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

**(1) ^{1}/_{2x} + ^{1}/_{3y} = 2**

^{1}/_{3x }+ ^{1}/_{2y} = ^{13}/_{6}

**(2) ^{2}/_{√x} + ^{3}/_{√y} = 2**

^{4}/_{√x }– ^{9}/_{√y} = -1

**(3) ^{4}/_{x} + 3y = 14**

^{3}/_{x} – 4y = 3

**(4) ^{5}/_{x-1} + ^{1}/_{y-2} = 2**

^{6}/_{x-1} – ^{3}/_{y-2} = 2

**(5) ^{(7x-2y)}/_{xy} = 5**

^{(8x+7y)}/_{xy} = 5

**(6) 6x + 3y = 6xy**

**2x + 4y = 5xy**

**(7) ^{10}/_{x+y }+ ^{2}/_{x-y} = 4**

^{15}/_{x+y }+ ^{5}/_{x-y} = -2

**(8) ^{1}/_{3x+y} + ^{1}/_{3x-y} = ^{3}/_{4}**

^{1}/_{2(3x+y)} – ^{1}/_{2(3x-y)} = ^{-1}/_{8}

Solution:

**(1) ^{1}/_{2x} + ^{1}/_{3y} = 2**

^{1}**/ _{3x }+ ^{1}/_{2y} = ^{13}/_{6}**

Let ^{1}/_{x }= p and ^{1}/_{y }= q, then the equation will be

^{p}/_{2} + ^{q}/_{3 }= 2 i.e., 3p + 2q – 12 = 0 —————(1)

^{p}/_{3} + ^{q}/_{2 }= ^{13}/_{6} i.e., 2p + 3q – 13 = 0 —————(2)

Apply cross multiplication method,

^{p}/_{[-26-(-36)]} = ^{q}/_{[-24-(-39)] }= ^{1}/_{9-4}

^{p}/_{10} = ^{q}/_{15} = ^{1}/_{5}

^{p}/_{10} = ^{1}/_{5 }; ^{q}/_{15} = ^{1}/_{5}

p = 2 and q = 3

p = ^{1}/_{x} = 2 and q = ^{1}/_{y} = 3

thus x = ^{1}/_{2} and y = ^{1}/_{3}

(2) ^{2}/_{√x} + ^{3}/_{√y} = 2

^{4}/_{√x }– ^{9}/_{√y} = -1

Let ^{1}/_{√x }= p and ^{1}/_{√y }= q, then the equation will be

2p + 3q_{ }= 2 —————(1)

4p – 9q = -1—————(2)

Multiplying equation (1) by 3,

6p + 9q = 6 ————–(3)

Adding equation (2) and (3),

10p = 5

p = ^{1}/_{2} ————–(4)

substitute the vale of p in equation (1) then,

2p + 3q_{ }= 2

2(^{1}/_{2}) + 3q = 2

1 + 3q = 2

3q = 1

q = ^{1}/_{3}

p = ^{1}/_{√x } = ^{1}/_{2} and q = ^{1}/_{√x} = ^{1}/_{3}

√x = 2 and √y = 3

thus x = 4 and y = 9

(3) ^{4}/_{x} + 3y = 14

^{3}/_{x} – 4y = 3

Let ^{1}/_{x }= p, then the equation will be

4p + 3y_{ }= 14 i.e., 4p + 3y – 14 = 0 —————(1)

3p- 4y = 23 i.e., 3p – 4y – 23 = 0 —————(2)

Apply cross multiplication method,

^{p}/_{[-69-56]} = ^{y}/_{[-(42)-(-92)] }= ^{1}/_{-16-9}

^{p}/_{-125} = ^{y}/_{50} = ^{-1}/_{25}

^{p}/_{125} = ^{-1}/_{25 }and ^{y}/_{50} = ^{-1}/_{25}

p = 5 and q = -2

p = ^{1}/_{x} = 5 and y = -2

thus x = ^{1}/_{5} and y = -2

(4)^{5}/_{x-1} + ^{1}/_{y-2} = 2

^{6}/_{x-1} – ^{3}/_{y-2} = 2

Put ^{1}/_{x-1} = p and ^{1}/_{y-2} = q in the given equation,

5p + q = 2 ———–(1)

