple in this case. Let's see how.

Graphical solution of two linear equations

Every point on the first line satisfies the first equation and so is a solution of the first equation.

Same way, every point on the second line satisfies the second equation and so is a solution of the second equation.

So it is easy to conclude that the point which lies on both the straight lines will satisfy both the linear equations and so the x, y value pair of this point will be the solution you are trying to find.

This point is the point of intersection of the two straight lines $P$ with coordinates (10, 6) as shown in the graph below.

ncert-class-10-maths-chapter-3-linear-eqns-1-3.jpg

Coordinates of point $P$ are,

$x=10$, and

$y=6$.

So answer to your problem is—price of a jotter pen is, Rs.6.

But how do you find the values of (x, y) coordinates for the point of intersection P?

Answer isby observation.

Obviously, this is not a reliable and accurate method for finding the solution of any pair of linear equations in two variables.

The points may involve coordinates with decimals of one, two, three or still larger granularity that would be very difficult to pin-point on a graph paper.

But don't worry. whenever you find a problem asking you to solve two linear equations in two variables, you will surely find that plotting the lines or finding the coordinates of the two points you can do easily.

Before going ahead further, we will answer the questionCan you find a unique solution to any pair of linear equations in two variables?

What do you think?

Conditions for getting a unique solution for a pair of linear equations in two variables

The answer lies in the characteristics "distinct" and "not parallel" in the specification of the two straight lines for getting a unique solution out of two linear equations.

You can get a unique solution for two linear equations in two variables only if the two equations represent straight lines that are distinct and not parallel to each other.

In our example problem, the two equations represented straight lines that were distinct and not parallel to each other.

There can be three possibilities for a pair of linear equations in two variables.

First possibility

The equations may be distinct with the corresponding straight lines having a unique point of intersection that will be the solution of the two equations.

Second possibility

The equations may have ratios of coefficients of variable $x$ and $y$ equal to each other. The corresponding parallel straight lines will never meet and no solution you can get for such a pair of equations. Example is,

$2x+5y=50$, and

$4x+10y=120$.

Multiplying coefficients of variables $x$ and $y$ of the first equation by same factor 2 you will get the two variable terms in the second equation. Basically the slopes of these two lines will be equal.

Observe that multiplying the constant term 50 of the first equation by 2 you get 100 and this is different from the constant term 120 of the second equation.

Also mark that in the second equation, you can factor out all the three terms by the common factor of 2. The resultant equation wil be reduced to,

$2x+5y=60$.

This is same as $4x+10y=120$ for all practical purposes.

The parallel lines are shown below graphically.

ncert-class-10-maths-chapter-3-linear-eqns-1-4.jpg

Third possibility

The third scenario is when the ratio of the coefficients of two variables as well as the constant terms become equal for the two equations.

An example is,

$2x+5y=60$, and

$4x+10y=120$.

Effectively if you multiply each of the three terms of the first equation by 2 you get the second equation.

We say—the two lines are coincident on each other. Basically the two equations represent the same straight line.

What will happen to the solution? You will have in your hands infinitely many solutions as each point on the two coincident lines is a solution, and by definition infinite number of points make up a line.

The graphical representation of two lines coincident upon each other is shown below.

ncert-class-10-maths-chapter-3-linear-eqns-1-5.jpg

With this knowledge gained now, you have to answer the following questions.

As these questions are on graphical representation of linear equations, instead of answers, the problems will be directly followed by solutions.

It will be good for you, if you first try to answer the questions before going through the solutions.

Problems to solve

Problem 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." Represent this situation algebraically and graphically.

Problem 2.

The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Problem 3.

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Solution to the problems

Problem 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." Represent this situation algebraically and graphically.

Solution 1

When you face a word problem as this one, you have three tasks,

  • First, you have to understand what are given and what are wanted,
  • Second, represent the problem in mathematical form of equations, graphs or any other required form of mathematical model, and
  • Third, Solve the problem using the mathematical representation of the problem.

So let's first take up the first task of understanding the problem from its word description.

A good practice is to right away define the variables for the quantities involved.

Just take care to use as few variables as possible.

In this way while you try to understand the problem, the mathematical model of the problem is automatically formed.

This is a problem of age of two persons and three time periods are involved—seven years back, present times and three years later in future.

The standard practice for this type of problem is to assume the variables as the ages at present times.

Advantage of this approach is—for years past, just subtract number of years from both ages, and for future years, add number of years to both ages.

Let us assume present age of Aftab and his daughter to be $a$ and $d$ years. These are our two variables.

By the first part of the statements, $a-7=7(d-7)$. Note that 7 years are reduced from both the ages (as time passes at same speed for both).

And by the second part of the statements, $a+3=3(d+3)$.

Both of the equations are linear, but for proper equation manipulation, you have to transform the equations in the standard form of $px+qy+r=0$, where $p$, $q$ are the two coefficients and $r$ the constant.

The first equation is,

$a-7=7(d-7)$,

Or, $a-7d+42=0$ in standard form.

The second equation is,

$a+3=3(d+3)$

Or, $a-3d-6=0$ in standard form.

For graphical or geometric representation we'll plot values of $a$ on x axis and values of $d$ on y axis.

For the first equation, at $d=0$, $a=-42$, that is, one point on the corresponding straight line is of coordinates (-42, 0).

And at $a=0$, $d=6$, that is, the second point on the corresponding straight line is of coordinates (0, 6).

