Pair of Linear Equations in Two Variables
Mathematics Class Tenth
NCERT Exrercise 3.6
Question: 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:
Given, `1/(2x) + 1/(3y) = 2` ---------- (i)
`1/(3x) + 1/(2y) = 13/6` ----------- (ii)
Let, `1/x=p` and `1/y=q`
Now, after substituting the values of `1/x` and `1/y` from equation (i)
`1/(2x) + 1/(3y) = 2`
`=> 1/2p+1/3q=2`
`=>(3p+2q)/6=2`
After cross multiplication, we get
`3p+2q=12` ---------- (iii)
Now, after substituting the values of `1/x` and `1/y` from equation (ii)
`1/(3x) + 1/(2y) = 13/6`
`=>1/3p+1/2q=13/6`
`=>(2p+3p)/6=13/6`
After cross multiplication, we get
`2p+3q = (13xx6)/6`
`=>2p+3q=13` ---------- (iv)
`=>2p = 13-3q`
`=>p = (13-3q)/2` ------- (v)
Now, after substituting the value of `p` in equation (iii), we get
`3xx(13-3q)/2+2q=12`
`=>((39-9q)+4q)/2=12`
After cross multiplication, we get
`=>39-9q+4q=12xx2=24`
`=>-5q=24-39=-15`
`=>q=(-15)/(-5)=3`
Now, by substituting the value of `q` in equation (v), we get
`p=(13-3xx3)/2`
`=>p=(13-9)/2=4/2`
`=>p=2`
Thus, here `p=2` and `q=3`
Here `p=2`
Thus, by putting the value of `p=1/x`
`1/x=2`
After cross multiplication, we get
`1=2x`
`=>x = 1/2`
And, since, `q=3`
Thus, by substituting the value of `q=1/y`
`1/y=3`
After cross multiplication, we get
`=>1=3y`
`=>y = 1/3`
Thus, `x=1/2` and `y=1/3` Answer
(ii) `2/sqrtx +3/sqrty=2`;
`4/sqrtx-9/sqrty=-1`
Given, `2/sqrtx +3/sqrty=2`;
`4/sqrtx-9/sqrty=-1`
Here, let `1/sqrtx = p` and 1/sqrty=q`
Thus, `2/sqrtx +3/sqrty=2`
`=>2p+3q=2` ---------- (i)
And, `4/sqrtx-9/sqrty=-1`
`=>4p-9q=-1` --------- (ii)
Now, after multiplying equation (i) by 2, we get
`2(2p+3q=2)`
`=>4p+6q=4` ---------- (iii)
Now, by subtracting equation (iii) from equation (ii) we get
`(4p-9q)-(4p+6q)=-1-4`
`=>\cancel(4p)-9q-\cancel(4p)-6q=-5`
`=>-15q=-5`
`=>q = (-5)/(-15)=1/3`
And by substituting value of `q` in equation, (ii), we get
`4p-9xx1/3=-1`
`=>4p-3=-1`
`=>4p=-1+3=2`
`=>p=2/4`
`=>p=1/2`
Now, by substituting `p=1/sqrtx`, we get
`1/sqrtx = 1/2`
After cross multiplication, we get
`sqrtx=2`
By squaring both sides, we get
`(sqrtx)^2=2^2`
`=>x = 4`
Now, since `q=1/3`
Thus, by substituting the value of `q=1/sqrty`
`1/sqrty = 1/3`
After cross multiplication, we get
`sqrty = 3`
After squaring both sides, we get
`(sqrty)^2=3^2`
`=>y = 9`
Thus, `x=4` and `y=9` Answer
(iii) `4/x+3y=14`
`3/x-4y=23`
Solution:
Given, `4/x+3y=14`
`3/x-4y=23`
Let, `1/x = p`.
