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

Question :

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

(i)  \(1\over 2x\) + \(1\over 3y\) = 2     and    \(1\over 3x\) + \(1\over 2y\) = 2

(ii)  \(2\over \sqrt{x}\) + \(3\over \sqrt{y}\) = 2       and    \(4\over \sqrt{x}\) – \(9\over \sqrt{y}\) = 2

(iii)  \(4\over x\) + 3y = 14       and     \(3\over x\) – 4y = 23

(iv)  \(4\over x\) + 3y = 14       and     \(6\over x – 1\) – \(3\over y – 2\) = 1

(v)  \(7x – 2y\over xy\)  = 5       and     \(8x + 7y\over xy\)  =  15

(vi)  6x + 3y = 6xy      and       2x + 4y = 5xy

(vii)  \(10\over x + y\) + \(2\over x – y\) = 4     and     \(15\over x + y\) – \(5\over x – y\) = -2

(viii)  \(1\over 3x + y\) + \(1\over 3x – y\) = \(3\over 4\)   and   \(1\over 2(3x + y)\) – \(1\over 2(3x – y)\) = \(-1\over 8\)

Solution :

(i)  Put \(1\over x\) = u    and    \(1\over y\) = v, then the equation becomes

\(1\over 2\)u + \(1\over 3\)v = 2    \(\implies\)    3u + 2v = 12            ………..(1)

\(1\over 3\)u + \(1\over 2\)v = \(13\over 6\)   \(\implies\)   2u + 3v = 13            ………….(2)

Multiply the equation (1) by 3 and equation (2) by 2, we get

9u + 6v = 36       ……(3)

4u + 6v = 26       ……..(4)

Subtract equation (4) from equation ( 3), we get

5u = 10      \(\implies\)    u = 2

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

18 + 6v = 36     \(\implies\)   v = 3

Now,  \(1\over x\) = u  \(\implies\)  x = \(1\over 2\)

and \(1\over y\) = v  \(\implies\)  y = \(1\over 3\)

(ii)  Put \(1\over \sqrt{x}\) = u    and    \(1\over \sqrt{y}\) = v, then the equation becomes

2u + 3v = 2         ……(1)

and    4u – 9u = -1           ……(2)

Multiply equation (1) by 3 and adding it with equation (2), we get

10u = 5      \(\implies\)   u = \(1\over 2\)

Put the value of u in equation (1), we get

3v = 1   \(\implies\)   v = \(1\over 3\)

Now,  \(1\over \sqrt{x}\) = u    \(\implies\)    x = 4

and \(1\over \sqrt{y}\) = v    \(\implies\)   y = 9

(iii)  The given linear equations are

\(4\over x\) + 3y = 14            …….(1)

and     \(3\over x\) – 4y = 23           …….(2)

Multiply equation (1) by 4 and equation (2) by 3, we get

\(16\over x\) + 12y = 56           …….(3)

\(9\over x\) – 12y = 69           …….(4)

Now, Add equation (3) and equation (3), we get

\(25\over x\) = 125     \(\implies\)   x = \(1\over 5\)

Put the value x in equation (1), we get

3y = -6  \(\implies\)  y = -2

Hence, x = \(1\over 5\)  and  y = -2.

(iv)  Let u = \(1\over x – 1\)  and  v = \(1\over y – 2\). Then, the given linear equations becomes,

5u + v = 2            ……….(1)

and    6u – 3v = 1           ………..(2)

Multiply equation (1) by 3 and adding it to equation (2), we get

21 u = 7        \(\implies\)    u = \(1\over 3\)

Put the value of u in equation (1), we get

\(5\over 3\) + v = 2     \(\implies\)   v = \(1\over 3\)

Now, u = \(1\over x – 1\)  \(\implies\)  x = 4

and v = \(1\over y – 2\)   \(\implies\)  y = 5

Hence, x = 4 and y = 5.

