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Question-1

Which of the following equations are linear equations?
        (i) (3/2)x + 4 = 2x  – 3
        (ii) 5y – 3 = 2y  + 4
        (iii) u + 4 = u2 – 4
        (iv) 3x + 2 = 5x – 7
        (v)  x2 + 2 = x + 1  
        (vi)  y – 3 = 3y + 4
        (vii) u + 1/u = 5
        (viii) (x – 1)2 = x2 – 2
        (ix) (x – 1)(x – 2) = 6
        (x)  (x – 2 )(x + 3) = x + 7

Solution:
(i) It is a linear equation.

(ii) It is a linear equation. 

(iii) It is not a linear equation because it is a quadratic equation. 

(iv) It is a linear equation.

(v) It is not a linear equation because it is a quadratic equation.

(vi) It is a linear equation. 

(vii) u + 1/u = 5
      ... u2 + 1 = 5u
      It is not a linear equation because it is a quadratic equation.

(viii) (x – 1)2 = x2 – 2
       x2 – 2x + 1 = x2 – 2
       ⇒ – 2x + 1 = – 2
       ⇒ – 2x + 1 + 2 = 0
       ... – 2x + 3 = 0
       It is a linear equation.

(ix) (x – 1)(x – 2) = 6
       ⇒ x2 + (-1 – 2)x + (-1)(-2) = 6
       ⇒ x2 – 3x + 2 = 6
      ⇒ x2 – 3x + 8 = 0
     It is not a linear equation because it is a quadratic equation.

(x) (x - 2 )(x + 3) = x + 7
     x2 + (-2 + 3)x + (-2)(3) = x + 7
     ⇒ x2 + x - 6 = x + 7
    ... x2 - 13 = 0
    It is not a linear equation because it is a quadratic equation.

Question-2

Name the quadrant in which the point lies:
        (a) A (1, 1)
        (b) B (2, 4)
        (c) C (-3, -10)
        (d) D (-1, 2)
        (e) E (1, -1)
        (f) F (-2, -4)
        (g) G (-3, 10)
        (h) H (1, -2)

Solution:

(a) A (1, 1) lies in the 1st quadrant,
(b) B (2, 4) lies in the 1st quadrant.
(c) C (-3, -10) lies in the 3rd quadrant.
(d) D (-1, 2) lies in the 2nd quadrant.
(e) E (1, -1) lies in the 4th quadrant.
(f)  F (-2, -4) lies in the 3rd quadrant.
(g) G (-3, 10) lies in the 2nd quadrant.
(h) H (1, -2) lies in the 4th quadrant.

Question-3

In the following equations, verify whether the given value of the variable is a solution of  the equation:  
        (i) x + 4 = 2x ; x = 4                        (ii) y – 7 = 3y + 8 ; y = 3       
        (iii) 3u + 2 = 2u + 7 ; u = 5             (iv) 2x – 3 = x /2 – 2 ; x = √2        
        (v) (5/2)x + 3 = 21/2 ; x = 3         (vi) 24 – 3(u – 2) = u + 8 ; u = -1  
        (vii) (x – 2) + (x + 3 ) = x + 8 ; x = 0
.

Solution:
(i) If we substitute x = 4, we get
           L.H.S = x + 4 = 4 + 4 = 8    
    and R.H.S = 2x = 2 x 4 = 8
       ... L.H.S = R.H.S.   
      Hence, 4 is a solution of x + 4 = 2x.

(ii) If we substitute y = 3, we get  
             L.H.S = y – 7 = 3 – 7 = -4  
      and R.H.S = 3(3) + 8 = 9 + 8 = 17
         ... L.H.S ≠ R.H.S.   
     Hence, 3 is not a solution of y – 7 = 3y + 8

(iii) If we substitute u = 5, we get
            L.H.S = 3u + 2 = 3(5) + 2 = 15 + 2 = 17
      and R.H.S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17      
         ... L.H.S = R.H.S.
      Hence, 5 is a solution of 3u + 2 = 2u + 7

(iv) If we substitute x = √2, we get
            L.H.S = 2 x – 3 = 2(√2) – 3 = 2√2 – 3
      and R.H.S = x/2 – 2 = √2/2 - 2
         ... L.H.S ≠ R.H.S.
      Hence, √2 is not a solution of 2x – 3 = x/2 – 2
 
(v) If we substitute x = 3, we get
           L.H.S = (5/2)x + 3 = (5/2)(3) + 3 = 15/2 + 3 = (15 + 6)/2 = 21/2  
     and R.H.S = 21/2
     ... L.H.S = R.H.S.
     Hence, 3 is a solution of (5/2)x + 3 = 21/2.

