# 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.