# Pair of Linear Equations in Two Variables : Exercise - 3.6 (Mathematics NCERT Class 10th) Like the video?  Subscribe Now  and get such videos daily!

Q.1       Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) ${1 \over {2x}} + {1 \over {3y}} = 2$

${1 \over {3x}} + {1 \over {2y}} = {{13} \over 6}$
(ii) ${2 \over {\sqrt x }} + {3 \over {\sqrt y }} = 2$
${4 \over {\sqrt x }} - {9 \over {\sqrt y }} = - 1$
(iii) ${4 \over x} + 3y = 14$
${3 \over x} - 4y = 23$
(iv) ${5 \over {x - 1}} + {1 \over {y - 2}} = 2$

${6 \over {x - 1}} - {3 \over {y - 2}} = 1$
(v) ${{7x - 2y} \over {xy}} = 5$
${{8x + 7y} \over {xy}} = 15$
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
(vii) ${{10} \over {x + y}} + {2 \over {x - y}} = 4$
${{15} \over {x + y}} - {5 \over {x - y}} = - 2$
(viii) ${1 \over {3x + y}} + {1 \over {3x - y}} = {3 \over 4}$
${1 \over {2\left( {3x + y} \right)}} - {1 \over {2\left( {3x - y} \right)}} = {{ - 1} \over 8}$

Sol.       (i) Taking ${1 \over x} = u$ and ${1 \over y} = v$. The given system of equations become
${1 \over 2}u + {1 \over 3}v = 2$
$\Rightarrow$ 3u + 2v = 12 ...(1)

${1 \over 3}u + {1 \over 2}v = {{13} \over 6}$
$\Rightarrow$ 2u + 3v = 13...(2)

Multiplying (1) by 3 and (2), we have
9u + 6v = 36 ...(3)
and, 4u + 6v = 26 ...(4)
Subtracting (4) from (3), we get
5u = 10
$\Rightarrow$ u = 2
Putting u = 2 in (3), we get
18 + 6v = 36
$\Rightarrow$ 6v = 18
$\Rightarrow$ v = 3
Now, u = 2
$\Rightarrow$ ${1 \over x} = 2$
$\Rightarrow$$x = {1 \over 2}$
and, v = 3
$\Rightarrow$ ${1 \over y} = 3$
$\Rightarrow$ $y = {1 \over 3}$
Hence, the solution is x = ${1 \over 2}$, $y = {1 \over 3}$.

(ii) The given system of equations is
${2 \over {\sqrt x }} + {3 \over {\sqrt y }} = 2$ and ${4 \over {\sqrt x }} - {9 \over {\sqrt y }} = - 1$
Putting $u = {1 \over {\sqrt x }}$ and $v = {1 \over {\sqrt y }}$ .Then , the given equations become
2u + 3v = 2 ...(1)
and, 4u – 9v = – 1 ...(2)
Multiplying (1) by 3, we get
6u + 9v = 6...(3)
Adding (2) and (3), we get
10u = 5
$\Rightarrow$ $u = {5 \over {10}} = {1 \over 2}$
Putting $u = {1 \over 2}$ in (1), we get
$2 \times {1 \over 2} + 3v = 2$
$\Rightarrow$ 3v = 1
$\Rightarrow$ $v = {1 \over 3}$
Now, $u = {1 \over 2}$
$\Rightarrow$ ${1 \over {\sqrt x }} = {1 \over 2}$
$\Rightarrow$ ${\sqrt x }$ = 2
$\Rightarrow$ x = 4

and, $v = {1 \over 3}$
$\Rightarrow$ ${1 \over {\sqrt y }} = {1 \over 3}$
$\Rightarrow$ ${\sqrt y }$ = 3
$\Rightarrow$ y = 9
Hence, the solution is x = 4, y = 9.

(iii) The given system of equations is
${4 \over x} + 3y = 14$...(1)
and, ${3 \over x} - 4y = 23$...(2)
Multiplying (1) by 4 and (2) by 3, we get
${{16} \over x} + 12y = 56$...(3)
and, ${9 \over x} - 12y = 69$...(4)
Adding (3) and (4), we get
${{25} \over x} = 125$
$\Rightarrow$ $x = {{25} \over {125}} = {1 \over 5}$
Putting $x = {1 \over 5}$ in (1), we get
$4 \times 5 + 3y = 14$
$\Rightarrow$ 3y = 14 – 20
$\Rightarrow$ 3y = – 6
$\Rightarrow$ y = – 2
Hence, the solution is $x = {1 \over 5}$, y = – 2.