(vi) If we substitute u = -1, we get
            L.H.S = 24 – 3{(-1) – 2}= 24 – 3(-1 –2) = 24 – (-9) = 24 + 9 = 33
      and R.H.S = u + 8 = -1 + 8 = 7
         ... L.H.S ≠ R.H.S.
      Hence, -1 is not a solution of 24 – 3(u – 2) = u + 8.

(vii) If we substitute x = 0, we get
              L.H.S = (x – 2) + (x + 3) = (0 – 2) + (0 + 3) = -2 + 3 = 1  
       and R.H.S = x + 8 = 0 + 8 = 8
           ... L.H.S ≠ R.H.S.
       Hence, 0 is not a solution of (x – 2) + (x + 3) = x + 8.

Question-4

Which of the following points lie on the x-axis?
        A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (-1, 0), F (0, -1), G (4, 0), H (0, 7)

Solution:
B (1, 0), D(0, 0), E(-1, 0) and G(4, 0) lies on the x-axis.

Question-5

Solve the following equations:
        (i) 3x + 3 = 15                     (ii) 2y + 7 = 19                (iii) x + 3 =
     
        (iv) √3x – 2 = 2√3 + 4          (v) 8u + = 3u +7         
        (vi) (√5 + 5)x + 4 = 2√5 + 8 (vii) 2x – (3x – 4) = 3x – 5    
        (viii) 2x + √2 = 3x – 4 – 3√2

Solution:
(i) 3x + 3 = 15
           3x = 15 – 3 (Transposing 3)
            3x = 12
              x = 12/3
          ... x = 4

(ii) 2y + 7 = 19
           2y = 19
  7
           2y = 12 (Transposing 7)
             y = 12/2
    ∴  y = 6

(iii) x + 3 =
           x =  
– 3 (Transposing 3)
        ⇒ x =  
      ⇒ x =
        ⇒ x = x
        ∴ x = 3
 
(iv)
√3x  – 2 = 2√3 + 4
      √3x
– 2 = 2√3 + 4
      √3x = 2√3 + 4 + 2      (Transposing -2)
     ⇒ 3x = 2√3 + 6
     ⇒x = 23 + 6
                   3
     ⇒x x       

     ⇒x =

     ⇒x

     ⇒x =

     ∴  x

(v) 8u + = 3u + 7

     ⇒8u - 3u = 7 -     (Transposing and 3u)

     ⇒5u =

     ⇒5u =

      ⇒5u =   

       ⇒u x

        u =

(vi) (√5 + 5)x + 4 = 2√5 + 8

      (√5 + 5)x = 2√ 5 + 8 - 4 (Transposing 4)

      (√5 + 5)x = 2√5 + 4

             ⇒x =

              ⇒x = x

              ⇒x =

              ⇒x

              ⇒x =
                       
              ⇒x

              ⇒x =

              ∴ x =

vii) 2x
  (3x   4) = 3x – 5
      2x
  3x + 4 = 3x – 5
      2x
 3x  3x  5  4 (Transposing 4 and 3x)
                   
 4x =  9
                 ∴ x = =

viii) 2x + √2 = 3 x
  4  3√2
      2x
  3x  4  3√2  √2 (Transposing 2 and 3x)
         ⇒
 x  4  4√2
         ⇒
 x  (4 + 4√2)
         ∴ x = 4 + 4√2

Question-6

Which of the following points lie on the x-axis?
         (A) (3, 4)
        (B) (7, 0)
        (C) (0, 8)
        (D) (-9, 0)
        (E)
        (F) (-3, -5)

Solution:
(B), (D), (E) lies on the x -axis.

Question-7

Which of the following points lie on the y-axis?
        A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (-1, 0), F (0, -1), G (4, 0), H (0, 7)

Solution:

C (0, 1), D (0, 0), F(0, -1) and H(0, 7) lies on the y-axis.

Question-8

Draw the graph of equation x – 2y = 4. Read a few solutions from the graph and verify the same by actual substitution. In each case, find the points where the line meets the two axes.

Solution:
The given equation is x – 2y = 4  

  => 2y =  x  – 4 | by transposition

=>  y  | Dividing both sides by 2


To draw the graph, we use the table of corresponding values of x  and y
 

x

4

2

0


y


0


-1


-2

 

Question-9

Plot the points A (2, 0), B (2, 2), C (0, 2) and join OA, AB, BC and CO. What figure do you obtain ?


Solution:


If we join the points OA, AB, BC and CO, the figure we obtain is a square.

