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

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Question 3: Solve using cross multiplication method.

(i) ax+by= a-b    and   bx-ay = a+b

Answer:

The given system of equations may be written as

ax+by -(a-b) = 0

bx-ay - (a+b)= 0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -(a+b) ] - [-a] \times [ -(a-b) ]} = \frac{-y}{[ a ] \times [ -(a+b) ] - [b] \times [ -(a-b) ]} = \frac{1}{[a] \times [-a]-[b] \times [b]}

\Rightarrow \frac{x}{-ab-b^2-a^2+ab} =  \frac{-y}{-a^2-ab+ab-b^2} = \frac{1}{-a^2-b^2}

\Rightarrow \frac{x}{-b^2-a^2} =  \frac{-y}{-a^2-b^2} = \frac{1}{-(a^2+b^2)}

\Rightarrow \frac{x}{-(a^2+b^2)} =  \frac{-y}{-(a^2+b^2)} = \frac{1}{-(a^2+b^2)}

\Rightarrow x = \frac{-(a^2+b^2)}{-(a^2+b^2)} = 1   and  y = - \frac{-(a^2+b^2)}{-(a^2+b^2)} = -1

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(ii) x+y = (a+b)    and   ax-by = a^2 - b^2

Answer:

The given system of equations may be written as

x+y - (a+b) = 0

ax-by -(a^2 - b^2)= 0

By Cross multiplication, we get

\frac{x}{[ 1 ] \times [ -(a^2 - b^2) ] - [-b] \times [ - (a+b) ]} = \frac{-y}{[ 1 ] \times [ -(a^2 - b^2) ] - [a] \times [ - (a+b) ]} =  \frac{1}{[1] \times [-b]-[1] \times [a]}

\Rightarrow \frac{x}{-a^2+b^2-ab-b^2} =  \frac{-y}{-a^2+b^2+a^2+ab} = \frac{1}{-b-a}

\Rightarrow \frac{x}{-a(a+b)} =  \frac{-y}{b(a+b)} = \frac{1}{-(a+b)}

\Rightarrow x = \frac{-a(a+b)}{-(a+b)} = a   and  y = - \frac{b(a+b)}{-(a+b)} = b

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(iii) ax+by = 1    and   bx+ay = \frac{2ab}{a^2+b^2}

Answer:

The given system of equations may be written as

ax+by - 1 = 0

bx+ay - \frac{2ab}{a^2+b^2} = 0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -\frac{2ab}{a^2+b^2} ] - [a] \times [ -1 ]} = \frac{-y}{[ a ] \times [ -\frac{2ab}{a^2+b^2} ] - [b] \times [ - 1 ]} =  \frac{1}{[a] \times [a]-[b] \times [b]}

\Rightarrow \frac{x}{-\frac{2ab}{a^2+b^2}+a} =  \frac{-y}{-\frac{2ab}{a^2+b^2}+b} = \frac{1}{a^2-b^2}

\Rightarrow \Big( \frac{x}{\frac{-2ab^2+a^3+ab^2}{a^2+b^2}} \Big) =  \Big( \frac{-y}{\frac{-2a^2b+b^3+a^2b}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}

\Rightarrow \Big( \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} \Big) =  \Big( \frac{-y}{\frac{b(b^2-a^2)}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}

\Rightarrow x =  \frac{a(a^2-b^2)}{a^2+b^2} \times \frac{1}{a^2-b^2}  = \frac{a}{a^2+b^2}

\Rightarrow y = - \frac{b(b^2-a^2)}{a^2+b^2}  \times \frac{1}{a^2-b^2} = \frac{b}{a^2+b^2} 

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(iv) \frac{a}{x} - \frac{b}{y} = 0    and   \frac{ab^2}{x} + \frac{a^2b}{y} = a^2+b^2 where x, y \neq 0

Answer:

Let \frac{1}{x} =u and \frac{1}{y} = v

The given system of equations may be written as

au-bv + 0 = 0

ab^2u + a^2bv - (a^2+b^2) = 0

By Cross multiplication, we get

\frac{u}{[ -b ] \times [ -(a^2+b^2) ] - [a^2b] \times [ 0 ]} = \frac{-v}{[ a ] \times [ -(a^2+b^2) ] - [ab^2] \times [ 0 ]} =  \frac{1}{[a] \times [a^2b]-[ab^2] \times [-b]}