Thus, by substituting `p=1/x` given, equations can be written as,
`4p+3y=14` ----------(i)
And, `3p-4y=23` ----------- (ii)
Now, from equation (i)
`4p=14-3y`
`=>p = (14-3y)/4`
Now, by substituting value of `p` in equation (ii), we get
`3xx(14-3y)/4-4y=23`
`=>((42-9y)-16y)/4=23`
After cross multiplication, we get
`42-9y-16y=23xx4=92`
`=>42-25y=92`
`=>-25y=92-42=50`
`=>y=50/(-25)=-2`
After substituting the value of `y` in equation (i), we get
`4p+3y=14` ----------(i)
`=>4p+3(-2)=14`
`=>4p-6=14`
`=>4p=14+6=20`
`=>p=20/4=5`
Now, by substituting value of `p=1/x`, we get
`=>1/x = 5`
After cross multiplication, we get
`5x=1`
`=>x=1/5`
Thus, `x=1/5` and `y=-2` Answer
(iv) `5/(x-1)+1/(y-2)=2`
`6/(x-1)-3/(y-2)=1`
Solution:
Given, `5/(x-1)+1/(y-2)=2` ---------- (i)
`6/(x-1)-3/(y-2)=1` ---------- (ii)
Let, substitute `1/(x-1)=p` and `1/(y-2) =q`
Therefore, from equation (i), we get
`5p+q=2` ---------- (iii)
And from equation (ii), we get
`6p-3q=1` ---------- (iv)
From equation (iii) `5p+q=2`
`=>q=2-5p`
Now, substituting the `q=2-5p` in equation (iv), we get
`6p - 3(2-5p) = 1`
`=>6p-6+15p=1`
`=>21p-6=1`
`=>21p = 1+6=7`
`=> p = 7/21=1/3`
Now, by substituting value of `p` in equation (iii), we get
`5p+q=2` ----(iii)
`=>5xx1/3+q=2`
`=>5/3+q=2`
`=>q=2-5/3`
`=>q=(6-5)3=1/3`
Now, since, `p=1/3`
Therefore, after substituting `p=1/(x-1)`, we get
`1/(x-1)=1/3`
After cross multiplication, we get
`=>3=x-1`
`=>x=3+1=4`
And, since, `q=1/3`
Thus, after substituting `q=1/(y-2)` we get
`1/(y-2)=1/3`
After cross multiplication, we get
`3=y-2`
`=>y=3+2=5`
Thus, `x=4` and `y=5` Answer
(v) `(7x-2y)/(xy) = 5`;
`(8x+7y)(xy)=15`
Solution:
Given, `(7x-2y)/(xy) = 5`;
`=>(7x)/xy-(2y)/(xy) = 5`
`=>7/y-2x/x =5` ---------- (i)
And given, `(8x+7y)(xy)=15`
`=>(8x)/(xy)+(7y)/(xy) = 15`
`=>8/y+7/x=15` -----------(ii)
Let, `1/x=p` and `1/y=q`
Thus, after substituting the values of `1/x` and `1/y`
From equation (i) we get,
`7q-2p=5` ------- (iii)
From equatin (ii), we get
`8q+7p=15` -------- (iv)
`=> 8q=15-7p`
`=>q=(15-7p)/8` ---------- (v)
Now, by substituting this value of `q` from equation (v) in equation (iii), we get
`7xx(15-7p)/8-2p =5`
`=>(105-49p-16p)/8=5`
After cross multiplication, we get
`105-49p-16p=8xx5`
`=>105-65p=40`
`=>105-40=65p`
`=>65=65p`
`=>p=65/65=1`
Now, by substituting the value of `p=1` in equation (v), we get
`q=(15-7xx1)/8`
`=>q=(15-7)/8`
`=>q=8/8=1`
Now, by substituting `q=1/y` we get
`1/y=1`
`=>y=1`
And, since, `p=1`
Thus, by substituting `p=1/x`, we get
`1/x=1`
`=>x=1`
Thus, `x=1` and `y=1` Answer
(vi) `6x+3y = 6xy`
`2x+4y=5xy`
Solution:
Given, `6x+3y = 6xy`
`=>(6x)/(6xy)+(3y)/(6xy)=1`
`=>1/y+1/(2x)=1` ------------ (i)
And, given, `2x+4y=5xy`
`=>(2x)/(xy)+(4y)/(xy)=5`
`=>2/y+4/x=5` --------- (ii)
Now, let `1/x=p` and `1/y=q`
After substituting the values of `p=1/x` and `q=1/y` in equation (i) we get
`q+1/2p=1` ----------- (iii)
`=>q=1-p/2`
`=>q=(2-p)/2` --------- (iv)
After substituting the values of `p=1/x` and `q=1/y` in equation (ii) we get
`2q+4p=5` ---------(v)
After substituting the value of `q=(2-p)/2` in above equation (v), we get
`2xx(2-p)/2+4p=5`
`=>(4-2p+8p)/2=5`
After cross multiplication, we get
`4-2p+8p=5xx2`
`=>4+6p=10`
`=>6p=10-4=6`
`=>p=6/6=1`
Now, by substituting the value of `p=1` in equation (v), we get
`2q+4xx1=5`
`=>2q+4=5`
`=>2q=5-4=1`
`=>q=1/2`
Now, by substituting `q=1/y` , we get
`1/y=1/2`
`=>y=2`
And since, `p=1`
Thus, by substituting, `p=1/x`, we get
`1/x=1`
`=>x = 1`
Thus, `x=1` and `y=2` Answer
