Page 5.21
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.
(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
(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
(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
(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
(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
(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 , $y=\dfrac{-4-0\times 1}{3}=-\dfrac{4}{3}$ ;
⇒$\left(0,-\dfrac{4}{3}\right);\left(1,-\dfrac{4}{3}\right)$
(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
(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 , $y=\dfrac{2\times 0-6}{-3}=\dfrac{-6}{-3}$
⇒ x= 0, y= 2; x= -3, y= 4; x= 3, y= 4; x= 6, y= -2
(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
(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.
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 :
(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 :
(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 :
(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 :
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 , $y=\dfrac{2\times 0-6}{-3}=\dfrac{-6}{-3}$
⇒ x= 0, y= 2; x= -3, y= 4; x= 3, y= 4; x= 6, y= -2
(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=6Sol :
(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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