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KC Sinha Mathematics Solution Class 9 Chapter 5 Linear Equations in Two Variable exercise 5.1

Page 5.21

Exercise 5


Type 1

Problems based on writing a linear equation in two variables x and y in the form ax-by+c=0 and problems on writing a given statement as a linear equations in two variables

WORKING RULE :

Use the following information wherever is required :

1. In order to write a given statement as a linear equation in two variables , identify the two variables and let these be x and y

2. Then, write the given condition in terms of x and y . The equation thus obtained will be the required equation.


Question 1 

Write each of the following equations in the Form ax + by + c = 0 and indicate the values of a , b and c in each case :

(i) 2x+3y=4.37

Sol :

⇒2x+3y-4.37=0

a=2 , b=3 , c=-4.37


(ii) x-4=\sqrt{3}y

Sol :

x-\sqrt{3}y-4=0

a=1 , b= -√3 , c= -4


(iii) 4=5x-3y

Sol :

⇒5x-3y-4

a=5 , b=-3 , c= -4


(iv) 2x=y

Sol :

⇒ 2x-y-0=0

a=2 , b=-1 , c=0

 


Question 2

Write each of the following as an equation in two variables :

(i) x=-5

Sol :

⇒x+5=0

⇒x+0+5=0

⇒x+0y+5=0

 

(ii) y=2

Sol :

⇒y-2=0

0+y-2=0

0x+y-2=0

 

(iii) 2x=3

Sol :

⇒2x-3=0

⇒2x+0-3=0

⇒2x+0y-3=0

 

(iv) 5y=2

Sol :

⇒5y-2=0

0+5y-2=0

0x+5y-2=0

 


Type 2

Problems based on examining whether given values are solution of the given equation or not.

WORKING RULE:

Put the given values in the given equation and simplify. If the given equation is satisfied by these values , then given values will be solution of given equation otherwise they will not


Question 3

Examine whether x = 2, ,y=1 , are solutions of the following equations or not ?

(i) 2x+5y=7

Sol :

⇒On putting x=2 and y=1

⇒2×2+5×1=7

⇒4+5=7

⇒9=7

NO


(ii) 2x-3y+7=8

Sol :

⇒On putting x=2 and y=1

⇒2×2-3×1+7=8

⇒4-3+7=8

⇒11-3=8

⇒8=8

YES


(iii) 3x+4y=9

Sol :

⇒On putting x=2 and y=1

⇒3×2+4×1=9

⇒6+4=9

⇒10=9

NO


(iv) 5x-7y=3

Sol :

⇒On putting x=2 and y=1

⇒5×2-7×1=3

⇒10-7=3

⇒3=3

YES


Question 4

Are x=-1 , y=3 solutions of the following equations or not ?

(i) 2x+5y=13

Sol :

⇒On putting x=-1 and y=3

⇒2×(-1)+5×3=13

⇒-2+15=13

⇒13=13

YES


(ii) 2x-3y=-11

Sol :

⇒On putting x=-1 and y=3

⇒2×(-1)-3×3=-11

⇒-2-9=-11

⇒-11=-11

YES


(iii) 5x+3y=4

Sol :

⇒On putting x=-1 and y=3

⇒5×(-1)+3×3=4

⇒-5+9=4

⇒4=4

YES


(iv) 2x+3y=41

Sol :

⇒On putting x=-1 and y=3

⇒2×(-1)+3×3=41

⇒-2+9=41

⇒7=41

NO


Question 5

In the following equation, f‌ind the values of a , so that x=1 , y=1 is its solution:

(i) 5x+3y=a

Sol :

⇒On putting x=1 and y=1

⇒5×1+3×1=a

⇒5+3=a

⇒a=8


(ii) ax-2y=10

Sol :

⇒On putting x=1 and y=1

⇒a×1-2×1=10

⇒a-2=10

⇒a=10+2

⇒a=12


(iii) ax+4y=18

Sol :

⇒On putting x=1 and y=1

⇒a×1+4×1=18

⇒a+4=18

⇒a=18-4

⇒a=14


(iv) 7ax+3ay=20

Sol :

⇒On putting x=1 and y=1

⇒7a×1+3a×1=20

⇒7a+3a=20

⇒10a=20

a=\dfrac{20}{10}

⇒a=2


Type 3

Problems based on solution of linear equation in two variables.

