SELINA Solution Class 9 Simultaneous (Linear) Equation (Including problems) Chapter 6 Exercise 6B

Question 1

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
13 + 2y = 9x
3y = 7x

Sol:

13 + 2y = 9x                                  ...(1)
3y = 7x                                          ...(2)

Multiplying equation no. (1) by 3 and (2) by 2, we get,
          39 + 6y = 27x                     ...(1)
+                6y = 14x                     ...(2)
        -      -         -        
                   39 = 13x
                     x = 3

From (2)
3y = 7x
∴ 3y = 7(3)
∴ y = ${\displaystyle \frac{21}{3}}$ = 7

Question 2

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
3x - y = 23
${\displaystyle \frac{x}{3}+\frac{y}{4}=4}$

Sol:

3x - y = 23                         ...(1)

${\displaystyle \frac{x}{3}+\frac{y}{4}=4}$
4x + 3y = 48                     ...(2)

Multiplying equation no. (1) by 3
9x - 3y = 69                      ....(3)

Adding equation (3) and (2)       
           9x - 3y = 69 
     +   4x + 3y = 48       
                 13x = 117
                     x = 9

From (1)
3(9) - y = 23
∴ 27 - y = 23
∴  y = 27 - 23
∴  y = 4

Question 3

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
${\displaystyle \frac{5y}{2}-\frac{x}{3}=8}$

${\displaystyle \frac{y}{2}+\frac{5x}{3}=12}$

Sol:

The given pair of linear equations are
${\displaystyle \frac{5y}{2}-\frac{x}{3}=8}$

⇒ ${\displaystyle -\frac{x}{3}+\frac{5y}{2}=8}$               .....(1) [ On Similifying ]

${\displaystyle \frac{y}{2}+\frac{5x}{3}=12}$

⇒ ${\displaystyle \frac{5x}{3}+\frac{y}{2}=12}$               .....(2) [ On Similifying ]

Multiply equation (1) by 5, we get

${\displaystyle -\frac{5x}{3}+\frac{25y}{2}=40}$         ......(3)
Adding equation (3) and (2)

          ${\displaystyle -\frac{5x}{3}+\frac{25y}{2}=40}$
    +      ${\displaystyle \frac{5x}{3}+\frac{y}{2}=12}$       
                                                 
                        ${\displaystyle \frac{26y}{2}=52}$
⇒ 13y = 52
⇒ y = 4

Substituting y = 4 in equation (1), We get
${\displaystyle -\frac{x}{3}+\frac{5\left(4\right)}{2}=8}$
⇒ ${\displaystyle -\frac{x}{3}=8-10}$
⇒ x = 6
∴ Solution is x = 6 and y = 4.

Question 4

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
${\displaystyle \frac{1}{5}(x-2)=\frac{1}{4}(1-y)}$
26x + 3y + 4 = 0

Sol:

${\displaystyle \frac{1}{5}(x-2)=\frac{1}{4}(1-y)}$
⇒ 4( x - 2 ) = 5( 1 - y )
⇒ 4x - 8 = 5 - 5y
⇒ 4x + 5y = 13                       .....(1)
26x + 3y = - 4                         .....(2)

Multiplying equation no. (1) by 3 and(2) by 5.
12x + 15y = 39                       .....(3)
130x + 15y = - 20                   .....(4)

Subtracting equation (4) from (3)
                12x + 15y = 39                      
        -     130x + 15y = - 20  
               -         -           +     
                        - 118 x = 59
                                 x = ${\displaystyle -\frac{59}{118}}$

                                 x = ${\displaystyle -\frac{1}{2}}$
From (1)
${\displaystyle 4(-\frac{1}{2})}$ + 5y = 13
5y = 13 + 2
y = 3

Question 5

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
y = 2x - 6
y = 0

Sol:

y = 2x - 6                  ...(1)
y = 0                        ....(2)

Adding equation (1) and (2)
             2x - y = 6
      +           y = 2   
                  2x = 6
                    x = 3
x = 3 and y = 0.

Question 6

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
${\displaystyle \frac{x-y}{6}=2(4-x)}$
2x + y = 3( x - 4 )

Sol:

The given pair of linear equations are
${\displaystyle \frac{x-y}{6}=2(4-x)}$

⇒ x - y = 12(4 - x)
⇒ x - y = 48 - 12x
⇒ 13x - y = 48                      ....(1) [ On simplifying ]

2x + y = 3( x - 4 )
⇒ 2x + y = 3x - 12
⇒ x - y = 12                         .....(2) [ On simplifying ]

Multiply equation (2) by 13, We get,
13x - 13y = 156                   .....(3)

Subtracting equation (1) from (3)
            13x - 13y = 156 
       -    13x -   y   =  48  
            -      +         -       
                    - 12y = 108
                         y = - 9
Substituting y = - 9 in equation (1), we get
13x - ( - 9) = 48
⇒ 13x = 39
⇒ x = 3
∴ Solution is x = 3 and y = - 9.

