Question 1:  Solve the pair of simultaneous equations using substitution method.

(i) 2x+3y = 12 and 3x+2y=18

Answer:

The given system of equation is:

2x+3y = 12 … … … … … (i)

3x+2y=18 … … … … … (ii)

From equation (i), we get

x = \frac{1}{2} (12-3y) \Rightarrow x =  6 - \frac{3}{2} y

Substituting x =  6 - \frac{3}{2} y in equation (ii), we get

3 ( 6 - \frac{3}{2} y) +2y=18

\Rightarrow 18 - \frac{9}{2} y = 18

\frac{9}{2} y = 0 \Rightarrow y = 0

Putting y = 0 in x =  6 - \frac{3}{2} y we get x = 6

Hence the solution of the given system of equations is x =6, y = 0

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(ii) \frac{b}{a} x + \frac{a}{b} y = a^2 + b^2 and x+y = 2ab

Answer:

The given system of equation is:

\frac{b}{a} x + \frac{a}{b} y = a^2 + b^2 … … … … … (i)

x+y = 2ab … … … … … (ii)

From equation (ii), we get

x = (2ab-y)

Substituting x = (2ab-y) in equation (i), we get

\frac{b}{a} (2ab-y) + \frac{a}{b} y = a^2 + b^2

\Rightarrow 2b^2 - \frac{b}{a} y + + \frac{a}{b} y = a^2 + b^2

\Rightarrow \frac{a^2-b^2}{ab} y = a^2 + b^2

\Rightarrow y = ab

Putting y = ab in x = (2ab-y) y we get x = ab

Hence the solution of the given system of equations is x =ab, y = ab

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(iii) \frac{x}{6} = y-6 and \frac{3x}{4} = 1+y

Answer:

The given system of equation is:

\frac{x}{6} = y-6 … … … … … (i)

\frac{3x}{4} = 1+y … … … … … (ii)

From equation (i), we get

y = \frac{x}{6} + 6

Substituting y = \frac{x}{6} + 6 in equation (ii), we get

\frac{3}{4} x = 1+ \frac{x}{6} + 6

\Rightarrow \Big( \frac{3}{4} - \frac{1}{6} \Big) x = 7

\Rightarrow x = 12

Putting x = 12 in y = \frac{12}{6} + 6 we get y = 8

Hence the solution of the given system of equations is x =12, y = 8

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(iv) \frac{x}{2} + \frac{2y}{3} = -1 and x - \frac{y}{3} = 3

Answer:

The given system of equation is:

\frac{x}{2} + \frac{2y}{3} = -1 … … … … … (i)

x - \frac{y}{3} = 3 … … … … … (ii)

From equation (ii), we get

x = \frac{y}{3} +3

Substituting x = \frac{y}{3} +3 in equation (i), we get

\frac{1}{2}\Big( \frac{y}{3} + 3 \Big) + \frac{2y}{3} = -1

\Rightarrow \frac{y}{6} + \frac{3}{2} + \frac{2y}{3} = -1

\Rightarrow y \Big( \frac{1}{6} + \frac{2}{3} \Big) = -1 - \frac{3}{2}

\Rightarrow y \frac{5}{6} = - \frac{5}{2} 

\Rightarrow y = -3

Putting y = -3 we get x = 3 + \frac{-3}{3} = 2

Hence the solution of the given system of equations is x =2, y = -3

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(v) 9-(x-4)= y + 7 and 2(x+y) = 4-3y

Answer:

The given system of equation is:

9-(x-4)= y + 7  \Rightarrow x+y = 6  … … … … … (i)

2(x+y) = 4-3y \Rightarrow 2x+5y =4  … … … … … (ii)

From equation (i), we get

x = 6-y

Substituting x =  6-y   in equation (ii), we get

2(6-y)+5y = 4

\Rightarrow 12 - 2y + 5y = 4

\Rightarrow y = - \frac{8}{3}

Putting y = - \frac{8}{3} in x =  6 - ( \frac{- 8}{3} )   we get x = \frac{26}{3}

Hence the solution of the given system of equations is x = \frac{26}{3} y = - \frac{8}{3}

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Question 2: Solve the system of equation using the method of elimination.