Question-10

Draw the graph of 2(x + 3) – 3 (y + 1) = 0. Read a few solutions from the graph and verify the same by actual substitution. In each case, find the points where the line meets the two axes.        


Solution:
The given equation is 2(x + 3) – 3(y + 1) = 0

3(y + 1) = 2(x + 3)

3y + 3 = 2x + 6 3y = 2x + 6 – 3      | by transposition

3y = 2x + 3 y =                | Dividing both sides by 3

To draw the graph, we use the table of corresponding values of x and y.
 

x 0 3 6
y 1 3 5




  
We plot the points (0, 1), (3, 3) and (6, 5)on the graph paper. Then we join the points by a ruler to get the line which is the graph of the given equation.     
   
Few solutions from the graph are (-3, -1),   

For (-3, -1) L.H.S. = 2 (-3 + 3) – 3 (-1 + 1) = 0 = R.H.S.

The solution (-3, -1) is verified

For L.H.S. = 2 - 3(0 + 1) = 3 – 3 = 0 = R.H.S

The solution is verified

The points where the given line meets the x-axis and the y-axis and respectively and (0, 1).

Question-11

Plot the points A (4, 4), B (-4, 4) and join OA, OB and BA. What figure do you obtain ?


Solution:


The figure obtained when we join OA, OB and BA is a triangle.

Question-12

Find a value for a so that each of the following equations may have x = l, y = l as a
Solution:
        (a) ax – 2y = 10
        (b) x – y = a


Solution:
(a) The given equation is ax – 2y = 10
If x = l, y = l is a solution of (1), then a(l) – 2(l) = 10
⇒ l(a – 2) = 10 ⇒ a – 2 = ( dividing both sides by l)
⇒ a = + 2

(b) The given equation is x – y = a.
If x = l, y = l is a solution of (1), then l – l = a
a = 0.

Question-13

Find out which of the following equations have x = 2, y = 1 as a
Solution:

       
a) 2x + 5y = 9
       
b) 5x +3y = 14
       
c) 2x + 3y = 7
       
d) 2x - 3y =1
       
e) 2x - 3y + 78
    
    f)  x + y + 4 = 0


Solution:
a) 2x + 5y = 9
    When
x = 2, 2 ×2+5y = 9
                               5
y = 9- 4
                               5
y = 5
                         
    y = 1
  x = 2, y = 1 is a solution of this equation.


b) 5
x + 3y = 14
    When
x = 2, 5×2 3y = 14
                           3
y = 14 10 = 4
                                   
 y =
 ∴ x = 2, y = 1 is not a solution of this equation.


c) 2
x+3y = 7
    When
x = 2, 2× 2 + 3y = 7
                                 3
y = 7 – 4
                                 3y = 3
                               y = 1
 ∴ x = 2 and y = 1 is the solution of this equation.


d) 2
x - 3y = 1
    When
x = 2,  2×2 – 3y = 1
                            -
3y = 14
                            
-3y = -3
                            y = 1                                    
  x = 2, y = 1 is a solution of this equation.


e) 2
x - 3y + 7=8
    When
x = 2, 2×2 3y + 7 = 8
                               -
3y = 8 - 4 - 7
                                
- 3y = - 3
                             
 y = 1
x =
2, y = 1 is a solution of this equation.


f)
x + y + 4 = 0
   When
x = 2, 2 + y + 4 = 0
                          y = 
- 4 - 2 = - 6                              
  x = 2, y = 1 is not a solution of this equation.

Question-14

Draw the graph for the equation –2x + y – 7 = 0. Check whether the point (-3, -2) is on the given line.


Solution:
–2x + y – 7 = 0
⇒ y = 2x + 7

x = 1

⇒ y = 2 + 7 = 9

x = 0

⇒ y = 2(0) + 7 = 0 + 7 = 7

x = -2

⇒ y = 2(-2) + 7 = -4 + 7 = 3

x

1

0

-2

y

9

7

3






The point (-3, -2) does not lie on the graph.

Question-15

In each case find the points where the line meets the two axes.
          (i) 2x + y = 6
          (ii)  x - 2y = 4
          (iii) 2(x - 1) + 3y = 4
          (iv) y - 3x = 9
          (v)  2(x + 3) - 3(y + 1) = 0
          (vi) (x - y) - y + 4 = 0


Solution:
(i) 2x + y = 6
    For the point on the x-axis the value of y = 0.
    Substituting for y = 0 in the equation,
    2x + 0 = 6      2x = 6      ⇒   x = 3     The point where the line 2x + y = 6 ill touch the x-axis is (3, 0).
    For the point on the y-axis the value of x = 0.
    Substituting for x = 0 in the equation,
    2 × 0 + y = 6                y = 6     The point where the line 2x+y=6 will touch the y-axis is (0, 6).