(v)  The given linear equations are :

\(7x – 2y\over xy\)  = 5       \(\implies\)   \(7\over y\) – \(2\over x\) = 5

and     \(8x + 7y\over xy\)  =  15         \(\implies\)   \(8\over y\) + \(7\over x\) = 15

Let u = \(1\over x\)  and  v = \(1\over y\). Then, the above equations becomes

7v – 2u  = 5              …….(1)

and  8v + 7u = 15        ………(2)

Multiply equation (1) by 7 and equation (2) by 2, we get

49 – 14u = 35        ……..(3)

and  16v + 14u = 30        ……..(4)

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

65v = 65   \(\implies\)   v = 1

Put the value of v in equation (1), we get

7 – 2u = 5      \(\implies\)    u = 1

Now, u = \(1\over x\)   \(\implies\)  x = 1

and v = \(1\over y\)    \(\implies\)  y = 1

Hence, x = 1, y = 1.

(vi)  The given linear equations are

6x + 3y = 6xy

and       2x + 4y = 5xy

On dividing above equations by xy, we get

\(3\over x\)  +  \(6\over y\)  = 6

and   \(4\over x\) + \(2\over y\) = 5

Put \(1\over x\) = u and \(1\over y\) = u, the above equation becomes,

3u + 6v = 6           ……..(1)

and  4u + 2v = 5           ……….(2)

Multiply equation (2) by 3 and subtracting it from equation (1), we get

-9u = -9   \(\implies\)   u = 1

Put the value of u in equation (1), we get

6v = 3  \(\implies\)   v = \(1\over 2\)

Now,  \(1\over x\) = u  \(\implies\)  x = 1

\(1\over y\) = v    \(\implies\)   y = 2

Hence,  x = 1 and y = 2.

(vii)  The given linear equations are

\(10\over x + y\) + \(2\over x – y\) = 4

and     \(15\over x + y\) – \(5\over x – y\) = -2

Put u = \(1\over x + y\)  and v = \(1\over x – y\), the given equations becomes

10u + 2v = 4      \(\implies\)   5u + v = 2          ………..(1)

and 15u – 5v = -2            ………(2)

Multiply equation (1) by 5 and adding it with equation (2), we get

40u = 8 \(\implies\) u = \(1\over 5\)

Put the value of u in equation (1), we get

v = 1

Now, u = \(1\over x + y\)  \(\implies\)   x + y = 5           …….(4)

and v = \(1\over x – y\)    \(\implies\)   x – y = 1             ……..(5)

Adding equation (4) and equation (5), we get

2x = 6 \(\implies\)  x = 3

Put the value of x in equation (4), we get

3 + y = 5    \(\implies\)  y = 2

Hence, x = 3 and y = 2.

(viii)  The given equation are

\(1\over 3x + y\) + \(1\over 3x – y\) = \(3\over 4\)

and   \(1\over 2(3x + y)\) – \(1\over 2(3x – y)\) = \(-1\over 8\)

Putting u = \(1\over 3x + y\) and v = \(1\over 3x – y\), then the equation becomes,

u + v = \(3\over 4\)          ……….(1)

and \(1\over 2\)u – \(1\over 2\)v = \(-1\over 8\)

\(\implies\)  u – v = \(-1\over 4\)          ……..(2)

Add equation (1) and equation (2), we get

2u = \(1\over 2\)   \(\implies\)   u = \(1\over 4\)

Put the value of u in equation (1), we get

\(1\over 4\) + v = \(3\over 4\)  \(\implies\)  v = \(1\over 2\)

Now, u = \(1\over 3x + y\)  \(\implies\)   3x + y = 4           ……..(3)

and  v = \(1\over 3x – y\)    \(\implies\)   3x – y = 2            ……..(4)

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

6x = 6   or  x = 1

Put the value x in equation (3), we get

3 + y = 4    or     y = 1

Hence, x = 1, y = 1.

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