(vii) `10/(x+y) +2/(x-y)=4`
`15/(x+y)-5/(x-y)=-2`
Solution: Let, `1/(x+y)=p` and `1/(x-y)=q`
Given, `10/(x+y) +2/(x-y)=4`
After substituting the value of `1/(x+y)=p` and `1/(x-y)=q`, we get
`10p+2q=4` ---------- (i)
`=>10p=4-2q`
`=>p=(4-2q)/10` --------(ii)
And given, `15/(x+y)-5/(x-y)=-2`
After substituting the value of `1/(x+y)=p` and `1/(x-y)=q`, we get
`15p-5q=-2` ---------- (iii)
Now, by substituting the value of `p=(4-2q)/10` from equation (ii), we get
`15xx(4-2q)/10-5q=-2`
`=>(60-30q-50q)/10=-2`
`=>(60-80q)/10=-2`
After cross multiplication, we get
`60-80q=-2xx10=-20`
`=>-80q=-20-60=-80`
`=>q=(-80)/(-80)=1`
Now, by substituting the value of `q=1` in equation (i), we get
`10p+2q=4` ---------- (i)
`=>10p+2xx1=4`
`=>10p=4-2`
`=>10p=2`
`=>p=2/10=1/5`
Now, since `p=1/(x+y)`
`:. 1/(x+y)=1/5`
After cross multiplication, we get
`=>5=x+y`
`=>x+y=5` ------ (iv)
And, since `q=1` and `q=1/(x-y)`
Therefore, `1/(x-y)=1`
After cross multiplication, we get
`x-y=1` ---------- (v)
By adding equation (iv) and (v), we get
`x+y+x-y=5+1`
`=>2x=6`
`=>x = 6/2=3`
Now, by substituting `x=3` in equation (v), we get
`3-y=1`
`=>-y=1-3=-2`
`=>y=2`
Thus, `x=3` and `y=2` Answer
(viii) `1/(3x+y)+1/(3x-y)=3/4`
`1/(2(3x+y))-1/(2(3x-y))=(-1)/8`
Solution:
Let, `1/(3x+y)=p` and `1/(3x-y)=q`
Given, `1/(3x+y)+1/(3x-y)=3/4`
Thus putting the values of `1/(3x+y)=p` and `1/(3x-y)=q`
`=>p+q=3/4`
After cross multiplication, we get
`=>4p+4q=3`----------(i)
`=>4p=3-4q`
`=>p=(3-4q)/4` ----------- (ii)
And given, `1/(2(3x+y))-1/(2(3x-y))=(-1)/8`
After putting the values of `1/(3x+y)=p` and `1/(3x-y)=q`
`p/2-q/2=(-1)/8`
`=>(p-q)/2=(-1)/8`
After cross multiplication, we get
`p-q=(-2)/8=(-1)/4`
Again after cross multiplication, we get
`4p-4q=-1` --------- (iii)
Now, by putting the value of `p=(3-4q)/4`, we get
`4xx(3-4q)/4-4q=-1`
`=>(12-16q-16q)/4=-1`
`=>(12-32q)/4=-1`
After cross multiplication, we get
`12-32q=-4`
`=>32q=12+4=16`
`=>q=16/32=1/2`
Now, by substituting the value of `q=1/2` in equation (i)
`=>4p+4q=3`----------(i)
`=>4p+4xx1/2=3`
`=>4p+2=3`
`=>4p=3-2=1`
`=>p=1/4`
Now, by substituting `p=1/(3x+y)`, we get
`1/(3x+y)=1/4`
After cross multiplication, we get
`3x+y=4` --------- (v)
And, since, `q=1/2`
Therefore, by substituting `q=1/(3x-y)`, we get
`1/(3x-y)=1/2`
After cross multiplication, we get
`3x-y=2` ---------- (vi)
By adding equation (v) and (vi), we get
`(3x+y)+(3x-y)=4+2`
`=>3x+y+3x-y=6`
`=>6x=6`
`=>x=6/6=1`
Now, by substituting `x=1` in equation (vi), we get
`3xx1-y=2`
`=>3-y=2`
`=>y=3-2=1`
Thus, `x=1` and `y=1` Answer
Question: 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 current.
Solution:
Let the speed of Ritu `=x\ km//h`
And let the speed of stream `=y\ km//h`
Thus, Speed of Ritu while rowing upstream `=x-y\ km//h`
And Speed of Ritu while rowing downstream `=x+y\ km//h`
Now, we know that, time × speed = distance
Thus, according to question, while rowing downstream
`2(x+y) = 20`
`=>x+y=20/10`
`=>x+y=10` ----------- (i)
And according to question, while rowing upstream
`2(x-y)=4`
`=>x-y=4/2`
`=>x-y=2` ------------- (ii)
Now, by adding equation (i) and (ii), we get
`(x+y)+(x-y)=10+2`
`=>x+y-x-y=12`
`=>2x=12`
`=>x=12/2=6`
Now by substituting value of `x=6` in equation (i), we get
`6+y=10`
`=>y=10-6=4`
Thus, speed of Ritu in still water `=6\ km//h` and speed of current `=4\ km//h` Answer
(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 women alone to finish the work, and also than taken by 1 man alone.