(ii) x - 2y = 4
     For the point on the x-axis the value of y= 0.
     Substituting for y= 0 in the equation,
     x - 2× 0 = 4           x = 4      The point where the line x - 2y = 4 will touch the x-axis is (4, 0).
     For the point on the y-axis the value of x = 0.
     Substituting for x = 0 in the equation,
     0 - 2y = 4      - 2y = 4          y == -2      The point where the line x = -2,y = 4 will touch the y-axis is (0, -2).


(iii) 2(x-1) + 3y = 4
 
     For the point on the x-axis the value of y= 0.
 
     Substituting for y= 0 in the equation,
      2(x-1) + 3 × 0 = 4         2x -2 + 0 = 4            Þ 2x = 4 + 2 = 6                   x = = 3       The point where the line 2(x - 1) + 3y = 4 will touch the x-axis is (3, 0).
      For the point on the y-axis the value of x = 0.
      Substituting for x = 0 in the equation,
      2(0 - 1) + 3y = 4          Þ -2 + 3y = 4                  3y = 4 + 2 = 6                     y = = 2       The point where the line 2(x-1) + 3y = 4 will touch the y-axis is (0, 2).


(iv) y - 3x = 9
 
     For the point on the x-axis the value of y = 0.
 
     Substituting for y = 0 in the equation,
 
     0 -3x = 9        -3x = 9
 
          x = -3       The point where the line y -3x = 9 will touch the x-axis is (-3, 0).
 
     
      The point on the y-axis the value of x = 0.
 
     Substituting for x = 0 in the equation,
 
             y -3×0 = 9             y = 9       The point where the line 2x + y= 0 will touch the y-axis is (0, 9).


(v) 2(x + 3) - 3(y + 1) = 0
 
     For the point on the x-axis the value of y = 0.
 
     Substituting for y = 0 in the equation,
 
     2(x + 3)-3(0 + 1) = 0              2x + 6 - 3 = 0                            2x = -3                               x = -
 
     The point where the line 2(x + 3)-3(y + 1)=0 will touch the x-axis is (-, 0).

 
     For the point on the y-axis the value of x = 0.
 
     Substituting for x = 0 in the equation, 
 
     2(0 + 3)-3(y + 1) = 0           0 + 6- 3y- 3 = 0                         -3y = -3                             y = 1       The point where the line 2(x + 3) - 3(y + 1) = 0 will touch the y-axis is (0, 1).


(vi) (x - y) -y + 4= 0
 
     For the point on the x-axis the value of y = 0.
 
     Substituting for y = 0 in the equation,
 
     (x - 0) - 0 + 4 = 0              x + 4 = 0                    x = -4       The point where the line (x - y) - y + 4 = 0 will touch the x-axis is (-4, 0).
      For the point on the y-axis the value of x= 0.
 
     Substituting for x = 0 in the equation,
 
     (0-y)-y+4 = 0           -y - y = -4            Þ -2y = -4                   y = 2       The point where the line (x - y) - y + 4 = 0 will touch the y-axis is (0, 2).

Question-16

Find at least 3 solutions for the following linear equation in two variables: 2x + 5y = 13


Solution:
2x + 5y = 13
Take x = 1,
2×1 + 5y = 13    5y = 13 - 2 = 11      \ y =
Take x = 2,
2×2 + 5y = 13    5y = 13 - 4 = 9     y =
Take x = 3,
2×3+5y=13   
 5y = 13 - 6 = 7  y = The three solutions of the given equation are x = 1, y = ; x = 2, y = and x = 3, y=.

Question-17

Draw the graph of the equation y = 3x – 4 and read off the value of  y when x = -1


Solution:
y = 3x – 4
 x = 1, 
y = 3 – 4 = -1

x = 0, 
⇒ y = 3(0) – 4 = -4

x = 2, 
⇒ y = 3(2) – 4 = 6 – 4 = 2

x

1

0

2

y

-1

-4

2




When x = -1 then y = 3(-1) – 4 = - 3 – 4 = - 7.

Question-18

Solve for x: 6(1 + x) + 52 - x = 98


Solution:
6(1+ x) + 52 - x =  98
6 + 6x+ 52 - x =  98
                    5x =  98 - 6 - 52
                    5x =  40
                      x =  8

Question-19

Find at least 3 solutions for the following linear equation in two variables: 5x + 3y = 4


Solution:
5x+3y = 4

  Take x = 1,
 5×1+3y = 4     3y = 4-5 = -1       y =

  Take x = 2,
5×2+3y = 4     3y = 4-10 = -6       y = = -2

   Take x = 3,
5×3+3y = 4       Þ 3y = 4-15 =-11        y = The three solutions of the given equation are x = 1, y = ; x = 2, y = -2 and x = 3, y =.

