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

Question 1

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
8x + 5y = 9
3x + 2y = 4

Sol:

8x + 5y = 9                                         ...(1)
3x + 2y = 4                                         ...(2)

8x + 5y = 9
∴ 5y = 9 - 8x
∴ y = ${\displaystyle \frac{9-8x}{5}}$                               ....(3)
Putting this value of y in (2)
${\displaystyle 3x+2\left[\frac{9-8x}{5}\right]=4}$
Multiplying by 5,
15x + 18 - 16x = 20
15x - 16x = 20 - 18
- x = 2
x = - 2
From (3) y = ${\displaystyle \frac{9-8x}{5}=\frac{9-8(-2)}{5}=\frac{25}{5}=5}$ 

y = 5   

Question 2

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
2x - 3y = 7
5x + y= 9

Sol:

2x - 3y = 7                                       ...(1)
5x + y = 9                                        ...(2)

5x + y = 9
∴ y = 9 - 5x                                     ...(3)
Putting this value of y in (1)
2x - 3 (9 - 5x) = 7
∴ 2x - 27 + 15x = 7
∴ 2x + 15x = 7 + 27
∴ 17x = 34
∴ x = 2
From (2)
y = 9 - 5(2) 
y = -1

Question 3

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
2x + 3y = 8
2x = 2 + 3y

Sol:

2x + 3y = 8                                       ...(1)
2x = 2 + 3y                                       ...(2)

2x = 2 + 3y
Putting this value of 2x in (1)
2 + 3y + 3y = 8
∴ 6y = 8 - 2
∴ 6y = 6
∴ y = 1

From (2) 2x = 2 + 3(1)
x = ${\displaystyle \frac{5}{2}}$
x = 2.5

Question 4

Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116

Sol:

The given pair of linear equations are
0.2x + 0.1y = 25                        ....(1)
2( x - 2 ) - 1.6y = 116                .....(2)

Consider equation (1)
0.2x + 0.1y = 25
⇒ 0.2x = 25 - 0.1y
⇒ x = ${\displaystyle \frac{25-0.1y}{0.2}}$              ....(3) 

Substitute the value of x from equation (3) in equation (2).
2( x - 2 ) - 1.6y = 116
⇒ ${\displaystyle 2[\frac{25-0.1y}{0.2}-2]}$- 1.6y = 116

⇒ ${\displaystyle 2\times \frac{25-0.14}{0.2}-2\times 2-1.6y=116}$

⇒ ${\displaystyle 0.2\times \frac{50-0.24}{0.2}-4-1.6y=116}$

⇒ ${\displaystyle 50-0.2y-0.8-0.32y=23.2}$

⇒ ${\displaystyle 26=0.52y}$

⇒ y = 50                                      ....(4)

Substitute the value of y from equation (4) in equation (3).
x = ${\displaystyle \frac{25-0.1y}{0.2}}$

⇒ x = ${\displaystyle \frac{25- 0.1\left(50\right)}{0.2}}$

⇒ x = ${\displaystyle \frac{25-5}{0.2}}$

⇒ x = 100
∴ Solution is x = 100 and y = 50.

Question 5

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
6x = 7y + 7
7y - x = 8

Sol:

6x = 7y + 7                                  ...(1)
7y - x = 8                                     ...(2)

7y - x = 8
∴ x = 7y - 8

Putting this value of x in (1)
6( 7y - 8 ) = 7y + 7
∴ 42y - 48 = 7y + 7
∴ 35y = 55
∴ y = ${\displaystyle \frac{11}{7}}$

From (2) 
${\displaystyle x=7\left(\frac{11}{7}\right)-8}$
x = 3

∴ x = 3, y = ${\displaystyle \frac{11}{7}}$.

Question 6

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
y = 4x - 7
16x - 5y = 25

Sol:

y = 4x - 7                                      ....(1)
16x - 5y = 25                                ....(2)

y = 4x - 7
Putting this value of y in (2)
∴ 16x - 5 (4x - 7) = 25
∴ 16x - 20x + 35 = 25
∴ - 4x = - 10
∴ x = ${\displaystyle \frac{5}{2}}$

From (1)
y = ${\displaystyle 4\left(\frac{5}{2}\right)-7}$
⇒ y = 10 - 7
⇒ y = 3
Solution is x = ${\displaystyle \frac{5}{2}}$ and y = 3.