\Rightarrow \frac{u}{a^2+b^3} =  \frac{-v}{-a^3-ab^2} = \frac{1}{a^3b+ab^3}

\Rightarrow u =  \frac{b(a^2+b^2)}{ab(a^2+b^2)} = \frac{1}{a}  \Rightarrow x = a

\Rightarrow -v =  \frac{-a(a^2+b^2)}{ab(a^2+b^2)} = \frac{-1}{b}  \Rightarrow y = b

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(v) 3x+2y+25=0    and   2x+y+10=0

Answer:

The given system of equations is

3x+2y+25=0

2x+y+10=0

By Cross multiplication, we get

\frac{x}{[ 2 ] \times [ 10 ] - [1] \times [ 25 ]} = \frac{-y}{[3] \times [ 10 ]- [ 2 ] \times [ 25 ]} =  \frac{1}{[3] \times [1]-[2] \times [2]}

\Rightarrow \frac{x}{-5} =  \frac{-y}{-20} = \frac{1}{-1}

\Rightarrow x =  \frac{-5}{-1} = 5

\Rightarrow y =  - ( \frac{-20}{-1} ) = -20

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(vi) ax+by = a^2    and   bx+ay=b^2

Answer:

The given system of equations may be written as

ax+by - a^2=0

bx+ay-b^2=0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -b^2 ] - [a] \times [-a^2 ]} = \frac{-y}{[b] \times [ -a^2 ]- [ a ] \times [ -b^2 ]} =  \frac{1}{[a] \times [a]-[b] \times [b]}

\Rightarrow \frac{x}{-b^3+a^3} =  \frac{-y}{-ba^2+ab^2} = \frac{1}{a^2-b^2}

\Rightarrow x =  \frac{a^3-b^3}{a^2-b^2} = \frac{(a-b)(a^2+ab+b^2)}{(a+b)(a-b)}  = \frac{a^2+ab+b^2}{a+b}  

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

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(vii) (a+2b)x+(2a-b)y=2    and   (a-2b)x+(2a+b)y=3

Answer:

The given system of equations may be written as

(a+2b)x+(2a-b)y - 2 = 0

(a-2b)x+(2a+b)y - 3 = 0

By Cross multiplication, we get

\frac{x}{[ 2a-b ] \times [ -3 ] - [2a+b] \times [ -2 ]} = \frac{-y}{[ a+2b ] \times [ -3 ] - [a-2b] \times [ - 2 ]} =  \frac{1}{[a+2b] \times [2a+b]-[a-2b] \times [2a-b]}

\Rightarrow \frac{x}{-6a+3b+4a+2b} =  \frac{-y}{-3a-6b+2a-4b} = \frac{1}{2a^2+5ab+2b^2 - 2a^2 +5ab - 2b^2}

\Rightarrow \frac{x}{-2a+5b} =  \frac{-y}{-a-10b} = \frac{1}{10ab}

\Rightarrow x =  \frac{-2a+5b}{10ab}  

\Rightarrow y = \frac{a+10b}{10ab}

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

Answer:

We can simplify the equation first:

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

\Rightarrow x ( \frac{a^2+b^2-ab}{a-b} ) - y ( \frac{a^2+b^2 +ab}{a+b} ) +0 = 0

\Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2}  ) +0 = 0

The given system of equations may be written as

\Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2}  ) +0 = 0

x+y-2a^2 = 0

By Cross multiplication, we get

\frac{x}{[ -\frac{a^3-b^3}{a^2-b^2} ] \times [-2a^2 ] - [0] \times [ 1 ]} = \frac{-y}{[ \frac{a^3+b^3}{a^2-b^2}] \times [ -2a^2 ] - [0] \times [ 1 ]} =  \frac{1}{[\frac{a^3+b^3}{a^2-b^2}] \times [1]-[-\frac{a^3-b^3}{a^2-b^2}] \times [1]}

\Rightarrow \frac{x}{\frac{(a^3-b^3)(2a^2)}{a^2-b^2} } =  \frac{-y}{\frac{(a^3+b^3)(-2a^2)}{a^2-b^2} } = \frac{1}{\frac{2a^3}{a^2-b^2} }

\Rightarrow \frac{x}{\frac{(a^3-b^3)(2a^2)}{a^2-b^2} } =  \frac{-y}{\frac{(a^3+b^3)(-2a^2)}{a^2-b^2} } = \frac{1}{\frac{2a^3}{a^2-b^2} }

\Rightarrow  \frac{x}{a^3-b^3} = \frac{y}{a^3+b^3} = \frac{1}{a}

\Rightarrow x =  \frac{a^3-b^3}{a}  

\Rightarrow y = \frac{a^3-b^3}{a}  

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(ix) bx+cy = a+b    and   ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}

Answer:

First simplify the equation:

ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}

\Rightarrow ax ( \frac{a+b-a+b}{a^2-b^2} ) + cy ( \frac{b+a-b+a}{b^2-a^2} ) = \frac{2a}{a+b}

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

\Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} = 1

The given system of equations may be written as

\Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} - 1 = 0

bx+cy-(a+b) = 0

By Cross multiplication, we get

\frac{x}{[ -\frac{c}{a-b} ] \times [-(a+b) ] - [-1] \times [ c ]} = \frac{-y}{[ \frac{b}{a-b}] \times [ -(a+b) ] - [-1] \times [ b ]} =  \frac{1}{[\frac{b}{a-b}] \times [c]-[-\frac{c}{a-b}] \times [b]}

\Rightarrow \frac{x}{\frac{c(a+b)}{a-b}+c } =  \frac{-y}{\frac{-b(a+b)}{a-b}+b } = \frac{1}{\frac{bc}{a-b} +\frac{bc}{a-b}   }

\Rightarrow \frac{x}{c(a+b)+c(a-b) } =  \frac{-y}{-b(a+b)+b(a-b) } = \frac{1}{2bc}

\Rightarrow  \frac{x}{2ca} = \frac{y}{2b^2} = \frac{1}{2bc}

\Rightarrow x =  \frac{2ca}{2bc} = \frac{a}{b}  

\Rightarrow y =  \frac{2b^2}{2bc} = \frac{b}{c}

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(x) (a-b)x+(a+b)y = 2a^2 - 2b^2    and   (a+b)(x+y) = 4ab

Answer:

First simplify the equations:

(a-b)x+(a+b)y = 2a^2 - 2b^2

\Rightarrow x = \frac{a+b}{a-b} y - 2 \frac{(a-b)(a+b)}{(a-b)} = 0

\Rightarrow x + \frac{a+b}{a-b} y -2(a+b) = 0

Also x + y - \frac{4ab}{a+b} = 0

By Cross multiplication, we get

\frac{x}{[ \frac{a+b}{a-b} ] \times [-\frac{4ab}{a+b}] - [-2(a+b)] \times [ 1 ]} = \frac{-y}{[ -2(a+b)]  \times [ 1 ] - [1] \times [ -\frac{4ab}{a+b}]} =  \frac{1}{[1] \times [1]-[\frac{a+b}{a-b}] \times [1]}

\Rightarrow \frac{x}{\frac{-4ab}{a-b}+2(a+b) } =  \frac{-y}{\frac{4ab}{a+b}-2(a+b) } = \frac{1}{1 -\frac{a+b}{a-b}   }

x = \frac{\frac{-4ab}{a-b}+2(a+b) }{1 -\frac{a+b}{a-b}}  = \frac{-4ab + 2(a^2-b^2)}{a - b - a - b}  = \frac{2ab-a^2+b^2}{b}

y = \frac{-2(a+b) + \frac{4ab}{a+b}}{1 - \frac{a+b}{a-b}} = \frac{-2(a+b)^2+4ab}{a-b-a-b} \times \frac{a-b}{a+b}

=  \frac{-2a^2-2b^2-4ab+4ab}{-2b}   \times \frac{a-b}{a+b}  = \frac{(a^2+b^2)(a-b)}{b(a+b)}

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(xi) \frac{a^2}{x} - \frac{b^2}{y} = 0   and   \frac{a^2b}{x} + \frac{b^2a}{y} = a+b where x, y \neq 0

Answer:

Let \frac{1}{x} = u and \frac{1}{y} = v

Therefore the given set of equations can be written as:

a^2u-b^2v+0 = 0

a^2bu + b^2av-(a+b) = 0

By Cross multiplication, we get

\frac{u}{[-b^2 ] \times [-(a+b)] - [0] \times [ b^2a]} = \frac{-v}{[ a^2]  \times [-(a+b) ] - [0] \times [ a^2b]} =  \frac{1}{[a^2] \times [-(a+b)-[-b^2] \times [a^2b]}

\Rightarrow \frac{u}{b^2(a+b)} =  \frac{-v}{-a^2(a+b) } = \frac{1}{a^3b^2+b^3a^2}

Therefore u = \frac{b^2(a+b)}{ab(a+b)} = \frac{b}{a} \Rightarrow x = \frac{a}{b}

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