Question-20

Find at least 3 solutions for the following linear equation in two variables: 2x + 3y = 4


Solution:
2x+3y = 4
Take x = 1,
2
×1+3y = 4
     
⇒  3y = 4 - 2 = 2
       
∴  y =

   Take x = 2,
2
× 2 + 3y = 4
   
 3y = 4 - 4 = 0
        
∴ y = 0

   Take x = 3,
2
× 3+3y = 4
      
⇒  3y = 4 - 6 = -2
      
⇒  y =
∴ The three solutions of the given equation are x = 1, y = ; x = 2, y = 0 and x = 3, y =.

Question-21

Solve for x
          (x -1) (x - 2) - (x - 3) (x - 4) = 24


Solution:
(x-1) (x - 2) - (x - 3) (x - 4) = 24
  x2 - 3x + 2 - (x2 - 7x + 12) = 24
     x2 - 3x + 2 - x2 + 7x - 12 = 24
                              4x - 10 = 24
                                     4x = 34
                                       x = 17/2

Question-22

Solve for x
          (x + 1) (x + 3) = (x + 4) (x – 1)


Solution:
(x + 1) (x + 3) = (x + 4) (x – 1)
    x2 + 4x +
3 =  x2 + 3x – 4
            4x +
3 3x – 4
                    x =  –
7

Question-23

Find at least 3 solutions for the following linear equation in two variables: 2x-3y= -11


Solution:
2x-3y = -11
 Take x = 1,
2×1-3y = -11   -3y = -11-2 = -13      y = =

   Take x = 2,
2×2 -3y = -11     -3y = -11-4
        -3y = -15         y = 5

 Take x = 3,
2×3-3y = -11    -3y = -11-6 = -17       y = The three solutions of the given equation are x = 1, y = ; x = 2, y = 5 and x = 3, y =.

Question-24

Solve for x
  √3u – 2 = 2 √3 + 4


Solution:
√3u – 2 = 2 √3+ 4
√3u = 2 √3+ 6
u = 2 + 2√3

Question-25

Find at least 3 solutions for the following linear equation in two variables: 2x-3y+7= 0


Solution:
2x-3y+7 = 0
Take x = 1,
  2×1 -3y + 7 = 0 -3y = -7-2 = -9            y = 3

 Take x = 2,
2 × 2 - 3y + 7 = 0         -3y = -4-7
            -3y = -11            y =
Take x = 3,
 2 × 3 - 3y+7 = 0
-3y = -7-6 = -13              Þ y = The three solutions of the given equation are x = 1, y = 3; x = 2, y = and x = 3, y =   .

Question-26

Draw the graph of y = 2x + 1, y = 2x and y = 2x - . Are these parallel?


Solution:
Given y = 2x + 1
If x = 0

⇒ y = 2(0) + 1 = 1

If x = 1

⇒ y = 2(1) + 1 = 2 + 1 = 3

If x = -2

⇒ y = 2(-2) + 1 = - 4 + 1 = - 3


x

0

1

-2

y

1

3

-3




Given y = 2x
If x = 0

⇒ y = 0

If x = 1

⇒ y = 2(1) = 2

If x = -2

⇒ y = 2(-2) = -4 


x

0

1

-2

y

0

2

-4




Given y = 2x -   
If x = 0

⇒ y = 2x -   = -  

If x = 1

⇒ y = 2x   = 2   = 3/2

If x = 
 2
⇒ y = 2x   =  – 4 –   =   9/2   


x

0

1

-2

y

-0.5

1.5

-4.5

 


 

y = 2x + 1, y = 2x and y = 2x – are parallel to each other.

Question-27

Find at least 3 solutions for the following linear equation in two variables: x + y + 4 = 0


Solution:
x + y + 4 = 0

Take x = 1,
1 + y + 4 = 0

y + 5 = 0
   y = -5

Take x = 2,
2 + y + 4 = 0    
y = -2-4 = -6     y = -6

Take x = 3,
3 + y + 4 = 0     y = -4-3 = -7

The three solutions of the given equation are x = 1, y = -5; x = 2, y = -6 and x = 3, y = -7.