WORKING RULE:

Use the following information whichever is required :

1. A linear equation in one variable has unique solution (one and only one solution)

2. A linear equation in two variables has inf‌initely many solutions.

3. Any number of solutions of a linear equation in two variables can be written.

For this :

(i) If linear equation in two variables be ax+by+c=0 , where a,b,c are real numbers and a≠0 , b≠0 then write y in terms of x

Here , by=-(ax+c) or y=-\dfrac{ax+c}{b}

(ii) Putting arbitrary real values of x, f‌ind the corresponding values of y. For convenience, putting y = 0 , find the value of x and putting x = 0, find the value of y

These solution will be of the form x= a, y = 0 and y = b.

(iii) We can f‌ind as many solutions as we need by putting arbitrary values of x and finding corresponding values of y.

All these pair of values of x and y will be solutions of given linear equation


Question 6

Find two solutions of each of following equations :

(i) 4x+3y=12

Sol :

⇒3y=12-4x

y=\dfrac{12-4x}{3}

When x=3 , y=\dfrac{12-4\times 3}{3}=\dfrac{12-12}{3}=0 ;

When x=1 , y=\dfrac{12-4\times 1}{3}=\dfrac{12-4}{3}=\dfrac{8}{3}

⇒(3,0); \left(1,\dfrac{8}{3}\right)

 

(ii) 2x+5y=0

Sol :

⇒5y=-2x

y=\dfrac{-2x}{5}

When x=0 , y=\dfrac{-2\times 0}{5}=0 ;

When x=1 , y=\dfrac{-2\times 1}{5}=\dfrac{-2}{5} ;

⇒(0,0)\left(1,-\dfrac{2}{5}\right)

 

(iii) 3y+4=0

Sol :

0x+3y+4=0

⇒3y=-4-0x

y=\dfrac{-4-0x}{3}

When x=0 , y=\dfrac{-4-0\times 0}{3}=-\dfrac{4}{3} ;

When x=1 , y=\dfrac{-4-0\times 1}{3}=-\dfrac{4}{3} ;

\left(0,-\dfrac{4}{3}\right);\left(1,-\dfrac{4}{3}\right)

 


Question 7 

Find at least three solutions of the following equations :

(i) 5x+3y=4

Sol :

⇒3y=4-5x

y=\dfrac{4-5x}{3}

When x=-1 , y=\dfrac{4-5\times (-1)}{3}=\dfrac{9}{3}=3

When x=2 , y=\dfrac{4-5\times 2}{3}=\dfrac{4-10}{3}=\dfrac{-6}{3}=-2

When x=-4 , y=\dfrac{4-5\times (-4)}{3}=\dfrac{4+20}{3}=8

⇒ x= -1, y= 3; x= 2, y= -2; x= -4, y= 8

 

(ii) 2x-3y=-11

Sol :

⇒-3y=-11-2x

y=\dfrac{-11-2x}{-3}

When x=-4 , y=\dfrac{-11-2\times (-4)}{-3}=\dfrac{-11+6}{-3}=\dfrac{5}{3}

When x=-1 , y=\dfrac{-11-2\times (-1)}{-3}=\dfrac{-11+2}{-3}=\dfrac{-9}{-3}=3

When x=2 , y=\dfrac{-11-2\times 2}{-3}=\dfrac{-11-4}{-3}=\dfrac{-15}{-3}=5

⇒ x= -4, y= 1; x= -1, y= 3; x= 2, y= 5

 

(iii) 2x+y=6

Sol :

⇒y=6-2x

When x=0 , y=6-2×0=6

When x=1 , y=6-2×1=6-2=4

When x=2 , y=6-2×2=6-4=2

⇒ x= 0, y= 6; x= 1, y= 4; x= 2, y= 2

 

(iv) 2x-3y=1

Sol :