Question 7

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
2x + 7y = 39
3x + 5y = 31

Sol:

2x + 7y = 39                           ...(1)
3x + 5y = 31                           ...(2)

2x + 7y = 39
∴ x = ${\displaystyle \frac{39-7y}{2}}$

Putting this value of x in (2)
${\displaystyle 3\left(\frac{39-7y}{2}\right)+5y=31}$

${\displaystyle 117-21y+10y=62}$

${\displaystyle -11y=-55}$

${\displaystyle y=5}$

From (1) x = ${\displaystyle \frac{39-7\left(5\right)}{2}}$

x = ${\displaystyle \frac{4}{2}}$
x = 2

Question 8

Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
1.5x + 0.1y = 6.2
3x - 0.4y = 11.2

Sol:

The given pair of linear equations are
1.5x + 0.1y = 6.2                  ...(1)
3x - 0.4y = 11.2                    ...(2)

Consider equation (1)
1.5x + 0.1y = 6.2
⇒ 1.5x = 6.2 - 0.1y

⇒ x = ${\displaystyle \frac{6.2-0.1y}{1.5}}$           ...(3)

Substitute the value of x from equation (3) in equation (2)
3x - 0.4y = 11.2

⇒ ${\displaystyle 3\left[\frac{6.2-0.1y}{1.5}\right]-0.4y=11.2}$

⇒ ${\displaystyle 2(6.2-0.1y)-0.4y=11.2}$

⇒ ${\displaystyle 12.4-0.2y-0.4=11.2}$

⇒ ${\displaystyle -0.6y=-1.2}$

⇒ y = 2                                  .....(4)

Substitute the value of y from equation (4) in equation (3)
x = ${\displaystyle \frac{6.2-0.1y}{1.5}}$

⇒ x = ${\displaystyle \frac{6.2-0.1\left(2\right)}{1.5}}$

⇒ x = ${\displaystyle \frac{(6.2-0.2)}{1.5}}$

⇒ x = 4
∴ Solution is x = 4 and y = 2.

Question 9

Solve the following pair of linear (Simultaneous ) equation using method of elimination by substitution :
2( x - 3 ) + 3( y - 5 ) = 0
5( x - 1 ) + 4( y - 4 ) = 0

Sol:

Given equations are
2( x - 3 ) + 3( y - 5 ) = 0                      ...(1)
5( x - 1 ) + 4( y - 4 ) = 0                      ...(2)

From (1), we get
2x - 6 + 3y - 15 = 0
⇒ 2x - 21 + 3y = 0
⇒ 2x = 21 - 3y

⇒ x = ${\displaystyle \frac{21-3y}{2}}$

From (2), we get
5( x - 1 ) + 4( y - 4 ) = 0
⇒ 5x - 5 + 4y - 16 = 0
⇒ 5x + 4y - 21 = 0                             ....(3)

Substituting x = ${\displaystyle \frac{21-3y}{2}}$ in (3), we get

${\displaystyle 5\left(\frac{21-3y}{2}\right)+4y-21=0}$

⇒ ${\displaystyle \frac{105-3y}{2}+4y-21=0}$

⇒ 105 - 15y + 8y - 42 = 0
⇒ -7y + 63 = 0
⇒ 7y = 63
⇒ y = 9

Substituting y = 9 in x = ${\displaystyle \frac{21-3y}{2}}$, we get

x = ${\displaystyle \frac{21-3\left(9\right)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3}$
∴ Solution is x = -3 and y = 9.

Question 10

Solve th following pair of linear (Simultaneous ) equation using method of elimination by substitution :
${\displaystyle \frac{2x+1}{7}+\frac{5y-3}{3}=12}$

${\displaystyle \frac{3x+2}{2}-\frac{4y+3}{9}=13}$   

Sol:

${\displaystyle \frac{2x+1}{7}+\frac{5y-3}{3}=12}$         (given)

⇒ ${\displaystyle \frac{3(2x+1)+7(5y-3)}{21}}$ = 12

⇒ 6x + 3 + 35y - 21 = 252
⇒ 6x + 35y - 18 = 252
⇒ 6x + 35y = 270
⇒ 6x = 270 - 35y
⇒ x = ${\displaystyle \frac{270-35y}{6}}$

${\displaystyle \frac{3x+2}{2}-\frac{4y+3}{9}=13}$          (given)

⇒ ${\displaystyle \frac{9(3x+2)-2(4y+3)}{18}=13}$

⇒ 27x + 18 - 8y - 6 = 234
⇒ 27x - 8y + 12 = 234
⇒ 27x - 8y = 222                       ....(1)

Substituting x = ${\displaystyle \frac{270-35y}{6}}$ in (1), we get

${\displaystyle 27\left(\frac{270-35y}{6}\right)}$ - 8y = 222

⇒ 7290 - 945y - 48y = 1332
⇒ - 993y = - 5958
⇒ y = 6

Substituting y = 6 in x = ${\displaystyle \frac{270-35y}{6}}$, we get

x = ${\displaystyle \frac{270-35\times 6}{6}=\frac{270-210}{6}=\frac{60}{6}=10}$

∴ Solution is x = 10 and y = 6.

