SOLUTIONS:
1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) ${1/(2x)+1/(3y) =2}$
${1/(3x)+1/(2y) =13/6}$
Solution
Let $1/x = u$ and $1/y=v$
Hence above equation becomes,
$1/2u+1/3v=2$ ....................(1)
and $1/3u+1/2v=13/6$ ..................(2)
now get rid of the fraction part by multiply eqn (1) and eqn (2) by 6.
Therefore , eqn (1) becomes, $3u+2v=12$ ...........(3)
and eqn (2) becomes , $2u+3v=13$ ................(4)
So, eqn (3) and (4) are the required linear equations and solve them using elimination method.
$3u+2v=12$ ×2 (multiplying by 2)
$2u+3v=13$ ×3 (multiplying by 3)
and subracting them.
Therefore $(6u+4v)-(6u+9v)=24-39$
$(-5v)=-15$
$ v=3$
and $2u+3v=13$ =>$2u+3×3=13$ => $2u=4$
$u=2$
Ans:So, $x=1/2$ and $y=1/3$
(ii) ${4/x +3y=14}$
${3/x-4y=23}$
Solution
Let $1/x = u$
Therefore the equations becomes
${4/x +3y=14}$ => ${4u +3y=14}$ ...........(1)
${3/x-4y=23}$ => ${3u-4y=23}$ .............(2)
Multiply eqn(1) by 4 and eqn (2) by 3 to eliminate $y$
${4u +3y=14}$ => ${16u +12y=56}$ ...........(3)
${3u-4y=23}$ => ${9u-12y=69}$ ..............(4)
Therefore adding eqn (3) and (4)
$(16u +12y) + (9u-12y) = 56+69$
$25u = 125$
$u = 125/25= 5$=> $x=1/5$
${4u +3y=14}$ => ${4×5 + 3y=14 }$
=> ${3y=14-20}$ => $y=-2$
(iii) ${2/√x+3/√y =2}$
${4/√x-9/√y= -1}$
Solution
Let $1/√x = u$ and $1/√y=v$
Therefore the eqn becomes;
${2u+3v =2}$ ...................(1)
${4u-9v= -1}$ .................(2)
Multiply eqn (1) by 3, => ${2u+3v =2}$× 3
$6u+9v=6$ ......................(3)
Solving eqn (2) and (3)
$(6u+9v)+(4u-9v)=6+(-1)$
$10u=5$ => $u={1/2}$
Since, $1/√x = u$, squaring both the side, => $1/x = u^2$
$1/x = (1/2)^2$
$x=4$
Substituting the value of $u$ in eqn (1) to get the value of $v$
${2u+3v =2}$ => ${2×(1/2)+3v =2}$
$v= 1/3$
Since, $1/√y=v$ ; squaring both the sides; => $1/y=v^2$
$1/y=(1/3)^2$
$y=9$
Ans: $x=4$ and $y=9$
(iv) ${5/(x-1) +1/(y-2) = 2}$
${6/(x-1) - 3/(y-2) = 1}$
Solution
Let $1/{x-1} = u $ and $1/{y-2}=v$
Therefore the eqn becomes;
${5u +v = 2}$ ...............(1)
${6u - 3v = 1}$ .............(2)
Multiply eqn(1) by 3, so ${5u +v = 2}$ ×3
=> ${15u +3v = 6}$ ........(3)
Therefore; $(15u +3v)+(6u - 3v )=(6+1)$
$21u=7$
$u=1/3$
Since $1/{x-1} = u $ , i.e $1/{x-1} = 1/3 $
i.e $x=4$
${5u +v = 2}$ =>${5×(1/3) +v = 2}$
$v=2-{5/3}$
$v=1/3$
$1/{y-2}=v$ =>$1/{y-2}=1/3$
$y-2=3$ => $y=5$
Ans: $x=4$ ;$y=5$
(v) ${(7x-2y)/(xy) = 5}$
${(8x+7y)/(xy)=15}$
Solution
Convert the above equation in the form of $1/x$ and $1/y$.
Therefore; ${(7x-2y)/(xy) = 5}$ => ${7x}/{xy} - {2y}/{xy} = 5$ => $7/y- 2/x=5$
and ${(8x+7y)/(xy)=15}$ => ${8x}/{xy} + {7y}/{xy} = 15 $ => $8/y+7/x=15$
Let $1/x=u$ and $1/y=v$
Therefore $7v-2u=5$ ............(1)
and $8v+7u=15$ ..............(2)
$7v-2u=5$ × 7 => $49v-14u=35$ ............(3)
$8v+7u=15$ × 2 => $16v+14u=30$ .............(4)
Therefore: $65v=65$ => $v=1$
Since,$1/y=v$ , therefore $y=1$
$7v-2u=5$ => $7× 1-2u=5$ => $2u=2$ => $u=1$
Since, $1/x=u$ => $x=1$
Ans:$x=1$ ; $y=1$
(vi)${6x + 3y = 6xy}$
${2x + 4y = 5xy}$
Solution
Divide the above eqn by $xy$
Therefore, above eqn becomes.
${6/y + 3/x = 6}$ .............(1)
${2/y + 4/x = 5}$ .............(2)
Let $1/x=u$ and $1/y=v$
Therefore $6v+3u=6$ ............(3)
$2v+4u=5$ ............(4)
multiply eqn (4) with 3 => $2v+4u=5$ × 3 => $6v+12u=15$ ..........(5)
Subaracting eqn (5) with (3), we get,
=> $(6v+12u)-(6v+3u) = (15-6)$
=> $9u=9$
=> $u=1$
Since $1/x=u$ ,
$x=1$
putting value of $u$ in eqn (3),
$6v+3u=6$ => $6v+3=6$
$v=1/2$
Since, $1/y=v$
$y=2$
Ans: $x=1$; $y=2$
(vii) ${10/(x+y)+2/(x-y)=4}$
${15/(x+y)-5/(x-y)= -2}$
Solution
Let $1/{x+y}=u$
and $1/{x-y}=v$
Therefore eqn becomes
$10u+2v=4$ ..........(1)
$15u-5v=-2$ ............(2)
Multiplying eqn (1) with 3 and eqn(2) with 2.