Question 7

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
3 - (x - 5) = y + 2
2 (x + y) = 4 - 3y

Sol:

3 - (x - 5) = y + 2           
∴ 3 - x + 5 = y + 2
∴ - x + 8 = y + 2 
∴ x + y = 6                             ....(1)

2( x + y ) = 4 - 3y
∴ 2x + 2y = 4 - 3y
∴ 2x + 5y = 4                        .....(2)

Multiplying equation no (1) by 2.
2x + 2y = 12                         .....(3)

Subtracting equation (2) from (3)
            2x + 2y = 12
       -    2x + 5y = 4  
           -    -         -     
                 - 3y = 8
                     y = - ${\displaystyle \frac{8}{3}}$

From (1)
x - ${\displaystyle \frac{8}{3}}$ = 6
⇒ x = ${\displaystyle \frac{26}{3}}$

Question 8

3 - (x - 5) = y + 2           
∴ 3 - x + 5 = y + 2
∴ - x + 8 = y + 2 
∴ x + y = 6                             ....(1)

2( x + y ) = 4 - 3y
∴ 2x + 2y = 4 - 3y
∴ 2x + 5y = 4                        .....(2)

Multiplying equation no (1) by 2.
2x + 2y = 12                         .....(3)

Subtracting equation (2) from (3)
            2x + 2y = 12
       -    2x + 5y = 4  
           -    -         -     
                 - 3y = 8
                     y = - ${\displaystyle \frac{8}{3}}$

From (1)
x - ${\displaystyle \frac{8}{3}}$ = 6
⇒ x = ${\displaystyle \frac{26}{3}}$

Sol:

2x - 3y - 3 = 0
⇒ 2x - 3y = 3                    .....(1)

${\displaystyle \frac{2x}{3}+4y+\frac{1}{2}}$ = 0
Multiply by 6,
${\displaystyle 6\times \frac{2x}{3}+6\times 4y+\frac{1}{2}\times 6=0\times 6}$
4x + 24y = - 3                   .....(2)

Multiplying equation no. (1) by 8
16x - 24y = 24                  .....(3)

Adding equation (3) and (2)
         16x - 24y = 24
      +  4x + 24y = - 3 
                  20x = 21
                      x = ${\displaystyle \frac{21}{20}}$

From (1)
∴ ${\displaystyle 2\left[\frac{21}{20}\right]}$ - 3y = 3
∴ - 3y = 3 - ${\displaystyle \frac{21}{20}}$
∴ y = ${\displaystyle -\frac{3}{10}}$

Question 9

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
13x+ 11y = 70
11x + 13y = 74

Sol:

13x + 11y = 70                        ...(1)
11x + 13y = 74                        ...(2)

Adding (1) and (2)
       13x + 11y = 70
+     11x + 13y = 74 
       24x + 24y = 144  
Dividing by 24,
   x + y = 6                             ....(3)

Subtracting (2) from (1)
       13x + 11y = 70
-     11x + 13y = 74  
       -     -          -       
       2x - 2y = - 4
Dividing by 2
         x - y = - 2                    ....(4)

Adding equation (3) and (4)
        x - y = - 2
   +  x + y = 6  
            2x = 4
              x = 2

From (3)
2 + y = 6
y = 4

Question 10

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
41x + 53y = 135
53x + 41y = 147

Sol:

41x + 53y = 135                             ...(1)
53x + 41y = 147                             ...(2)

Adding equation (1) and (2)
        41x + 53y = 135
   +   53x + 41y = 147  
         94x + 94y = 282
Dividing by 94,
            x + y = 3                             ....(3)
Subtracting equation (2) from (1)
        41x + 53y = 135
 -      53x + 41y = 147   
        -      -          -          
        - 12x + 12y = - 12
Dividing by 12,
          - x + y = -1                          ....(4)
Adding (3) and (4)
            x + y = 3
    +   - x + y = -1 
                 2y = 2
                   y = 1

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

Question 11

If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x.

Sol:

2x + y = 23                        ...(1)
4x - y = 19                         ...(2)

Adding equation (1) and (2) we get,
         2x + y = 23
    +   4x - y = 19 
               6x = 42
                 x = 7

From (1)
2x + y = 23
⇒ 2(7) + y = 23
⇒ 14 + y = 23
⇒ y = 23 - 14
y = 9

∴ x - 3y = 7 - 3(9) = -20
and 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31.     

Question 12

If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.