(iv) Let $u = {1 \over {x - 1}},v = {1 \over {y - 2}}$. Then, the given system of equations becomes
5u + v = 2 ...(1)
and, 6u – 3v = 1 ...(2)
Multiplying (1) by 3, we get
15u + 3v = 6 ...(3)
Adding (2) and (3), we get
21u = 7
$\Rightarrow$$u = {1 \over 3}$
Putting $u = {1 \over 3}$ in (1), we get
${5 \over 3} + v = 2$
$\Rightarrow$ $v = 2 - {5 \over 3} = {{6 - 5} \over 3} = {1 \over 3}$
Now, $u = {1 \over 3}$
$\Rightarrow$ ${1 \over {x - 1}} = {1 \over 3}$
$\Rightarrow$ x – 1 = 3
$\Rightarrow$ x = 4
$v = {1 \over 3}$
$\Rightarrow$ ${1 \over {y - 2}} = {1 \over 3}$
$\Rightarrow$ y – 2 = 3
$\Rightarrow$ y = 5
Hence, the solution is x = 4, y = 5.

(v) The given system of equations is
${{7x - 2y} \over {xy}} = 5$
$\Rightarrow$ ${7 \over y} - {2 \over x} = 5$
and, ${{8x + 7y} \over {xy}} = 15$
$\Rightarrow$ ${8 \over y} + {7 \over x} = 15$
Let u = ${1 \over x}$, v = ${1 \over y}$. Then, the above equations become
7 – 2u = 5 ...(1)
and, 8v + 7u = 15 ...(2)
Multiplying (1) by 7 and (2), we get
49v – 14u = 35 ...(3)
and, 16v + 14u = 30
Adding (3) and (4), we get
65v = 65
$\Rightarrow$ v = 1
Putting v = 1 in (1), we get
7 – 2u = 5
$\Rightarrow$ – 2u = – 2
$\Rightarrow$ u = 1
Now, u = 1
$\Rightarrow$ ${1 \over x} = 1$
$\Rightarrow$ x = 1
and, v = 1
$\Rightarrow$ ${1 \over y} = 1$
$\Rightarrow$ y = 1
Hence, the solution is x = 1, y = 1.

(vi) The given system of equations is 6x + 3y = 6xy and 2x + 4y = 5xy, where x and y are non -zero.
Since $x \ne 0,y \ne 0,$ , we have xy $\ne$ 0.
On dividing each one of the given equations by xy we get
${3 \over x} + {6 \over y} = 6$ and ${4 \over x} + {2 \over y} = 5$
Taking ${1 \over x} = u$ and ${1 \over y} = v$, the above equations become
3u + 6v = 6...(1)
and, 4u + 2v = 5...(2)
Multiplying (2) by 3, we get
12u + 6v = 15 ...(3)
Subtracting (1) from (3), we get
9u = 15 – 6 = 9
$\Rightarrow$ u = 1
Putting u = 1 in (1), we get
3×1 + 6v = 6
$\Rightarrow$ 6v = 6 – 3 = 3
$\Rightarrow$ $v = {1 \over 2}$
Now, u = 1
$\Rightarrow$ ${1 \over x} = 1$
$\Rightarrow$ x = 1
and, $v = {1 \over 2}$
$\Rightarrow$ ${1 \over y} = {1 \over 2}$
$\Rightarrow$ y = 2
Hence, the given system of equations has one solution x = 1, y = 2.

(vii) The given system of equations is
${{10} \over {x + y}} + {2 \over {x - y}} = 4$
${{15} \over {x + y}} - {5 \over {x - y}} = - 2$
Putting $u = {1 \over {x + y}}$ and $v = {1 \over {x - y}}$. Then, the given equations become
10u +2v = 4
$\Rightarrow$ 5u + v = 2 ...(1)
and, 15u – 5v = – 2...(2)
Multiplying (1) by 5, we get
25u + 5v = 10 ...(3)
Adding (2) and (3), we get
40u = 8
$\Rightarrow$ $u = {1 \over 5}$
Putting $u = {1 \over 5}$ in (1), we get
$5\left( {{1 \over 5}} \right) + v = 2$
$\Rightarrow$ v = 2 – 1 = 1
Now, $u = {1 \over 5}$
$\Rightarrow$ x + y = 5 ...(4)
v = 1
$\Rightarrow$ ${1 \over {x - y}} = 1$
$\Rightarrow$ x – y = 1
Adding (4) and (5), we get
2x = 6
$\Rightarrow$ x = 3
When x = 3, then from (4), we get
3 + y = 5
$\Rightarrow$ y = 5 – 3 = 2
Hence, the given system of equations has one solution
x = 3, y = 2