6p – 3q = 1 ————-(2)

Multiplying equation (1) by 3,

15p + 3q = 6

Adding (2) and (3)

21p = 7

p = ^{1}/_{3}

put the value of p in (1)

5 x ^{1}/_{3} + q = 2

q = 2 – ^{5}/_{3} = ^{1}/_{3}

p = ^{1}/_{x-1} = ^{1}/_{3}

x-1 = 3

x = 4

q = ^{1}/_{y-2} = ^{1}/_{3}

y – 2 = 3

y = 5

Thus, x = 4 and y = 5

(5) ^{(7x-2y)}/_{xy} = 5

^{(8x+7y)}/_{xy} = 5

^{7}/_{y} – ^{2}/_{x} = 5 ——-(1)

^{8}/_{y} + ^{7}/_{x} = 15 ——-(2)

Putting ^{1}/_{x} = p and ^{1}/_{y }= q in the given equation,

-2p + 7q = 5 i.e., -2p + 7q – 5 = 0 —————(3)

7p + 8q = 15 i.e., 7p + 8q – 15 = 0 ———-(4)

By cross multiplication method

^{p}/_{[-105-(-40)} = ^{q}/_{[-35-30]} = ^{1}/_{[-16-49]}

^{p}/_{-65} = ^{q}/_{-65} = ^{1}/_{-65}

^{p}/_{-65} = ^{1}/_{-65} and ^{q}/_{-65} = ^{1}/_{-65}

p = 1 and q = 1

^{1}/_{x} = p = 1 and ^{1}/_{y }= q = 1

x =1 and y = 1

(6) 6x + 3y = 6xy

^{6}/_{y} + ^{3}/_{x} = 6 ————-(1)

2x + 4y = 5xy

^{2}/_{y} + ^{4}/_{x} = 5 ————-(2)

Putting ^{1}/_{x} = p and ^{1}/_{y }= q

3p + 6q – 6 = 0

4p + 2q – 5 = 0

By cross section method

^{p}/_{[-30-(-12)] } = ^{q}/_{[-24-(-15)]} = ^{1}/_{6-24}

^{p}/_{-18 } = ^{q}/_{-9} = ^{1}/_{-18}

^{p}/_{-18 } = ^{1}/_{-18} and ^{q}/_{-9} = ^{1}/_{-18}

p = 1 and q = ^{1}/_{2}

^{1}/_{x} = p = 1 and ^{1}/_{y }= q = ^{1}/_{2}

x = 1 and y = 2

(7) ^{10}/_{x+y }+ ^{2}/_{x-y} = 4

^{15}/_{x+y }+ ^{5}/_{x-y} = -2

Put ^{1}/_{x+y }= p and ^{1}/_{x-y} = q in the equations

10p + 2q = 4 i.e., 10p + 2q – 4 = 0 ———(1)

15p – 5p = -2 i.e., 15p – 5p + 2 = 0 ————–(2)

By cross multiplication we get

^{p}/_{4-20} = ^{q}/_{-60-(-20)} = ^{1}/_{-50-30}

^{p}/_{-16 }= ^{q}/_{-80} = ^{1}/_{-80}

^{p}/_{-16 } = ^{1}/_{-80} and ^{q}/_{-80} = ^{1}/_{-80}

p = ^{1}/_{5} and q = 1

^{1}/_{x+y }= p = ^{1}/_{5} and ^{1}/_{x-y} = q = 1

x + y = 5 ———-(3)

x – y = 1 ————-(4)

add (3) and (4)

2x = 6

x = 3 ————(5)

Substitute the value of x in (3)

