Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y
Case I. Cost of 5 pencils = 5x
Cost of 7 pens = 7y
<>According to question,
5x + 7y = 50
Case II. Cost of 7 pencils = 7x
Cost of 5 pens = 5y
According to question,
7x + 5y = 46
Thus, we have
Thus, we have following table :
Fig. 3.5.
When we plot the graph of the given equation, we find that both the lines intersect at the point (3, 5). So, x = 3,y = 5 is the required solution of the pair of linear equation.
Hence, the cost of 1 pencil be Rs. 3 cost of 1 pen be Rs. 5.
Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
Case I. Cost of 3 bats = 3x
Cost of 6 balls = 6y
According to question,
3x + 6y = 3900
Case II. Cost of I bat = x
Cost of 3 more balls = 3y
According to question,
x + 3y = 1300
So, algebraically representation be
3x + 6y = 3900
x + 3y = 1300
Graphical representation :
We have, 3x + 6y = 3900
⇒ 3(x + 2y) = 3900
⇒ x + 2y = 1300
⇒ a = 1300 - 2y
Thus, we have following table :
We have, x + 3y = 1300
⇒ x = 1300 - 3y
Thus, we have following table :
When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.
(i) 10 students of Class X took part in 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.
(i) Let the number of boys be x and number of girls be y.
Case I. x + y = 10 ...(i)
Case II. y = x + 4
⇒ x - y = -4 ...(ii)
We have, x + y = 10
⇒ x = 10 - y
Thus, we have following table :
Fig. 3.4.
We have, x - y = -4
⇒ x = y - 4
Thus we have following table :
When we plot the graph of the given equation, we find that both the lines intersect at me point (3, 7). So.r = 3,y = 7 is the required solution of the pair of linear equation.
Hence, the number of boys be 3 and the number of girls be 7, who took part in quiz.
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.
Let present age of Aftab be x years and present age of his daughter be y years.
Case I. Seven years ago,
Age of Aftab = (x - 7) years
Age of his daughter = (y - 7) years
According to question :
(x - 7) = 7 (y - 7)
⇒ x - 7 = 7y - 49
⇒ x - 7y = -42
Case II.
Three years later,
Age of Aftab = (x + 3) years
Age of his daughter = (y + 3) years
Accoring to questions,
x + 3 = 3 (y + 3)
⇒ x + 3 = 3y + 9
⇒ x — 3y = 6
So, algebraic expression be
x - 7y = -42 ...(i)
x - 3y = 6 ...(ii)
Graphical representation
For eq. (i), we have
x - 7y = -42
⇒ x — 7y — 42
Thus, we have following table :
From eqn. (ii), we have
x -3y = 6
⇒ x = 3y + 6
Thus, we have following table
When we plot the graph of equations. We find that both the lines intersect at the point (42, 12). Therefore, x = 42, y = 12 is the solution of the given system of equations.
Fig. 3.1.
Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y. So, algebraic representation
<>2x + y = 160
4x + 2y = 300
⇒ 2x + y = 150
Graphical representation, we have
2x + y = 160
⇒ y = 160 - 2x
We have,
2x + y = 150
⇒ y = 150 - 2x
When we plot the graph of the equation we find that two lines do not intersect i.e. they are parallel.
Fig. 3.3.