(viii) Taking $u = {1 \over {3x + y}}$ and $v = {1 \over {3x - y}}$. The give system of equations becomes
u + v $= {3 \over 4}$...(1)
${1 \over 2}u - {1 \over 2}v = {{ - 1} \over 8}$
$\Rightarrow$ u – v = ${{ - 1} \over 4}$...(2)
Adding (1) and (2), we get
2u = ${3 \over 4} - {1 \over 4}$
$\Rightarrow$ $2u = {2 \over 4} = {1 \over 2}$
$\Rightarrow$ $u = {1 \over 4}$
Putting $u = {1 \over 4}$ in (1), we get
${1 \over 4} + v = {3 \over 4}$
$\Rightarrow$ $v = {3 \over 4} - {1 \over 4} = {2 \over 4} = {1 \over 2}$
Now, $u = {1 \over 4}$
$\Rightarrow$ ${1 \over {3x + y}} = {1 \over 4}$
$\Rightarrow$ 3x + y = 4...(3)
and, $v = {1 \over 2}$
$\Rightarrow$ 3x – y = 2...(4)
$\Rightarrow$${1 \over {3x - y}} = {1 \over 2}$
Adding (3) and (4), we get
Putting x = 1 in (3), we get
3x + y = 4
$\Rightarrow$ y = 4 – 3 = 1
Hence, the solution is x = 1, y = 1.

Q.2      Formulate the following problems as a pair if 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 be 1 man alone.
(iii) Roohi travels 300 km to her home partly be train and partly be bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes linger. find the speed of the train and the bus separately.
Sol.       (i) Let the speed of the boat in still water be x km/hr. and the speed of the stream by y km/hr. Then,
Speed upstream = (x – y) km/hr
Speed downstream = (x + y) km/hr
Time taken to cover 20 km downstream = 2 hours
$\Rightarrow$ ${{20} \over {x + y}} = 2$
$\Rightarrow$ x +y = 10...(1)
Time taken to cover 4 km upstream = 2 hours
$\Rightarrow$ ${4 \over {x - y}} = 2$
$\Rightarrow$ x – y = 2...(2)
Adding (1) and (2), we get
2x = 12
$\Rightarrow$ x = 6
Putting x = 6 in (1), we get
6 + y = 10
$\Rightarrow$ y = 10 – 6 = 4
Hence, the speed of boat in still water = 6km/hr
and, speed of stream = 4 km/hr

(ii) Let 1 woman can finish the embroidery in x days and 1 man can finish the embroidery in y days.
Then,
1 woman's 1 day's work = ${1 \over x}$
1 man's 1 day's work = ${1 \over y}$
Therefore, ${2 \over x} + {5 \over y} = {1 \over 4}$ and ${3 \over x} + {6 \over y} = {1 \over 3}$
Putting ${1 \over x} = u$ and ${1 \over y} = v$, these equations become
2u + 5v = ${1 \over 4}$...(1)
and, 3u + 6v = ${1 \over 3}$...(2)
Multiplying (1) by 3 and (2) by 2 and subtracting, we get
$3v = {1 \over {12}}$
$\Rightarrow$ $v = {1 \over {36}}$
Putting $v = {1 \over {36}}$ in (1), we get
$2u = \left( {{1 \over 4} - {5 \over {36}}} \right) = {{9 - 5} \over {36}}$
$= {4 \over {36}} = {1 \over 9}$
$\Rightarrow$$u = {1 \over {18}}$
Now, $u = {1 \over {18}}$
$\Rightarrow$${1 \over x} = {1 \over {18}}$
$\Rightarrow$ x = 18
and,
$v = {1 \over {36}}$
$\Rightarrow$ ${1 \over y} = {1 \over {36}}$
$\Rightarrow$ y = 36
Thus, I woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.

(iii) Let the speed of train be c km/hr and speed of bus be y km/hr.
Then, ${{60} \over x} + {{240} \over y} = 4$ ...(1)
and, ${{100} \over x} + {{200} \over y} = 4 + {{10} \over {60}} = {{25} \over 6}$ ...(2)
Let ${1 \over x} = u$ and ${1 \over y} = v$, than eqns. (1) and (2) becomes
60u +240v = 4 ...(3)
100u + 200v = ${{25} \over 6}$ ... (4)
Multiplying eqn. (3) by 5 and eqn. (4) by 6, we get $\Rightarrow$ $u = {{ - 5} \over { - 300}} = {1 \over {60}}$
From eqn. (3)
60 × ${1 \over {60}}$ + 240 × v = 4

$\Rightarrow$ $v = {3 \over {240}} = {1 \over {80}}$
Since, x = ${1 \over u} = 60$ and y = ${1 \over v} = 80$
Therefore, Speed of train is 60 km/h and speed of bus is 80 km/h.