NCERT Solutions for Class 10th Maths Chapter 3 : Linear Equations
In this page we have NCERT Solutions for Class 10th Maths:Linear Equations for
Exercise 3.1 on page 44. Hope you like them and do not forget to like , social_share
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A pair of linear equations in two variables can be represented, and solved, by the:
(i) graphical method
(ii) algebraic method How to draw the graph of the linear equation
Suppose the equation given is
$ax+by+c=0 \; , \; a \neq 0 \; and \; b \neq 0$
Find the value of y for x=0
$y=\frac {-c}{b}$
This point will lie on Y –axis. And the coordinates will be $(0,\frac{-c}{b})$
Find the value of x for y=0
$x=\frac {-c}{a}$
This point will lie on X –axis. And the coordinates will be $(\frac {-c}{a}, 0)$
Now we can draw the line joining these two points
If the points from above does not corresponds to integers, we can choose points where values are integers
NCERT Solutions for Class 10th Maths Chapter 3 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.” (Isn’t this interesting?) Represent this situation algebraically and graphically. Solution 1:
Let age of Aftab is represented by x
And age of daughter is represented by y
Then Seven years ago,
Age of Aftab = $x -7$
Age of daughter= $y-7$
As per the question Aftab was seven times as old as her daughter then were so we get
$(x - 7) = 7 (y - 7)$
$x - 7y = -42$ ----(1)
Now Three years from now ,
Age of Aftab = $x +3$
Age of daughter = $y+3$
As per question Aftab shall be three times as old as her daughter will be so we get
$x + 3 = 3 (y + 3)$
$ x -3y = 6$ ----- (2) Algebraic representation of situation
$x - 7y = -42$ ----(1)
$ x -3y = 6$ ----- (2)
To represent it graphically, we need to plot these equation on the XY coordinate system.
We can choose two values of each equation and plot it on XY coordinate system
$x - 7 y = - 42$
or
$y=\frac {(x+42)}{7}$
Now
$x -3 y = 6$
$y=\frac {(x-6)}{3}$
We can plot these two lines on XY coordinate system using the two point derived above
Question 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:
Let Cost of one bat = Rs x
Let Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
$3 x + 6y = 3900$ ---- (1)
Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So we get
$x + 2 y = 1300$ ---(2) Algebraic representation of situation
$3 x + 6y = 3900$ ---- (1)
$x + 2 y = 1300$ ---(2)
To represent it graphically, we need to plot these equation on the XY coordinate system.
We can choose two values of each equation and plot it on XY coordinate system
$3 x + 6y = 3900$
or
$y=\frac {(3900-3x)}{6}$
$x +2 y = 1300$
$y=\frac {(1300-x)}{2}$
Question 3.
The cost of 2 kg of apples and 1kg 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 Rs300. . Represent this situation algebraically and geometrically. Solution 3:
Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
$2 x + y = 160$ ----- (1)
Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
so we get
$4x + 2y= 300$ --- (2) Algebraic representation of situation
$2 x + y = 160$ ----- (1)
$4x + 2y= 300$ --- (2)
To represent it graphically, we need to plot these equation on the XY coordinate system.
We can choose two values of each equation and plot it on XY coordinate system
$2x +y = 160$
or
$y=160-2x$
$4x + 2y= 300 $
$y=\frac {(300-4x)}{2}$
Summary
NCERT Solutions for Class 10 Maths Chapter 3 Linear Equations Exercise 3.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download Linear equation Exercise 3.1 assignment as pdf
This chapter 3 has total 7 Exercise 3.1 ,3.2,3.3,3.4,3.5,3.6 and 3.7. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below