Question-28

Which of the following equations are linear?
          (i) 5x2 + 4x + 1 = 0
          (ii) x3 – 1 = 0
          (iii) x3 + 1 = 0
          (iv) (x – 5) (x – 7) = 8
          (v) 6x = 42
          (vi) = 9
          (vii) x + 10 = 17
          (viii) x – 9 = 2
          (ix) (x + 2)2 = x2 + 16
          (x) (x – 2) (x + 5) = x + 14


Solution:
(v), (vi), (vii), (viii) and (ix) are in linear equations.

Question-29

Find four solutions for the following equation: 12x + 5y= 0


Solution:
12x+5y = 0

     Take x = 1,
 12×1 + 5y = 0
 Þ 5y+12 = 0      5y = -12       y =

Take x = 2,
12×2 + 5y = 0 24+5y = 0     5y = -24       y =

     Take x = 3,
12×3 + 5y = 0
      36+5y = 0      5y = -36       y =
   Take x = 4,
12×4 + 5y = 0
    48+5y = 0      
5y = -48        y = The four solutions of the given equation are x = 1, y = ; x = 2, y = , x = 3, y = and x = 4, y = .

Question-30

Draw the graphs of 3x - 2 = 0 and 2y – 1 = 0. Do these lines intersect? What relationship do you observe between these lines?


Solution:
3x - 2 =  0
        x =  2/3 = 0.7
 2y – 1 = 0
         y = 1/2






Line y = ½ and x = 2/3 are perpendicular to each other.

Question-31

Find four solutions for the following equation: 5x-3y = 0


Solution:
5x-3y = 0

Take x = 1,
5×1 - 3y = 0 5-3y = 0   -3y = -5      y =

    Take x = 2,
  5×2 - 3y = 0 10 - 3y = 0    -3y = -10       y =
           y =
  Take x = 3,
 5×3 - 3y = 0
15-3y = 0     3y = 15       \ y = 5
 Take x = 4,
5×4 -3y = 0
20-3y = 0    -3y = -20       y = The four solutions of the given equation are x = 1, y = ; x = 2, y = , x = 3, y = 5 and x = 4, y = .

Question-32

Solve for x(√2 + 2) x + 4 = 2√2 + 8
                    


Solution:
  (2 + 2 )x =  22 + 8 – 4
(2 + 2 )x = 22 +  4
(
2 + 2 )x = 2(2 + 2)
            x = 2

Question-33

Find four solutions for the following equation: 2(x-1)+3y = 4


Solution:
2(x-1)+3y = 4

     Take x = 1,
2(1-1)+3y = 4   0+3y = 4      3y = 4        y =

     Take x = 2,
2(2-1)+3y = 4   2+3y = 4      3y = 4-2
               = 2        y =

     Take x = 3,
2(3-1)+3y = 4
   4+3y = 4       3y = 4-4
                = 0        y = 0
     Take x = 4,
2(4-1)+3y = 4   
6+3y = 4        Þ 3y = 4-6
               = -2         
y = The four solutions of the given equation are x = 1, y = ; x = 2, y = , x = 3, y = 0 and x = 4, y = .

Question-34

Solve for x :
          =


Solution:
   =
(x + 1)(x - 1) = (x + 4)(x + 2)
          x2 - 1 = x2 + 6x + 8
              - 1 = 6x + 8
               6x = -9
                 x = -

Question-35

Find four solutions for the following equation: 2x-3(y-2) = 1


Solution:
2x-3(y-2) = 1
     
Take x = 1,
2×1-3(y-2) = 1   2-3y+6 = 1 -3y = 1-8 = -7            y =            y =

Take x = 2,
2×2 -3(y-2) = 1   4-3y+6 = 1          -3y = 1-10
                  = -9           y = = 3

      Take x = 3,
2×3 -3(y-2) = 1
  Þ 6-3y+6 = 1        -3y = 1-12

                  = -11           y = =   
                 
      Take x = 4,
2× 4-3(y-2) = 1   8-3y+6 = 1          3y = 1-14
                 = -13    y = = The four solutions of the given equation are x = 1, y = ; x = 2, y = 3, x = 3, y = and x = 4, y = .