⇒-3y=1-2x

y=\dfrac{1-2x}{-3}

When x= -1 , y=\dfrac{1-2\times (-1)}{-3}=\dfrac{1+2}{-3}=\dfrac{3}{-3}=-1

When x=2 , y=\dfrac{1-2\times 2}{-3}=\dfrac{1-4}{-3}=\dfrac{-3}{-3}=1

When x=-4 , y=\dfrac{1-2\times (-4)}{-3}=\dfrac{1+8}{-3}=\dfrac{9}{-3}=-3

⇒ x= -1, y= -1; x= 2, y= 1; x= -4, y= -3

 


Question 8 

Find four solutions of the equations :

(i) x+2y=6

Sol :

⇒2y=6-x

y=\dfrac{6-x}{2}

When x=8 , y=\dfrac{6-8}{2}=\dfrac{-2}{2}=-1

When x=2 , y=\dfrac{6-2}{2}=\dfrac{4}{2}=2

When x= -2 , y=\dfrac{6-(-2)}{2}=\dfrac{8}{2}=4

When x= 0 , y=\dfrac{6-0}{2}=\dfrac{6}{2}=3

⇒ x= 8, y= -1; x= 2, y= 2; x= -2, y= 4; x= 0, y= 3

 

(ii) x+y=0

Sol :

⇒y=-x

When x=1 , y=-(1)=-1

When x=2 , y=-(2)=-2

When x=3 , y=-(3)=-3

When x=4 , y=-(4)=-4

⇒ x= 1, y= -1; x= 2, y= -2; x= 3, y= -3; x= 4, y= -4

 

(iii) 2x-3(y-2)=0

Sol :

⇒2x-3y+6=0

⇒-3y=2x-6

y=\dfrac{2x-6}{-3}

When x=0 , y=\dfrac{2\times 0-6}{-3}=-6

⇒ x= 0, y= 2; x= -3, y= 0; x= 3, y= 4; x= 6, y= 6

 

(iv) 2(x-1)+3y=4

Sol :

⇒2x-2+3y=4

⇒3y=4+2-2x

y=\dfrac{6-2x}{3}

When x=0 , y=\dfrac{6-2\times 0}{3}=\dfrac{6}{3}=2

When x=3 , y=\dfrac{6-2\times 3}{3}=\dfrac{6-6}{3}=0

When x=-3 , y=\dfrac{6-2\times (-3)}{3}=\dfrac{6+6}{3}=4

When x=6 , y=\dfrac{6-2\times 6}{3}=\dfrac{6-12}{3}=-2

⇒ x= 0, y= 2; x= 3, y= 0; x= -3, y= 4; x= 6, y= -2

 

(v) x=0

Sol :

 

⇒ x= 0, y= 1; x= 0, y= 2; x= 0, y= 3; x= 0, y= 4

 

(vi) y=0

Sol :

⇒0x+y+0=0

⇒y=-0x

⇒y=0x

When x=1 , y=0x=0×1=0

When x=2 , y=0x=0×2=0

When x=3 , y=0x=0×3=0

When x=4 , y=0x=0×4=0

⇒ x= 1, y= 0; x= 2, y= 0; x= 3, y= 0; x= 4, y= 0

 


Question 9 

Find solutions of the form x=a, y=0 and x=0 , y=b of the following pair of linear equations. Do they have any such common solution ?

(i) 5x+3y=15 and 5x+2y=10

Sol :

Given equation : 5x+3y=15

⇒3y=15-5x

y=\dfrac{15-5x}{3}

When x=0 , y=\dfrac{15-5\times 0}{3}=\dfrac{15}{3}=5

When x=3 , y=\dfrac{15-5\times 3}{3}=\dfrac{15-15}{3}=0

⇒x=0 , y=5 ; x=3 , y=0 ..(i)

 

Given equation : 5x+2y=10

⇒2y=10-5x

y=\dfrac{10-5x}{2}

When x=0 , y=\dfrac{10-5\times 0}{2}=\dfrac{10}{2}=5

When x=2 , y=\dfrac{10-5\times 2}{2}=\dfrac{10-10}{2}=0

⇒x=0 , y=5 ; x=2 , y=0 ..(ii)