3 + y = 5

y = 2

Thus x = 3 and y = 2

(8) ^{1}/_{3x+y} + ^{1}/_{3x-y} = ^{3}/_{4}

^{1}/_{2(3x+y)} – ^{1}/_{2(3x-y)} = ^{-1}/_{8}

Put ^{ 1}/_{3x+y} = p and ^{1}/_{3x-y} = q

p + q = ^{3}/_{4} ————–(1)

^{p}/_{2} – ^{q}/_{2 }= ^{-1}/_{8} ————-(2)

Adding (1) and (2)

2p = ^{3}/_{4} – ^{1}/_{4}

2p = ^{1}/_{2}

p = ^{1}/_{4}

Substituting the value of p in (2)

^{1}/_{4} – q = –^{1}/_{4}

q = ^{1}/_{4} + ^{1}/_{4} = ^{1}/_{2}

^{1}/_{3x+y} = p = ^{1}/_{4}

3x + y = 4 ————-(3)

^{1}/_{3x-y} = q = ^{1}/_{2}

3x – y = 2 —————(4)

add (3) and (4)

6x = 6

x = 1 ———–(5)

Substituting in (3)

3(1) + y = 4

y = 1

Hence x = 1 , y = 1

**Formulate the following problems as a pair of equations, and hence find their solutions:**

**(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. **

**(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. **

**(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.**

Solution:

(i)Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing Upstream = x – y km/h

Downstream = x + y km/h

According to question,

2(x+y) = 20

x + y = 10 ———–(1)

2(x-y) = 4

x – y = 2————(2)

Adding equation (1) and (2),

2x = 12

x = 6

Putting this in equation (1), we obtain y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii)Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in 1 day = ^{1}/_{x}

Work done by a man in 1 day = ^{1}/_{y}

According to the question,

4(^{2}/_{x} + ^{5}/_{y}) = 1

^{2}/_{x} + ^{5}/_{y} = ^{1}/_{4}

3(^{3}/_{x} + ^{6}/_{y}) = 1

^{3}/_{x} + ^{6}/_{y} = ^{1}/_{3}

Put ^{1}/_{x} = p and ^{1}/_{y} = q

2p + 5q = ^{1}/_{4} i.e., 8p + 20q = 1

3p + 6q = ^{1}/_{3} i.e., 9p + 18q = 1

by cross multiplication

^{p}/_{-20-(-18)} = ^{q}/_{-9-(-8)} = ^{1}/_{-144-180}

^{p}/_{-2 }= ^{q}/_{-1} = ^{1}/_{-36}

^{p}/_{-2 } = ^{1}/_{-36} and ^{q}/_{-1} = ^{1}/_{-36}

p = ^{1}/_{18} and q = ^{1}/_{36}

^{1}/_{x }= p = ^{1}/_{18} and ^{1}/_{y} = q = ^{1}/_{36}

x = 18 and y = 35

Hence, number of days taken by a woman = 18

Number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

^{60}/_{u }+ ^{240}/_{v} = 4 ——–(1)

^{100}/_{u }+ ^{200}/_{v} = ^{25}/_{6} ——–(2)

Put ^{1}/_{u} = p and ^{1}/_{v} = q

60 p + 240q = 4 ———–(3)

100p + 200q = ^{25}/_{6} i.e., 600p + 2400q = 40 ————(4)

Multiply equation (3) by 10

600p + 2400q = 40 ———–(5)

Subtracting equation (4) from (5)

1200q = 15

q = ^{15}/_{1200} = ^{1}/_{80} ————–(6)

Substitute in equation (3)

60p + 3 = 4

60p = 1

p = ^{1}/_{60}

p = ^{1}/_{u} = ^{1}/_{60 } and q = ^{1}/_{u} = ^{1}/_{80 }

u = 60 km/h and v = 8km/h

Hence, speed of train = 60 km/h

Speed of bus = 80 km/h

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## 1 thought on “Linear equations in two variables – Exercise 3.6 – Class 10”

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