Question-36

Solvefor x:
           3(2u + 3) – 2 (1 – u) + 1 = 0


Solution:
3(2u + 3) – 2 (1 – u) + 1 = 0
       6u + 9 – 2 + 2u + 1 = 0
                            8u + 8 = 0
                              u + 1 = 0
                                   u = -1

Question-37

Find four solutions for the following equation: 2(x+3)-3(y+1)=0


Solution:
2(x+3)-3(y+1) = 0
           Take x = 1,
2(1+3)-3(y+1) = 0    2× 4-3y-3 = 0 -3y = 0-8+3 = -5                y =                y =

           Take x = 2,
2(2+3)-3(y+1) = 0     Þ 2× 5-3y-3 = 0       10-3y-3 = 0              -3y = -10+3
                     = -7                 y ==

Take x = 3,
2(3+3)-3(y+1) = 0    2× 6-3y-3 = 0
 Þ -3y=-12+3 =-9               y =

                      = 3
           Take x = 4,
2(4+3)-3(y+1) = 0    2× 7-3y-3 = 0             -3y = -14+3

                        = -11        y = = The four solutions of the given equation are x = 1, y = ; x = 2, y = , x = 3, y = 3 and x = 4, y = .

Question-38

Solve for y :
          6(7y  - 2) + 4y  = 5(y  +1)


Solution:
6(7y - 2) + 4y   =  5(y  + 1)
42y  - 12 + 4y   =  5y  + 5
42y  + 4y  - 5y  =  5 + 12
               41y  =  17
                    y =  17/41

Question-39

Find four solutions for the following equation: (x-4)-y+4=0


Solution:
(x-4)-y+4 = 0
     Take x = 1,
(1-4)-y+4 = 0
 or-3-y+4 = 0
        or-y = -1         y = 1

    Take x = 2,
(2-4)-y+4 = 0
or -2-y+4 = 0
       or -y = -2         y = 2

    Take x = 3,
(3-4)-y+4 = 0
 or-1-y+4 = 0
      or –y = -3         y = 3

    Take x = 4,
(4-4)-y+4 = 0
 or 0-y+4 = 0
      or –y = -4         y = 4

The four solutions of the given equation are x = 1, y = 1; x = 2, y = 2; x = 3, y = 3 and x = 4, y = 4.

Question-40

Find four solutions for the following equation: x + y = 0


Solution:
x + y = 0
Take x =1, 1+y = 0
               ∴ y = -1
Take x = 2, 2+y = 0
               ∴y = -2
Take x = 3, 3+y = 0
               ∴ y = -3
Take x = 4, 4+y = 0
               ∴ y = -4
∴The four solutions of the given equation are x = 1, y = -1; x = 2, y = -2; x = 3, y = -3 and x = 4,y = -4.

Question-41

Find four solutions for the following equation: x - y = 0


Solution:
x –  y = 0
Take x = 1, 1 – y = 0  


                   y = 1
Take x = 2, 2
– y = 0
                 y = 2
Take x = 3, 3
– y = 0
                ∴ y = 3
Take  x = 4, 4
–  y = 0
                ∴ y = 4

∴ The four solutions of the given equation are x = 1, y = 1; x = 2, y = 2; x = 3, y = 3 and x = 4, y = 4.

Question-42

Find four solutions for the following equation: x = 0.


Solution:
The equation x = 0, y can have any values.
For x = 0, y can have values 1,2,3,4,………

 The four solutions of the given equations are x = 0, y = 1; x = 0, y = 2; x = 0, y = 3; and x = 0, y = 4.

Question-43

Find the solutions of the form x = a, y = 0 and x = 0, y = b for each of the following pairs of equations. 
          Do they have any common such solution?
          (i)   3x + 2y = 6 and 5x - 2y =10
          (ii)  5x + 3y = 15 and 5x + 2y =10
          (iii) 9x + 7y = 63 and x - y =10


Solution:
(i) 3x+2y = 6 and 5x-2y = 10
     Put y = 0, 3x+ 2×0
= 6 
     or 3x = 6
         ∴ x = 2

 Put x = 0 in 3x+2y = 6
                   3 × 0+2 × y = 6
                                   2y = 6
                                   y = 3
∴ The solutions of the equation 3x+2y = 6 is x = 2, y = 0 and x = 0, y = 3.
     5x-2y = 10
   Take y = 0, 5x-2 × 0 = 10
      or 5x = 10
          ∴ x = 2.
     Put x = 0 in 5x-2y = 10
    5 × 0-2y = 10,
         ⇒ -2y = 10
            ∴ y = -5,
     The solutions of the equation 5x-2y = 10 are x = 2, y = 0; x = 0, y = -5.
 ∴ The solution x = 2, y = 0 is common to both the equations.


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

        5x+3y = 15
     Put y = 0 in 5x+3 × 0 = 15
     or 5x = 15
           ∴ x = 3

     Put x = 0 in 5x+3y = 15
     5 × 0+3y = 15
             ⇒ 3y = 15
               ∴ y = 5

The solutions of the equation 5x+3y = 15 are x = 3, y = 0 and x = 0, y = 5.
The solution x = 0, y = 5 is common to both the equations.
Put y = 0 in 5x+2y = 10
     5x+2 × 0 = 10
             ⇒ 5x = 10
              x = 2

     Put x = 0 in 5x+2y = 10
     5 × 0+2y = 10
                2y = 10
                   y = 5
 The solutions of the equation 5x+2y = 10 are x = 2, y = 0 and x = 0, y = 5.