(i)   \frac{2}{x} + \frac{2}{3y} = \frac{1}{6}    and   \frac{3}{x} + \frac{2}{y} =0

Answer:

Taking \frac{1}{x} = u and \frac{1}{y} = v , the given system of equations become

12u + 4v = 2 … … … … … (i)

3u+2v= 0   … … … … … (ii)

Now multiplying equation (ii) by 4 and subtracting it from equation (i) we get

12u+4v=1

\underline{ (-) \hspace{0.5cm} 12u+8v=0}  

-4v=1

\Rightarrow v = - \frac{1}{4}

Substituting v = - \frac{1}{4} in (ii) we get

3u+ 2 (- \frac{1}{4} ) = 0 \Rightarrow u = \frac{1}{6}

Therefore x = \frac{1}{u} = 6 and y = \frac{1}{v} = -4

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(ii)   8v-3u=5uv    and   6v-5u=-2uv

Answer:

If we put v=0 in either of the equations, we get u = 0 . Similarly if we put u=0 in either of the equations, we get v = 0 . Therefore u=0 and v=0 is one of the solutions that satisfies the two equation. However, for u \neq 0 and v \neq 0 , we follow the following approach.

Dividing each of the given equations by uv we get:

\frac{8}{u} - \frac{3}{v} = 5 … … … … … (i)

\frac{6}{u} - \frac{5}{v} = -2 … … … … … (ii)

Taking \frac{1}{u} = x and \frac{1}{v} = y , the given system of equations become

8x - 3y = 5 … … … … … (iii)

6x - 5y= -2   … … … … … (iv)

Now multiplying equation( (i) by 3 and equation (ii) by 4   we get

24x - 9y = 15   … … … … … (v)

24x - 20y= -8   … … … … … (vi)

Subtracting (vi) from (v) we get

24x - 9y = 15

\underline{ (-) \hspace{0.5cm} 24x - 20y= -8}  

11y = 23

\Rightarrow y = \frac{23}{11}

Substituting y = \frac{23}{11} in (iii) we get

8x - 3( \frac{23}{11} ) = 5 \Rightarrow x = \frac{31}{22}

Therefore u = \frac{1}{x} = \frac{22}{31} and v = \frac{1}{y} = \frac{11}{23}

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(iii)   \frac{1}{2(2x+3y)} + \frac{12}{7(3x-2y)} = \frac{1}{2}    and   \frac{7}{2x+3y} + \frac{4}{3x-2y} = 2

Answer:

Taking \frac{1}{2x+3y} = u and \frac{1}{3x-2y} = v , the given system of equations become

\frac{1}{2} u + \frac{12}{7} v = \frac{1}{2} \Rightarrow 7u + 24v = 7 … … … … … (i)

7u + 4v= 2   … … … … … (ii)

Subtracting (ii) from (1) we get

 7u + 24v = 7

\underline{ (-) \hspace{0.5cm} 7u + 4v= 2}  

20v = 5

\Rightarrow v = \frac{1}{4}

Substituting v = \frac{1}{4} in (i) we get

7u + 24( \frac{1}{4} ) = 7 \Rightarrow u = \frac{1}{7}

Therefore

\frac{1}{2x+3y} = u \Rightarrow 2x+3y = 7   … … … … … (iii)

\frac{1}{3x-2y} = v \Rightarrow 3x-2y = 4   … … … … … (iv)

Now multiplying equation( (iv) by 2 and equation (iii) by 3   we get

6x+9y=21   … … … … … (v)

6x-4y=8   … … … … … (vi)

Subtracting (vi) from (v) we get

6x+9y=21

\underline{ (-) \hspace{0.5cm} 6x-4y=8}  

13y = 13

\Rightarrow y = 1

Substituting y =1 in (v) we get

6x+9(1)=21 \Rightarrow x = 2

Therefore x =2, y = 1 is the solution for the given system of equations.

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(iv)   0.4x+0.3y=1.7    and   0.7x-0.2y=0.8

Answer:

The given system of equation is:

0.4x+0.3y=1.7 … … … … … (i)

0.7x-0.2y=0.8 … … … … … (ii)

Now multiplying equation( (ii) by 30 and equation (i) by 20   we get

8x+6y=34   … … … … … (iii)

21x-6y=24   … … … … … (iv)

Adding (iv) and (iii) we get

8x+6y=34

\underline{ (+) \hspace{0.5cm} 21x-6y=24}  

29x = 58

\Rightarrow x = 2

Substituting x =2 in (iii) we get

8(2)+6y=34 \Rightarrow y = 3

Therefore x =2, y = 3 is the solution for the given system of equations.