Question 11

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution:
3x + 2y =11
2x - 3y + 10 = 0

Sol:

3x + 2y = 11
⇒ 3x = 11 - 2y
⇒ x = ${\displaystyle \frac{11-2y}{3}}$                          ...(1)
And,
2x - 3y + 10 = 0
⇒ 2x${\displaystyle \left(\frac{11-2y}{3}\right)-3y+10}$ = 0

⇒ ${\displaystyle \frac{22-4y}{3}-3y}$ = - 10

⇒ ${\displaystyle \frac{22-4y-9y}{3}}$ = - 10

⇒ 22 - 13y = - 30
⇒ 13y = 52
⇒ y = 4

Substituting the value of y in (1), we have

${\displaystyle x=\frac{11-2\left(4\right)}{3}=\frac{11-8}{3}=\frac{3}{3}=1}$

∴ Solution is x = 1 and y = 4.

Qustion 12

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution :
2x - 3y + 6 = 0
2x + 3y - 18 = 0

Sol:

2x - 3y + 6 = 0
⇒ 2x = 3y - 6
⇒ x = ${\displaystyle \frac{3y-6}{2}}$                     ...(1)
And,
2x + 3y - 18 = 0
⇒ 2${\displaystyle \left(\frac{3y-6}{2}\right)}$+ 3y = 18          ...[From(1)]

⇒ 3y - 6 + 3y = 18
⇒ 6y = 24
⇒ y = 4
Substituting the value of y in (1), we have
x = ${\displaystyle \frac{3\times 4-6}{2}=\frac{12-6}{2}=\frac{6}{2}=3}$
∴ Solution is x = 3 and y = 4.

Question 13

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution:
${\displaystyle \frac{3x}{2}-\frac{5y}{3}+2=0}$

${\displaystyle \frac{x}{3}+\frac{y}{2}=2\frac{1}{6}}$

Sol:

${\displaystyle \frac{x}{3}+\frac{y}{2}=2\frac{1}{6}}$

${\displaystyle \frac{x}{3}+\frac{y}{2}=\frac{13}{6}}$

⇒ ${\displaystyle \frac{2x+3y}{6}=\frac{13}{6}}$

⇒ 2x + 3y = 13
⇒ 2x = 13 - 3y
⇒ ${\displaystyle x=\frac{13-3y}{2}}$                     ....(1)
And,

${\displaystyle \frac{3x}{2}-\frac{5y}{3}+2=0}$

⇒ ${\displaystyle \frac{3}{2}\left(\frac{13-3y}{2}\right)-\frac{5y}{3}=-2}$

⇒ ${\displaystyle \frac{39-9y}{4}-\frac{5y}{3}=-2}$

⇒ ${\displaystyle \frac{117-27y-20y}{12}=-2}$

⇒ ${\displaystyle \frac{117-47y}{12}=-2}$

⇒ 117 - 47y = - 24
⇒ 47y = 141
⇒ y = 3
Substituting the value of y in (1), we have
${\displaystyle x=\frac{13-3\times 3}{2}=\frac{13-9}{2}=\frac{4}{2}=2}$

∴ Solution is x = 2 and y = 3.

Question 14

Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
${\displaystyle \frac{x}{6}+\frac{y}{15}=4}$

${\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}}$ 

Sol:

${\displaystyle \frac{x}{6}+\frac{y}{15}=4}$     

⇒ ${\displaystyle \frac{5x+2y}{30}=4}$

⇒ 5x + 2y = 120
⇒ 5x = 120 - 2y

⇒ x = ${\displaystyle \frac{120-2y}{5}}$                   ....(1)

And,
${\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}}$

⇒ ${\displaystyle \frac{1}{3}(x-\frac{y}{4})=\frac{19}{4}}$

⇒ ${\displaystyle \frac{1}{3}(\frac{120-2y}{5}-\frac{y}{4})=\frac{19}{4}}$

⇒ ${\displaystyle \frac{480-8y-5y}{20}=\frac{57}{4}}$ 

⇒ ${\displaystyle \frac{480-13y}{20}=\frac{57}{4}}$

⇒ 480 - 13y = 285
⇒ 13y = 195
⇒ y = 15
Substituting the value of y in (1), we have

${\displaystyle x=\frac{120-2\times 15}{5}=\frac{120-30}{5}=\frac{90}{5}=18}$
∴ Solution is x = 18 and y = 15