(iii) 9x+7y = 63 and x - y = 10

      Put y = 0 in 9x+7y = 63
                   9x+7 × 0 = 63
                           ⇒  9x = 63
                                x = 7
      Put x = 0 in 9x+7y = 63
                   9 × 0+7y = 63
                          ⇒ 7y = 63
                                y = 9

      The solutions of the equation 9x+7y = 63 are x = 7, y = 0 and x = 0, y = 9.
      Put y = 0 in x - y = 10
                                x - 0 = 10
                                x = 10

      Put x = 0 in x- y = 10
                             0 - y = 10
                                   y = -10

      The solutions of the equation x - y = 10 are x = 10, y = 0 and x = 0, y = -10.
      There is no solution common to both the equations.        

Question-44

 

Find the value of 'a' so that the following equation may have x = 1, y = 1 as a
Solution: 3x + ay = 6


Solution:
           3x + ay = 6
Take x = 1, y = 1,
     3×1 + a×1 = 6
           or 3 + a = 6
         ∴  a = 3

 

Question-45

 

Find the value of 'a' so that the following equation may have x = 1, y = 1 as a
Solution: ax - 2y = 10


Solution:
Given ax - 2y = 10
     Take x = 1, y = 1,
⇒ a×1 - 2×1 = 10
       or a - 2 = 10
      ∴ a = 12

 

Question-46

 

Find the value of 'a' so that the following equation may have x = 1, y = 1 as a
Solution: 5x + 3y = a


Solution:
Given, 5x + 3y = a
   Take x = 1, y = 1
  5 ×1 + 3 ×1 = a
         or 5 + 3 =  a
          ∴ a = 8.

 

Question-47

 

Find the value of 'a' so that the following equation may have x = 1, y = 1 as a
Solution: 5x + 2ay = 3a


Solution:
Given 5x + 2ay = 3a
        Take x = 1, y = 1
5×1 + 2 × a × 1 = 3 a
             or 5 + 2a = 3a
        3a - 2a = 5
             ∴ a = 5.

 

Question-48

 

Find the value of 'a' so that the following equation may have x = 1,  y = 1 as a
Solution: 9ax + 12ay = 63


Solution:
                Given 9ax + 12ay = 63
               Take x = 1, y = 1
9 × a × 1 + 12 × a × 1 = 63
                      9a + 12a = 63
                              21a = 63
                    a = 3.

 

Question-49

 

Find the value of a so that the equation x - y = a  may have x = 1, y = 1 as a solution


Solution: x - y = a
Given x = 1, y = 1 is a solution of the equation  x - y = a

          ⇒ 1 - 1 = a
            a = 0

 

Question-50

Draw the graph of the following equation: x = 2


Solution:
The line x = 2 is parallel to the Y axis and passes through the point (2, 0).
  

Question-51

Draw the graph of the following equation: y = 3


Solution:
The line y = 3 is parallel to the X- axis and passes through the point (0, 3).

Question-52

Draw the graph of the following equation: x = -1


Solution:
The line x = - 1 is parallel to the Y-axis and passes through the point (-1, 0).
 

Question-53

Draw the graph of the following equation: y = -3


Solution:
The line y = - 3 is parallel to the X -axis and passes through the point (0, - 3).
      

Question-54

Draw the graph of the following equation: 2 y + 5 = 0


Solution:
2 y  + 5 = 0
2 y  = -5
y  = - 5/2       

Question-55

Draw the graph of the following equation: 3x - 2 = 0


Solution:
 3x - 2 = 0 
 3x   = 2
x   = 2/3
      

Question-56

Draw the graph of the following equation: x + y = 0


Solution:
 x + y = 0

X 0 1 2
Y 0 -1 -2




 The line x + y = 0 passes through the origin O (0, 0), A = (1, - 1) and B (2 - 2).
 
 
 

Question-57

Draw the graph of the following equation: x - y = 0


Solution:
x - y = 0

X 0 1 2
Y 0 1 2




The line x - y = 0 passes through the origin O (0, 0),  A (1, 1) and B (2, 2).

Question-58

Draw the graph of the following equation: x = 0


Solution:
The line x = 0 lies on the y-axis.

Question-59

Draw the graph of the following equation: y = 0


Solution:
The line y = 0 lies on the x -axis.





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