$30u+6v=12$
$30u-10v= - 4$
Subracting the above eqn's
=> $(30u+6v)-(30u-10v) = (12)-(-4)$
=> $16v=16$
$v=1$
Putting the value of $v$ in eqn (1)
$10u+2v=4$ => $10u+2×1 =4$
$10u=2$
$u=1/5$
Since , $1/{x+y}=u$ =>$1/{x+y}=1/5$ => $x+y=5$ ............(3)
and $1/{x-y}=v$ => $1/{x-y}=1$ => $x-y=1$ ...............(4)
Solving eqn (3) and (4)
we get, $2x=6$ => $x=3$
and $y=2$
Ans: $x=3$ , $y=2$
(viii) ${1/(3x+y)+1/(3x-y) = 3/4}$
${1/(2(3x+y))-1/(2(3x-y)) = -1/8}$
Solution
Let $1/{3x+y} =u$ and $1/{3x-y} = v$
Therefore $u+v=3/4$ ............(1)
and $u/2 -v/2= -1/8$ ...........(2)
Multiplying eqn (2) with 2, we get
$u/2 -v/2= -1/8$ ×2 => $u -v= -1/4$ ........(3)
Solving eqn (1) and (3), we get
$2u= {3/4-1/4}$
$2u=2/4$
$u=1/4$
putting value of $u$ in eqn (1), we get
$1/4+v=3/4$ => $v=3/4-1/4$
$v=1/2$
Since, $1/{3x+y} =u$ i.e $1/{3x+y} =1/4$ => $3x+y=4$ .........(4)
and $1/{3x-y} = v$ i.e $1/{3x-y} = 1/2$ => $3x-y=2$ ...........(5)
Solving eqn (4) and (5)
$6x=6$ => $x=1$
and $3x+y=4$ => $y=1$
Ans: $x=1$ ; $y=1$
2.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.
Solution
Let the speed of rowing in still water be $x$ km/h
Let the speed of current be $y$ km/h
Relative downstream speed= $x+y$ km/h
Relative upstream speed= $x-y$ km/h
Therefore $S=D/T$ => $x+y$ = $20/2$ i.e $10$ ...........(1)
and $x-y$ = $4/2$ = $2$ ..............................(2)
Solving eqn (1) and (2)
$2x=12$ => $x=6$ km/h
and $y=10-x$ => $y=4$ km/h
Ans: speed of rowing in still water = 6 km/h and speed of current = 4 km/h
(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.
Solution .
Let $n$ be the no. of days taken by 1 woman to finish the work,
and $m$ be the no. of days taken by 1 man to finish the work.
Therefore, in 1 day $1/n$th of the work is finished by 1 woman
and, in 1 day $1/m$th of the work is finished by 1 man
Since given:2 women and 5 men can together finish an embroidery work in 4 days
Therefore in one day, amount of work done is,
2 women will do $2/n$th of the work in 1 days and
5 men will do $5/m$th of the work in 1 day.
Therefore in 1 day, they will do ${2/n}+{5/m}$th of the work.
Since they complete the job in 4 days, therefore in 1 day, $1/4$th of the work is finished.
Therefore, ${2/n}+{5/m}$ = $1/4$ ...............(1)
Similiarly, 3 women and 6 men can finish it in 3 days;
i.e $3/n + 6/m = 1/3$ ...............(2)
Let $1/n = x$ and $1/m=y$
Therefore eqn (1) and (2) becomes
$2x+5y=1/4$ ..............(3) multiplying by 3 => $6x+15y=3/4$
$3x+6y=1/3$ .................(4) multiplying by 2 => $6x+12y=2/3$
Therefore $3y= (3/4-2/3)$ = $1/12$
$y=1/36$
Since $1/m=y$ =>$1/m=1/36$ =>$m=36$ days
and $2x+5y=1/4$ => $2x=1/4 -5(1/36)$ =$(1/9)$
$x=1/18$
Since $1/n = x$ => $1/n = 1/18$ => $n=18$ days.
(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 the 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 separate.
Solution
Let the speed of the train be $x$ km/h
Let the speed of the bus be $y$ km/h
Since Time =$D/S$.
Total time taken= $4=60/x + 240/y$ ............(1)
and $25/6$ = $100/x$ +$ 200/y$ ....................(2)
Let $1/x=u$ and $1/y=v$
Therefore;
$60u+240v=4$ ...................(3)
$100u+200v=25/6$ ...............(4)
$u=(4-240v)/60$ => $(1/15-4v)$..............(5)
Subsituting $u$ in eqn (4)
$100u+200v=25/6$=> $100(1/15-4v)+200v=25/6$
$100/15 -400v + 200v =25/6$
$200v=(100/15-25/6)$
$200v={(100×6 )-(15×25)}/{15×6}$ = ${(600)-(375)}/{15 × 6}$
$v=225/{15×6×200}$ = $1/80$
=> $y=80$
$u= (1/15-4×{1/80})$ =>$(1/15-1/20)$
$u=1/60$ => $x=60$
Ans: Hence, speed of the train is $60$ km/h and speed of the bus is $80$ km/h