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(v)   \frac{x}{2} + y = 0.8    and   \frac{7}{x+\frac{y}{2}} = 10

Answer:

The given system of equation is:

\frac{x}{2} + y = 0.8 \Rightarrow x + 2y = 1.6 … … … … … (i)

\frac{7}{x+\frac{y}{2}} = 10 \Rightarrow 2x+y = 1.4 … … … … … (ii)

Now multiplying equation( (i) by 2 and equation (ii) by 1   we get

2x+4y=3.2   … … … … … (iii)

2x+y=1.4   … … … … … (iv)

Subtracting (iv) and (iii) we get

2x+4y=3.2

\underline{ (-) \hspace{0.5cm} 2x+y=1.4}  

3y = 1.8 

\Rightarrow y = 0.6

Substituting y = 0.6 in (i) we get

x = 1.6 - 2(0.6) = 0.4

Therefore x =0.4, y = 0.6 is the solution for the given system of equations.

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(vi)   \frac{x}{3} + \frac{y}{4} = 11    and   \frac{5x}{6} - \frac{y}{3} = -7

Answer:

The given system of equation is:

\frac{x}{3} + \frac{y}{4} = 11 \Rightarrow 4x+3y=132 … … … … … (i)

\frac{5x}{6} - \frac{y}{3} = -7  \Rightarrow 5x-2y=-42 … … … … … (ii)

Now multiplying equation( (i) by 2 and equation (ii) by 3   we get

8x+6y = 264   … … … … … (iii)

15x-6y = -126   … … … … … (iv)

Adding (iv) and (iii) we get

8x+6y = 264

\underline{ (+) \hspace{0.5cm} 15x-6y = -126}  

23x = 138 

\Rightarrow x = 6

Substituting x = 6 in (i) we get

3y = 132 -4(6) -= 132 - 24 = 108 \Rightarrow y = 36

Therefore x =, y = 36 is the solution for the given system of equations.

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(vii)   3x- \frac{y+7}{11} + 2 = 10    and   2y + \frac{x+11}{7} = 10

Answer:

The given system of equation is:

3x- \frac{y+7}{11} + 2 = 10 \Rightarrow 33x-y=95 … … … … … (i)

2y + \frac{x+11}{7} = 10 \Rightarrow x+14y = 59 … … … … … (ii)

Now multiplying equation( (i) by 14 and equation (ii) by 1   we get

462x-14y=1330   … … … … … (iii)

x+14y=59   … … … … … (iv)

Adding (iv) and (iii) we get

462x-14y=1330

\underline{ (+) \hspace{0.5cm} x+14y=59}  

463 x = 1389 

\Rightarrow x = 3

Substituting x = 3 in (i) we get

y = 33(3) - 95 = 99 - 95 = 4

Therefore x =3, y = 4 is the solution for the given system of equations.

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(viii)   \frac{1}{2x} - \frac{1}{y} = -1    and   \frac{1}{x} + \frac{1}{2y} = 8

Answer:

Taking \frac{1}{x} = u and \frac{1}{y} = v , the given system of equations become

u - 2v = -2 … … … … … (i)

2u+v=16   … … … … … (ii)

Now multiplying equation (i) by 2 and subtracting (ii) from equation (i) we get

2u-4v=-4

  \underline{ (-) \hspace{0.5cm} 2u+v=16}  

                   -5v=-20

\Rightarrow v = 4

Substituting v = 4 in (ii) we get

u = 2(4)-2 = 6

Therefore x = \frac{1}{6} and y = \frac{1}{4}

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(ix)   \frac{6}{x+y} = \frac{7}{x-y} + 3    and   \frac{1}{2(x+y)} = \frac{1}{3(x-y)}

Answer:

Taking \frac{1}{x+y} = u and \frac{1}{x-y} = v , the given system of equations become

6u-7v=3 … … … … … (i)

3u -2v =0   … … … … … (ii)