From (i) and (ii) , we can say that

Yes common solution is x=0 , y=5

 

(ii) x+y=3 and 2x+5y=12

Sol :

Given equation : x+y=3

⇒y=3-x

When x=0 , y=3-0=3

When x=3 , y=3-3=0

⇒x=0 , y=3 ; x=3 , y=0 ..(i)

 

Given equation : 2x+5y=12

⇒5y=12-2x

y=\dfrac{12-2x}{5}

When x=0 , y=\dfrac{12-2\times 0}{5}=\dfrac{12}{5}=2.4

When x=6 , y=\dfrac{12-2\times 6}{5}=\dfrac{0}{5}=0

⇒x=0 , y=2.4 ; x=6 , y=0 ..(i)

From (i) and (ii) , we can say that

No

 

(iii) 2x+3y=1 and x-y=1

Sol :

Given equation : 2x+3y=1

⇒3y=1-2x

y=\dfrac{1-2x}{3}

When x=0 , y=\dfrac{1-2\times 0}{3}=\dfrac{1}{3}

When x=1/2 , y=\dfrac{1-2\times \dfrac{1}{2}}{3}=\dfrac{1-1}{3}=0

⇒x=0 , y= 1/3 ; x= 1/2 , y=0 ..(i)

 

Given equation : x-y=1

⇒-y=1-x

⇒y=x-1

When x=0 , y=0-1=-1

When x=1 , y=1-1=0

⇒x=0 , y=-1 ; x=1 , y=0..(ii)

From (i) and (ii) , we can say that

NO

 

(iv) 3x-9y=6 and 2x-6y=8

Sol :

Given equation : 3x-9y=6

⇒-9y=6-3x

y=\dfrac{6-3x}{-9}

When x=2 , y=\dfrac{6-3\times 2}{-9}=\dfrac{6-6}{-9}=0

When x=0 , y=\dfrac{6-3\times 0}{-9}=\dfrac{6}{-9}=\dfrac{-2}{3}

⇒x=2 , y=0 ; x=0 , y=-2/3..(i)

 

Given equation :2x-6y=8

⇒-6y=8-2x

y=\dfrac{8-2x}{-6}

When x=0 , y=\dfrac{8-2\times 0}{-6}=\dfrac{8}{-6}=\dfrac{-4}{3}

When x=4 , y=\dfrac{8-2\times 4}{-6}=\dfrac{8-8}{6}=0

⇒x=0 , y=-4/3 ; x=4 , y=0..(ii)

From (i) and (ii) , we can say that

NO

 


Type 4

Problems based on graph of a linear equation in two variables.

WORKING RULE:

1. Write y

2.

3.

4.

5.

6.

7.

8.


Question 10 

(i) Draw the graph of x+2y=6

Sol :

(ii) Draw the graph of x+y=7

Sol :


(iii) Draw the graph of x-y=7

Sol :


(iv) Draw the graph of 2x+y=3

Sol :


Question 11 

Draw the graph of the following equations and find the coordinates of the points where the cuts the axes :

(i) 3x+2y=6

Sol :


(ii) 2x+3y=12

Sol :


(iii) y+3x=9

Sol :


(iv) (x-4)-y+4=0

Sol :

 


(v) 4x-5y=20

Sol :


(vi) x+y=0

Sol :

 


Question 12

From the choices given below choose the equation whose graphs have been given:

(a) For figure (i):

(i) x+y=0

(ii) y=2x

(iii) y=x

(iv) y=2x+1

 

(b) For figure (ii):

(i) x+y=0

(ii) y=2x

(iii) y=2x+4

(iv) y=x-4

 

(c) For figure(iii):

(i) x+y=0

(ii) y=2x

(iii) y=2x+1

(iv) y=2x-4

Sol :

 


Question 13 

Draw the graph of the following :

(i) x=3

Sol :

 

 

(ii) x=-2

Sol :

 

 

(iii) y=2

Sol :

 

 

(iv) y=-3

Sol :

 

 

(v) x-3=0

Sol :

 

 

(vi) 2x-3=0

Sol :

 

 

 

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