Sol:

10 y = 7x - 4
- 7x + 10y = - 4                     ...(1)
12x + 18y = 1                        ...(2)

Multiplying equation no. (1) by 12 and (2) by 7.
- 84x + 120y = - 48               ....(3)
84x + 126y = 7                      ....(4)

Adding equation (3) and (4)
        - 84 + 120y = -48
   +     84 + 126y =   7    
                   246y = - 41
                         y = - ${\displaystyle \frac{1}{6}}$

From (1)
- 7x + 10${\displaystyle (-\frac{1}{6})}$ = - 4
- 7x = - 4 + ${\displaystyle \frac{5}{3}}$ 
- 7x = ${\displaystyle -\frac{7}{3}}$
    x = ${\displaystyle \frac{1}{3}}$

∴ 4x + 6y = ${\displaystyle 4\left(\frac{1}{3}\right)+6(-\frac{1}{6})=\frac{4}{3}-\frac{6}{6}=\frac{4}{3}-1=\frac{1}{3}}$

∴ 8y - x = ${\displaystyle 8(-\frac{1}{6})-\frac{1}{3}=-\frac{4}{3}-\frac{1}{3}=-\frac{5}{3}}$

Question 13.1

Solve for x and y : 
${\displaystyle \frac{y+7}{5}=\frac{2y-x}{4}+3x-5}$

${\displaystyle \frac{7-5x}{2}+\frac{3-4y}{6}=5y-18}$

Sol:

The given pair of linear equation are
${\displaystyle \frac{y+7}{5}=\frac{2y-x}{4}+3x-5}$
⇒ 55x + 6y = 128                    ....(1)[ On simplifying ]

${\displaystyle \frac{7-5x}{2}+\frac{3-4y}{6}=5y-18}$
⇒  15x + 34y = 132                  ....(2)[ On simplifying ]

Multiply equation (1) by 3 and equation (2) by 11, we get :
 165x + 18y = 384                    ....(3)
165x + 374y = 1452                 .....(4)

Subtracting (4) from (3)
     165x + 18y = 384
-    165x + 374y = 1452  
      -       -             -         
              - 356y = - 1068
                  y = 3
Substituting y = 3 in equation (1), we get
55x + 6(3) = 128
⇒ 55x = 110
⇒ x = 2
∴ Solution is x = 2 and y = 3.

Question 13.2

The given pair of linear equation are
${\displaystyle \frac{y+7}{5}=\frac{2y-x}{4}+3x-5}$
⇒ 55x + 6y = 128                    ....(1)[ On simplifying ]

${\displaystyle \frac{7-5x}{2}+\frac{3-4y}{6}=5y-18}$
⇒  15x + 34y = 132                  ....(2)[ On simplifying ]

Multiply equation (1) by 3 and equation (2) by 11, we get :
 165x + 18y = 384                    ....(3)
165x + 374y = 1452                 .....(4)

Subtracting (4) from (3)
     165x + 18y = 384
-    165x + 374y = 1452  
      -       -             -         
              - 356y = - 1068
                  y = 3
Substituting y = 3 in equation (1), we get
55x + 6(3) = 128
⇒ 55x = 110
⇒ x = 2
∴ Solution is x = 2 and y = 3.

Sol:

The given pair of linear equations are
4x = 17 - ${\displaystyle \frac{x-y}{8}}$
⇒ 33x - y = 136                    ...(1)[ On Simplifying ]

2y + x = 2 + ${\displaystyle \frac{5y+2}{3}}$
⇒ 3x + y = 8                        ...(2)[ On Simplifying ]

Multiply equation (2) by 11, we get,
33x + 11y = 88                   ....(3)

Subtracting equation (1) from (3)
   33x + 11y = 88
-  33x - y = 136    
     -    +      -         
           12y = - 48
              y = - 4
Substituting y = - 4 in equation (1), we get :
33x - ( - 4 ) = 136
⇒  33x = 132
⇒  x = 4
∴ Solution is x = 4 and y = - 4.

Question 14

Find the value of m, if x = 2, y = 1 is a solution of the equation 2x + 3y = m.

Sol:

Let x = 2 and y = 1 be a solution of the equation.
2x + 3y = m
⇒ 2(2) + 3(1) = m
⇒ 4 + 3 = m
⇒ m = 7
∴ If x = 2 and y = 1 is the solution of the equation
2x + 3y = m then the value of m is 7.

