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Pair of Linear Equation in Two Variables Exercise 3.1

Hi Friends and Champs!

Before to start next exercise, we need to know some basic terms:

Consistent pair of linear equations: A pair of linear equations in two variables which has a solution.

Dependent pair of linear equations: A pair of linear equations which has infinite many solutions.

Inconsistent pair of linear equations: A pair of linear equations which has no solution.

Linear General equation of a line: ax+by+c=0

let we have two linear equations in two variables:

a1 x + b1 y + c1 = 0 -------(1)

a2 x + b2 y + c2 = 0--------(2)

comparing both equation (1) and (2)

If   a1/a2 = b1/b2=c1/c2    Lines are Coincident with infinite solution.

If   a1/a2 ≠ b1/b2                Lines are intersecting with exactly one solution.

If   a1/a2 = b1/b2 ≠ c1/c2    Lines are parallel with no solution.  

                                                                        Exercise- 3.1

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

Solution: 

(1)  Let there are "x" no of boys and "y" no of girls,

        then x + y  = 10  ------------(1)

        and y - x = 4

        => x - y = - 4 --------------(2)

From equation (1) y = 10 - x and from equation (2) y = x+4

let's make table for both equations:

 
and graph for the above table is

  

 Clearly it is seen that both lines are intersecting at point (3,7)

So number of boys are  = 3 

and number of girls are =7

(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.

Solution: 

Let cost of one pencil = x

and cost of one pen = y

=> 5x + 7y = 50

=> y = (50 - 5x) / 7 --------------(1)

and 7x + 5y = 46

=> y = ( 46 - 7x ) / 5 ---------------(2)

let's make table and graph for both equations.

      





  


both lines are intersecting at y = 5 and x = 3

so cost of one pen is 5

and cost of one pencil is 3.

2. On comparing the ratios  find out a1/a2 , b1/b2 ,c1/c2   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

Solution : 

(i) 5x – 4y + 8 = 0 7x + 6y – 9 = 0

 a1/a2 = 5/7

 b1/b2 =-4/6

 c1/c2 =-8/-9

so a1/a2 ≠ b1/b2≠c1/c2

lines are intersecting at exactly one point.

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

a1/a2 = 9/18=1/2

b1/b2 =-3/6=1/2

c1/c2 = 12/24=1/2

so a1/a2 = b1/b2= c1/c2

lines are coincident with infinite solution.

iii) 6x – 3y + 10 = 0 2x – y + 9 = 0

a1/a2 = 6/2=3

 b1/b2 = -3/-1=3

c1/c2 = 10/9

so a1/a2 = b1/b2≠ c1/c2

lines are parallel with no solution.

3. On comparing the ratios , a1/a2 , b1/b2 ,c1/c2 find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2 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) 4/3x + 2y =8 2x + 3y = 12

Solution :



5. 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.

Solution :


 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

Solution:

(i) 3x + 2y - 7=0


 (ii) 4x + 6y -10 =0



(iii) 6x + 9y - 24 =0


 7. 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.

Solution:


 


 

 






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