Now multiplying equation (ii) by 2 and subtracting it from equation (i) we get

6u-7v=3

      \underline{ (-) \hspace{0.5cm} 3u -2v =0}  

                   -3v= 3

\Rightarrow v = -1

Substituting v = -1 in (ii) we get

u = \frac{2}{3} (-1) = - \frac{2}{3}

Therefore x+y = - \frac{3}{2}  … … … … … (iii)

x-y = -1  … … … … … (vi)

Adding (iii) and (iv) we get 2x = - \frac{5}{2} \Rightarrow x = - \frac{5}{4}

from (iv) y = x+1 = - \frac{5}{4} + 1 = - \frac{1}{4}

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(x)   \frac{5}{x+y} - \frac{2}{x-y} = -1    and   \frac{15}{x+y} + \frac{7}{x-y} = 10

Answer:

Taking \frac{1}{x+y} = u and \frac{1}{x-y} = v , the given system of equations become

5u-2v=-1 … … … … … (i)

15u+7v=10   … … … … … (ii)

Now multiplying equation (i) by 3 and subtracting (ii) from equation (i) we get

15u-6v=-3

      \underline{ (-) \hspace{0.5cm} 15u+7v=10}  

                   -13v=-13

\Rightarrow v = 1

Substituting v = 1 in (ii) we get

u = \frac{1}{5} (2-1) = \frac{1}{5}

Therefore x+y = 5   … … … … … (iii)

x-y = 1  … … … … … (vi)

Adding (iii) and (iv) we get 2x = 6 \Rightarrow x = 3

from (iv) y = x-1 = 3-1 = 2

Therefore x = 3, y = 2

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(xi)   \frac{1}{2(x+2y)} + \frac{5}{3(3x-2y)} = -\frac{3}{2}    and   \frac{5}{4(x+2y)} - \frac{3}{5(3x-2y)} = \frac{61}{60}

Answer:

Taking \frac{1}{x+2y} = u and \frac{1}{3x-2y} = v , the given system of equations become

3u+10v=-9 … … … … … (i)

75u-36v=61   … … … … … (ii)

Now multiplying equation (i) by 25 and subtracting (ii) from equation (i) we get

75u+250v=-225

      \underline{ (-) \hspace{0.5cm} 75u-36v=61}  

                   286v=-286

\Rightarrow v = -1

Substituting v = -1 in (ii) we get

3u = -9 -10(-1) = 1 \Rightarrow u = \frac{1}{3}

Therefore x+2y = 3   … … … … … (iii)

3x-2y = -1  … … … … … (vi)

Adding (iii) and (iv) we get 4x = 2 \Rightarrow x = \frac{1}{2}

from (iv) y = \frac{1}{2} (3 - \frac{1}{2} ) = \frac{5}{4}

Therefore x = \frac{1}{2} , y = \frac{5}{4}

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(xii)   \frac{2}{3x+2y} + \frac{3}{3x-2y} = \frac{17}{5}    and   \frac{5}{3x+2y} + \frac{1}{3x-2y} = 2

Answer:

Taking \frac{1}{3x+2y} = u and \frac{1}{3x-2y} = v , the given system of equations become

10u+15v=17 … … … … … (i)

5u+v=2   … … … … … (ii)

Now multiplying equation (ii) by 2 and subtracting (ii) from equation (i) we get

10u+15v=17

      \underline{ (-) \hspace{0.5cm} 10u+2v=4}  

                   13v = 13

\Rightarrow v = 1

Substituting v = 1 in (ii) we get

5u = 2-1= 1 \Rightarrow u = \frac{1}{5}

Therefore 3x+2y = 5   … … … … … (iii)

3x-2y = 1  … … … … … (vi)

Adding (iii) and (iv) we get 6x = 6 \Rightarrow x = 1

from (iii) 2y = 5-3(1) = 2 \Rightarrow y = 1

Therefore x = 1, y = 1

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(xiii)   \frac{5}{x+1} - \frac{2}{y-1} = \frac{1}{2}    and   \frac{10}{x+1} + \frac{2}{y-1} = \frac{5}{2}

Answer:

Taking \frac{1}{x+1} = u and \frac{1}{y-1} = v , the given system of equations become

10u-4v=1 … … … … … (i)

20u+4v=5   … … … … … (ii)