Question 15

10% of x + 20% of y = 24
3x - y = 20

Sol:

10% of x + 20% of y = 24
⇒ 0.1x + 0.2y = 24                  .....(1) [ On Simplyfying ]
3x - y = 20                               .....(2)

Multiply equation (2) by 0.2, We get :
 0.6x - 0.2y = 4                        ......(3)

Adding equation (3) and (1)
   0.6x - 0.2y = 4
+ 01x + 0.2y = 24 
              0.7x = 28
                  x = 40
Substituting x = 40 in equation (1), We get
0.1(40) + 0.2y = 24
⇒ 0.2y = 20
⇒ y = 100
∴ Solution is x = 40 and y = 100.

Question 16

The value of expression mx - ny is 3 when x = 5 and y = 6. And its value is 8 when x = 6 and y = 5. Find the values of m and n.

Sol:

The value of expression mx - ny is 3 when x = 5 and y = 6.
⇒ 5m - 6n = 3                              .....(1)

The value of expression mx - ny is 8 when x = 6 and y = 5.
⇒ 6m - 5n = 8                              ....(2)

Multiply equation (1) by 6 and equation (2) by 5, We get:
30m - 36n = 18                            ....(3)
30m - 25n = 40                            .....(4)

Subtracting equation (4) from (3)
   30m - 36n = 18
-  30m - 25n = 40 
    -       +        -     
           - 11n = - 22
                 n = 2
Substituting n = 2 in equation (1), we get
5m - 6(2) = 3
⇒ 5m = 15
⇒ m = 3
∴ Solution is m = 3 and n = 2.

Question 17

Solve :
11(x - 5) + 10(y - 2) + 54 = 0
7(2x - 1) + 9(3y - 1) = 25

Sol:

11( x - 5 ) + 10( y - 2 ) + 54 = 0             (given)
⇒ 11x - 55 + 10y - 20 + 54 = 0
⇒ 11x + 10y - 21 = 0
⇒ 11x + 10y = 21                                 ....(1)

7( 2x - 1 ) + 9(3y - 1) = 25                     (given)
⇒ 14x - 7 + 27y - 9 = 25
⇒ 14x + 27y - 16 = 25
⇒ 14x + 27y = 41                               .....(2)

Multiplying equation (1) by 27 and equation (2) by 10, we get,
297x + 270y = 567                            ....(3)
140x + 270y = 410                            .....(4)

Subtracting equation (4) from equation (3), we get
157x = 157
⇒ x = 1

Substituting x = 1 in equation (1), we get,
11 x 1 + 10y = 21
⇒  10y = 10
⇒  y = 1
∴ Solution set is x = 1 and y = 1.

Question 18

Solve :
${\displaystyle \frac{7+x}{5}-\frac{2x-y}{4}=3y-5}$

${\displaystyle \frac{5y-7}{2}+\frac{4x-3}{6}=18-5x}$

Sol:

${\displaystyle \frac{7+x}{5}-\frac{2x-y}{4}=3y-5}$          .....(Given)

⇒ 4( 7 + x ) - 5( 2x - y ) = 20( 3y - 5 )
⇒ 28 + 4x - 10x + 5y = 60y - 100
⇒  - 6x - 55y = - 128                            ......(1)

${\displaystyle \frac{5y-7}{2}+\frac{4x-3}{6}=18-5x}$        ......(Given)
⇒ 3(5y - 7) + 4x - 3 = 6( 18 - 5x )
⇒ 15y - 21 + 4x - 3 = 108 - 30x
⇒ 34x + 15y = 132                              .......(2)

Multiplying equation (1) by 34 and equation (2) by 6, We get
- 204x - 1870y = - 4352                     .....(3)
204x + 90y = 792                              ......(4)

Adding equation (3) and (4), We get
       - 204x - 1870y = - 4352                     .....(3)
  +     204x + 90y   = 792         
                - 1780y = -3560
                       ⇒ y = 2

Substituting y = 2 in equation (1), We get
- 6x - 55 x 2 = - 128
⇒ - 6x - 110 = - 128
⇒ - 6x = - 18
⇒ x = 3
∴ Solution is x = 3 and y = 2.

Question 19

Solve :
${\displaystyle 4x+\frac{x-y}{8}=17}$

${\displaystyle 2y+x-\frac{5y+2}{3}=2}$

Sol:

${\displaystyle 4x+\frac{x-y}{8}=17}$                     (Given)
⇒ 32x + x - y = 136
⇒ 33x - y = 136                             ......(1)

${\displaystyle 2y+x-\frac{5y+2}{3}=2}$               (Given)
⇒ 6y + 3x - 5y - 2 = 6
⇒ 3x + y = 8                                 .......(2)

Adding equations (1) and (2), we get
          33x - y = 136 
  +        3x + y = 8     
                 36x = 144
                     x = 4

Substituting x = 4 in equation (2), We get
3 x 4 + y = 8
⇒ 12 + y = 8
⇒ y = 8 - 12
⇒ y = - 4
∴ Solution is x = 4 and y = - 4