You will get the straight line corresponding to linear equation $a-7d+42=0$ by joining these two points and extending the line on both directions. It is shown below.

ncert-class-10-maths-chapter-3-linear-eqns-1-6.jpg

Same way, for the second equation $a-3d-6=0$ the two points on the line are, (6, 0) and (0, -2). The corresponding straight line is shown below.

ncert-class-10-maths-chapter-3-linear-eqns-1-7.jpg

And when you plot the two lines together you will get the following graphical representation.

ncert-class-10-maths-chapter-3-linear-eqns-1-8.jpg

The two lines intersect at point P with values of $a=42$, which is Aftab's present age, and $d=12$, which is his daughter's present age.

7 years ago Aftab's daughter was 5 years old and Aftab 35 years old, 7 times the age of his daughter's age.

And 3 years from now, Aftab's daughter will be 15 and Aftab will be 45, three times the age of his daughter's age.

Both the given conditions are satisfied by the solution. This is called verification of the solution. It makes you confident about the correctness of your solution.

Though these values were not asked for, you got the solution of the pair of linear equations also. You just have to draw the lines on the same x-y coordinate axes.

Problem 2

The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Solution 2.

You have to assume here the prices of a bat and a ball remains same on both occasions of purchases.

Let $x$ be the price (in Rs.) of a bat and $y$ be the price (in Rs.) of a ball.

For the first purchase the linear equation for the total price is then,

$3x+6y=3900$.

The first step that you have to take is to examine the equation and eliminate any common factor of all three terms. In this case you can divide the equation by 3 to transform the equation to its minimal form.

$x+2y=1300$.

It is always easier to deal with this minimal form of an equation.

The two points on the corresponding straight line for $y=0$ and $x=0$ are, (1300, 0) and (0, 650).

To draw the corresponding straight line first you have to calibrate the units of both x-axis and y-axis to $1=100$. In this way 13 will represent 1300, and 6.5 will represent 650.

Now you face a difficulty of pin-pointing 650 on the y-axis. The point falls between two marked values. How to draw the line then?

These are the cases when you will examine the equation and look for any convenient value of $y$ and $x$ on the line (satisfying the equation) that you can mark on the graph paper confidently and accurately. In this case with a little thought you can easily discover the second convenient point (300, 500) satisfying the equation.

The corresponding geometrical representation of the straight line will be,

ncert-class-10-maths-chapter-3-linear-eqns-1-9.jpg

For the second purchase the total cost of 1 bat and 3 balls has been Rs.1300. So the algebraic equation will be,

$x+3y=1300$.

The two points on the corresponding straight line for $y-0$ and $x=0$ are, (1300, 0) and (0, $\frac{1300}{3}$).

Again the second point is infeasible to pin-point with accuracy. But with a bit of examination of this second equation this time also you discover the second convenient point (400, 300) satisfying the equation.

The line through these two points plotted on the same x-y coordinate system as above you will get the following situation,

ncert-class-10-maths-chapter-3-linear-eqns-1-10.jpg

The second straight line will pass through the point (1300, 0) on x-axis which will be the intersection point of the two lines. The solution will be the awkward situation of Free balls.

Problem 3.

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Solution 3.

Let us assume $x$ and $y$ to be the cost per kg of apples and grapes respectively.

For the first purchase, the total cost of 2 kg of apples and 1 kg of grapes will then be represented by,

$2x+y=160$.

For $y=0$ and $x=0$ the two points on the straight line corresponding to the algebraic equation are, (80, 0) and (0, 160).

The straight line joining these two points and extended on both directions is represented by,

ncert-class-10-maths-chapter-3-linear-eqns-1-11.jpg

Assume the prices of 1 kg of apple and 1 kg of grapes after 1 month are $p$ and $q$ (in Rs.). Mark that the prices may not be same with the prices on earlier occasion. This is where your real world awareness comes into play. It is always better to be safe.

The algebraic equation representing the total price of 4 kg of apples and 2 kg of grapes after a month is,

$4p+2q=300$.

The two points on the corresponding straight line on x-y axes corresponding to values of $p$ and $q$ respectively are, (75, 0) for $q=y=0$ and (0, 150) for $p=x=0$. Here values of $p$ are plotted on x-axis and that of $q$ on y-axis.

Joining the two lines and extending on both directions you will get the corresponding straight line as,

ncert-class-10-maths-chapter-3-linear-eqns-1-12.jpg

As we have thought, the prices of apples and grapes have indeed changed after 1 month. Can you say why?

Just multiply all three terms of the first equation by 2. You will get,

$4x+2y=320$.

It means—if the prices remained unchanged after 1 month, total price of 4 kg of apple and 2 kg of grapes would have been Rs.320.

But actually it is Rs.300. So total price is reduced.

You can say for sure that the prices have changed, but you won't be able to say which price has changed and by how much.

Nothing to worry though. You were only to form the equations and plot them which you have done.

Interesting, isn't it?


NCERT Solutions for Class 10 Maths

Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths on Real numbers part 3, HCF and LCM by factorization and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 2, Euclid’s division algorithms, HCF and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions

Chapter 3: Linear Equations

NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection

NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form

NCERT solutions for class 10 maths Chapter 3 Linear equations 5 Algebraic solution by Cross Multiplication

NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination

NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution

NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions

NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.

Chapter 4: Quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization

Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 2 Ratio values for selected angles

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities



Chapter 8: Introduction to Trigonometry, only solutions to selected problems

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1