Solution:
Let the number of days taken by one women to complete one work `=x`
Therefore, work done by a women in 1 day `=1/x`
And let number of days taken by a man to complete one work `=y`
Thus, work done by man in 1 day = `1/y`
Given, in question,
Days taken by 2 women and 5 men = 4
Therefore,
`4(2/x+5/y)=1`
`=>2/x+5/y=1/4` ---------- (i)
And as given, in question,
Number of days taken by 3 women and 6 men = 3
Therefore,
`3(3/x+6/y)=1`
`=>3/x+6/y=1/3` --------- (ii)
Now, let `1/x=p` and `1/y=q`.
Thus, after putting the values of `1/x` and `1/y`, we get
`3p+6q=1/3`
After cross multiplication, we get
`3(3p+6q)=1`
`=>9p+18q=1` ---------- (iii)
`=>9p=1-18q`
`=>p=(1-18q)/9` -------- (iv)
After putting the values of `1/x` and `1/y`, we get from equation (i)
`=>2/x+5/y=1/4` ---------- (i)
`=>2p+5q=1/4`
After cross multiplication, we get
`4(2p+5q)=1`
`=>8p+20q=1` ------ (v)
After substituting the value of `p=(1-18q)/9` from equation (iv), we get
`8xx(1-18q)/9+20q=1`
`=>(8-144q+180q)/9=1`
`=>(8+36q)9=1`
`=>8+36q=9`
`=>36q=9-8=1`
`=>q=1/36` ---------- (vi)
Now, by substituting the value of `q=1/y`, we get
`1/y=1/36`
After cross multiplication, we get
`y=36`
Now, by substituting, the value of ` q=1/36` in equation (v)
`=>8p+20q=1` ------ (v)
`=>8p+20xx(1/36)=1`
`=>8p+20/36=1`
`=>8p+5/9=1`
`=>8p=1-5/9`
`=>8p=(9-5)/9`
`=>8p=4/9`
`=>p=4/(9xx8)`
`=>p=1/(9xx2)`
`=>p=1/18`
Now, by substituting the value of `p=1/x`, we get
`1/x=1/18`
After cross multiplication, we get
`x=18`
Thus, number of days taken by a women alone to finish the work = 18
And the number of days taken by a man alone to finish the work = 36
Thus, `18` days and `36` days Answer
(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 separately.
Solution:
Let, the speed of train `=t\ km//h`
And the speed of bus `=b\ km//h`
Given, total distance =300 km
Case : 1
Time taken = 4 hour
Distance covered by train = 60km
Thus, distance covered by bus = 300 km –60 km=240km
Now, we know that, time = distance / speed.
Thus, `4=60/t+240/b`
Let, `1/t=p` and `1/b=q`
Thus, by substituting, the values of `1/t` and `1/b`, we get
`60p+240q=4` ------------ (i)
Case: 2
Time taken = 10 minutes longer = 4 hours + 10/60 hours
`=4+1/6= (24+1)/6=25/6` hours
Distance travel by train = 100 km
Thus, distance travel by bus = 300 km – 100 km = 200 km
Now, we know that, time = distance / speed.
Thus, `25/6 = 100/t+200/b`
Now, by substituting `1/t=p` and `1/b=q`, we get
`100p+200q=25/6`
After cross multiplication, we get
`6(100p+200q)=25`
`=>600p+1200q=25` ------------- (ii)
Now, equation (i) × 10
`=>10(60p+240q=4)`
`=>600p+2400q=40` ------- (iii)
Now, after subtracting equation (ii) from equation (iii), we get
`(600p+2400q)-(600p+1200q)=40-25`
`=>\cancel(600p)+2400q-\cancel(600p)-1200q=15`
`=>1200q = 15`
`=>q=15/1200=1/80` ---------- (iv)
Now, by substituting the value of `q=1/b`, we get
`1/b=1/80`
`=>b=80`
Now, by substituting the value of `q=1/80` in equation (i) `60p+240q=4`, we get
`60p+240xx1/80=4`
`=>60p+3=4`
`=>60p=4-3=1`
`=>p=1/60`
Now, by substituting the value `p=1/t`, we get
`1/t=1/60`
After cross multiplication, we get
`t=60`
Here we have `t = 60` and `b = 60`.
Thus, speed of train = 60 km/h and speed of bus = 80 km/h Answer
Reference: