Chapter 3: Pair of Linear Equations in Two Variables

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Pair of linear equations in two variables CBSE NCERT Solutions

Question:

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)   frac{1}{2x} + frac{1}{3y} = 2

     frac{1}{3x} + frac{1}{2y} = frac{13}{6}

 

(ii)   frac{2}{sqrt{x}} + frac{3}{sqrt{y}} = 2

        frac{4}{sqrt{x}} - frac{9}{sqrt{y}} = -1

 

(iii)    frac{4}{x} + 3y = 14

          frac{3}{x} - 4y = 14

       

(iv)   frac{5}{x-1} + frac{1}{y-2} = 2

        frac{6}{x-1} - frac{3}{y-2} = 1

 

(v)  7x - 2yxy = 57x - 2yxy = 5

      8x + 7yxy = 158x + 7yxy  = 15

 

(vi) 6x + 3y = 6xy

       2x + 4y = 5xy

 

(vii)   frac{10}{x+y} + frac{2}{x-y} = 4

          frac{15}{x+y} - frac{5}{x-y} = -2

 

(viii)    frac{1}{3x+y} + frac{1}{3x-y} = frac{3}{4}

          frac{1}{2(3x+y)} - frac{1}{2(3x-y)} = - frac{1}{8}

    

Answer:

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

      frac{6}{x-1} - frac{3}{y-2} = 1            … (2)

Let  1/x = p and  1/y = q

Putting the values of  1/x  = p and  1/y = q in equation (1) and equation (2), we get

\frac{p}{2} + \frac{q}{3} = 2  and  \frac{p}{3} + \frac{q}{2} = \frac{13}{6}

 Multiplying both equations by 6, we get 

3p + 2q = 12 and 2p + 3q = 13

3p + 2q – 12 = 0    ...........… (3)

and 2p + 3q – 13 = 0 ..........… (4)

 

 

\frac{p}{(2)(-13)-(3)(-12)} = \frac{q}{(-12)(2)-(-13)(3)} = \frac{1}{(3)(3)-(2)(2)}

\Rightarrow \frac{p}{-26+36}=\frac{q}{-24+39}=\frac{1}{9-4}

\Rightarrow \frac{p}{10}=\frac{q}{15}=\frac{1}{5}

 = 2 and q = 3

But 1/x = p and  1/y = q

Putting the value of p = 2 and q = 3 in 1/x = p and  1/y = q respectively, we get

=  1/2 and y =   1/3

 

(ii)       frac{2}{sqrt{x}} + frac{3}{sqrt{y}} = 2      … (1)

          frac{4}{sqrt{x}} - frac{9}{sqrt{y}} = -1    … (2)

Let  \frac{1}{\sqrt{x}} = p  and   \frac{1}{\sqrt{y}} = q

Putting  \frac{1}{\sqrt{x}} = pand   \frac{1}{\sqrt{y}} = q in equation (1) and equation (2), we get

2p + 3q = 2   … (3)

4p − 9q = −1 … (4)

Multiplying equation (3) by 2 and subtracting it from equation (4), we get

4p − 9q  – 2 (2p + 3q) = -1 - 2(2)

4p − 9q  − 4p − 6q = -1 -4 

− 15q = - 5

 =  \frac{-5}{-15} = \frac{1}{3}

Putting the value of q in equation (3), we get

2p + 1 = 2

2p = 1

 =  \frac{1}{3}

Putting the values of p = \frac{1}{3} and q = \frac{1}{3}   in  \frac{1}{\sqrt{x}}  = p and   \frac{1}{\sqrt{y}} = q respectively, we get

   \frac{1}{\sqrt{x}} = \frac{1}{2}     and   \frac{1}{\sqrt{y}} = \frac{1}{3}

  \frac{1}{x} = \frac{1}{4}   and  \frac{1}{y} = \frac{1}{9}

 = 4 and y = 9

 

(iii)   \frac{4}{x} + 3y = 14 … (1)

        \frac{3}{x} − 4y = 23 … (2)

Let   1/x = p

we get

4p + 3y = 14 … (3)

3p − 4y = 23 … (4)

Multiplying equation (3) by 3 and equation (4) by 4, we get

3 (4p + 3y - 14 =0 ) and 4(3p − 4y – 23 = 0)

12p + 9y – 42 = 0 … (6)      12p − 16y – 92 = 0 … (7)

Subtracting equation (7) from equation (6), we get

9y − (−16y) – 42 − (−92) = 0

25y + 50 = 0

 = −5025 – 5025 = −2

Putting the value of y in equation (4), we get

4p + 3 (−2) = 14

4p – 6 = 14

4p = 20

   = 5

Putting the value of p in equation (3), we get

1/x  = 5

 x =  1/5

Therefore, x = 1/5  and y = −2

(iv)      frac{5}{x-1} + frac{1}{y-2} = 2       … (1)

           frac{6}{x-1} - frac{3}{y-2} = 1        … (2)

Let  \frac{1}{x-1}  = p and  \frac{1}{y-2} = q    

Putting  \frac{1}{x-1}  = p and  \frac{1}{y-2}  = q in (1) and (2), we get

5p + q = 2

5p + q – 2 = 0 … (3)

And, 6p − 3q = 1

6p − 3q – 1 = 0 … (4)

Multiplying equation (3) by 3 and adding it to equation (4), we get

3 (5p + q − 2) + 6p − 3q – 1 = 0

15p + 3q – 6 + 6p − 3q – 1 = 0

21p – 7 = 0

 = 1/3

Putting p = 1/3  in equation (3), we get

5 \left ( \frac{1}{3} \right ) + q - 2 = 0

5 + 3q = 6

3q = 6 5 = 1

 = 1/3

Putting the values of p = 1/3 and q = 1/3  in \frac{1}{x-1}  = p and \frac{1}{y-2}  = q respectively, we get

  \frac{1}{x-1} = \frac{1}{3}   and   \frac{1}{y-2} = \frac{1}{3} 

3 = − 1 and 3 = y – 2

 = 4 and y = 5

(v) 7x − 2y = 5xy … (1)

      8x + 7y = 15xy … (2)

Dividing both the equations by xy, we get

\frac{7}{y} - \frac{2}{x} = 5 ....(3)

\frac{8}{y} + \frac{7}{x} = 15 ...(4)

Let  1/x = p and  1/y = q

Putting    = p and   = q in equation (3) and equation (4), we get

7q − 2p = 5 … (5)

8q + 7p = 15 … (6)

From equation (5), we get

  2p = 7q – 5

 =  \frac{7q - 5}{2}

Putting value of p =  \frac{7q - 5}{2}  in (6), we get

8q + 7 \left ( \frac{7q - 5}{2} \right ) = 15

16q + 49q – 35 = 30

65q = 30 + 35 = 65

 = 1

Putting value of q in (5), we get

7 (1) − 2p = 5

2p = 2

 = 1

Putting the value of p = 1 and q = 1 in p = 1/x  and q =  1/y  respectively, we get x = 1 and y = 1

(vi) 6x + 3y − 6xy = 0 … (1)

       2x + 4y − 5xy = 0 … (2)

Dividing both the equations by xy, we get

\frac{6}{y} + \frac{3}{x} - 6 = 0    ....(3)

\frac{2}{y} + \frac{4}{x} - 5 = 0    ....(4)

Let   1/x = p and   1/y = q 

Putting  1/x = p and  1/y = q  in equation (3) and equation (4), we get

6q + 3p – 6 = 0 … (5)

2q + 4p – 5 = 0 … (6)

From (5),

3p = 6 − 6q

 = 2 − 2q

Putting this in equation (6), we get

2q + 4 (2 − 2q) – 5 = 0

2q + 8 − 8q – 5 = 0

− 6q = − 3  = 1/2

Putting the value of q in (p = 2 – 2q), we get

= 2 – 2(\frac{1}{2}) = 2 – 1 = 1

Putting the values of p = 1 and q = 1/2 in  1/x = p and   1/y = q respectively, we get x = 1 and y = 2.

(vii)       frac{10}{x+y} + frac{2}{x-y} = 4           … (1)                           

                                    

               frac{15}{x+y} - frac{5}{x-y} = -2       …(2)

Let  \frac{1}{x+y}  = p and  \frac{1}{x-y}  = q 

Putting \frac{1}{x+y}  = p and  \frac{1}{x-y} = q  in equation (1) and equation (2), we get

10p + 2q = 4 … (3)

15p − 5q = −2 … (4)

From equation (3),

2q = 4 − 10p

 = 2 − 5p … (5)

Putting q = 2 − 5p  in equation (4), we get

15p – 5 (2 − 5p) = −2

15p – 10 + 25p = −2

40p = 8  =   1/5

Putting the value of p = 1/5  in equation (5), we get

q = 2 - 5 \left ( \frac{1}{5} \right )= 2 - 1 = 1

Putting values of p = 1/5 and q = 1 in \frac{1}{x+y}  = p and  \frac{1}{x-y} = q respectively, we get,   \frac{1}{x+y} = \frac{1}{5}    and  \frac{1}{x+y} =1

 + y = 5 … (6) and x – y = 1 … (7)

Adding equation (6) and equation (7), we get

2x = 6  = 3

Putting x = 3 in equation (7), we get

3 – y = 1

 = 3 – 1 = 2

Therefore, x = 3 and y = 2

 

(viii)      frac{1}{3x+y} + frac{1}{3x-y} = frac{3}{4}                           … (1)

             frac{1}{2(3x+y)} - frac{1}{2(3x-y)} = - frac{1}{8}           … (2)

Let  \frac{1}{3x+y}  = p and  \frac{1}{3x-y}  = q    

Putting \frac{1}{3x+y}  = p and \frac{1}{3x-y}  = q in (1) and (2), we get

+ q =  \frac{3}{4} and   \frac{p}{2} - \frac{q}{2} = \frac{-1}{8}  

4p + 4q = 3 … (3) and 4p − 4q = − 1 … (4)

Adding equation (3) and equation (4), we get

8p = 2  = 1/4  

Putting value of p in equation (3), we get

4 \left ( \frac{1}{4} \right ) + 4q = 3

1 + 4q = 3

4q = 3 – 1 = 2

 = 1/2  

Putting the value of p = 1/4  and q = 1/2 in  \frac{1}{3x+y}  = p and \frac{1}{3x-y}  = q respectively, we get

   \frac{1}{3x+y} = \frac{1}{4}    and  \frac{1}{3x-y} = \frac{1}{2}  

3x + y = 4 … (5) and 3x – y = 2 … (6)

Adding equation (5) and equation (6), we get

6x = 6  = 1

Putting x = 1 in equation (5), we get

3 (1) + y = 4

 = 4 – 3 = 1

Therefore, x = 1 and y = 1

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