Adding (i) and (ii)  we get

10u-4v=1

      \underline{ (+) \hspace{0.5cm} 20u+4v=5}  

                   30u = 6 

\Rightarrow u = \frac{1}{5}

Substituting u = \frac{1}{5} in (i) we get

4v = 10(\frac{1}{5})-1 = 2-1 = 1 \Rightarrow v = \frac{1}{4}

Therefore x+1 = 5 \Rightarrow x = 4

y-1 = 4  \Rightarrow y = 5 

Therefore x = 4, y = 5

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(xiv)   x-y+z=4 , x-2y-2z=9    and   2x+y+3z=1

Answer:

Given system of equations is

x-y+z=4 … … … … … (i)

x-2y-2z=9 … … … … … (ii)

2x+y+3z=1 … … … … … (iii)

From (i) z = 4 - x +y

Substituting this in (ii) and (iii) we get

3x-4y = 17   … … … … … (iv)

x-4y = 11   … … … … … (v)

Subtracting (v) from (iv) we get

3x-4y = 17

      \underline{ (-) \hspace{0.5cm} x-4y = 11}  

                   x= 3 

From (v) 4y = 3 - 11 = -8 \Rightarrow y = -2

Therefore z = 4 -3 + (-2) = -1

Hence x = 3, y = -2 \ and \  z = -1

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(xv)   x-y+z=4 , x+y+z=2    and   2x+y-3z=0

Answer:

Given system of equations is

x-y+z=4 … … … … … (i)

x+y+z=2 … … … … … (ii)

2x+y-3z=0 … … … … … (iii)

From (i) y = x+z-4

Substituting this in (ii) and (iii) we get

2x+2z=6   … … … … … (iv)

3x-2z=4   … … … … … (v)

Adding (v) from (iv) we get

2x+2z=6

      \underline{ (+) \hspace{0.5cm} 3x-2z=4}  

                   5x= 10 

Therefore x = 2

From (v) 2z = 6-2(2) = 4 \Rightarrow z = 1

Therefore y = 2+1-4 = -1

Hence x = 2, y = -1 \ and \  z = 1

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(xvi)   \frac{10}{x+y} + \frac{2}{x-y} = 4    and   \frac{15}{x+y} - \frac{9}{x-y} = -2

Answer:

Taking \frac{1}{x+y} = u and \frac{1}{x-y} = v , the given system of equations become

10u+2v=4 … … … … … (i)

15u-9v=-2   … … … … … (ii)

Multiplying (i) by 3 and (ii) by 2 we get

30u+6v=12 … … … … … (iii)

30u-18v=-4   … … … … … (iv)

subtracting (iv) from (iii)

30u+6v=12

      \underline{ (-) \hspace{0.5cm} 30u-18v=-4}  

                   24v=16

\Rightarrow v = \frac{2}{3}

Substituting v = \frac{2}{3} in (ii) we get

10u = 4 - 2 ( \frac{2}{3} ) = \frac{8}{3} \Rightarrow u = \frac{4}{15}

Therefore x+y = \frac{15}{4} … … … … … (iii)

x-y = \frac{3}{2} … … … … … (vi)

Adding (iii) and (iv) we get 2x = \frac{21}{4} \Rightarrow x = \frac{21}{8}

from (iv) y = \frac{21}{8} - \frac{3}{2} = \frac{9}{8}

Hence x = \frac{21}{8} and y = \frac{9}{8}

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(xvii)   \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}    and   \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = - \frac{1}{8}

Answer:

Taking \frac{1}{3x+y} = u and \frac{1}{3x-y} = v , the given system of equations become

4u+4v=3 … … … … … (i)

4u-4v=-1   … … … … … (ii)

Adding (i) and (ii)

4u+4v=3

      \underline{ (-) \hspace{0.5cm} 4u-4v=-1}  

                   8u=2

\Rightarrow u = \frac{1}{4}

Substituting u = \frac{1}{4} in (ii) we get

v = \frac{3}{4} - \frac{1}{2} = \frac{1}{2}

Therefore 3x+y = 4   … … … … … (iii)

3x-y = 2  … … … … … (vi)

Adding (iii) and (iv) we get 6x = 6 \Rightarrow x = 1

from (iv) y = 3(1) - 2 = 1

Exercise 7(a) continued …

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