Linear Equations
NCERT Exercise 3.6
Part 1
Question 1: Solve the following pairs of equations by reducing them to a pair of linear equations:
(a) `(1)/(2x)+(1)/(3y)=2` and `(1)/(3x)+(1)/(2y)=(13)/(6)`
Answer: First equation:
`(1)/(2x)+(1)/(3y)=2`
Or, `(3y+2x)/(6xy)=2`
Or, `2x+3y=12xy`
Second equation:
`(1)/(3x)+(1)/(2y)=(13)/(6)`
Or, `(2y+3x)/(6xy)=(13)/(6)`
Or, `6(2y+3x)=12xx6xy`
Or, `18x+12y=78xy`
Multiplying the first equation by 4 and subtracting the resultant from second equation, we get;
`18x + 12y – 8x – 12y = 78xy – 48xy`
Or, `10x = 30xy`
Or, `1 = 3y`
Or, `y = 1/3`
Substituting the value of x in first equation, we get;
`2x+3y=12xy`
Or, `2x+3xx1/3=2x\xx1/3`
Or, `2x+1=4x`
Or, `2x=1`
Or, `x=1/2`
Hence, `x=1/2` and `y=1/3`
(b) `(2)/(sqrtx)+(3)/(sqrty)=2` and `(4)/(sqrtx)-(9)/(sqrty)=-1`
Answer: First equation:
`(2)/(sqrtx)+(3)/(sqrty)=2`
Or, `(3sqrtx+2sqrty)/(sqrt(xy))=2`
Or, `3sqrtx+2sqrty=2sqrt(xy)`
Second equation:
`(4)/(sqrtx)-(9)/(sqrty)=-1`
Or, `(4sqrty-9sqrtx)/(sqrt(xy))=-1`
Or, `4sqrty-9sqrtx=-sqrt(xy)`
Or, `9sqrtx-4sqrty=sqrt(xy)`
From first and second equation, it is clear;
`3sqrtx+2sqrty=2(9sqrtx-4sqrty)`
Or, `3sqrtx+2sqrty=18sqrtx-8sqrty`
Or, `3sqrtx+2sqrty+8sqrty=18sqrtx`
Or, `10sqrty=15sqrtx`
Or, `sqrtx=(10)/(15)sqrty=(2)/(3)sqrty`
Substituting the value of √x in any of the above equation, we get;
`3sqrtx+2sqrty=2sqrt(xy)`
Or, `3xx(2)/(3)sqrty+2sqrty=2xx(2)/(3)sqrty\xx\sqrty`
Or, `2sqrty+2sqrty=(4)/(3)y`
Or, `4sqrty=(4)/(3)y`
Or, `sqrty=4÷(4)/(3)=3`
Substituting the value of √y in second equation, we get;
`9sqrtx-4sqrty=sqrt(xy)`
Or, `9sqrtx-4xx3=3sqrtx`
Or, `9sqrtx-3sqrtx=12`
Or, `6sqrtx=12`
Or, `sqrtx=2`
Hence, `x=4` and `y=9`
(c) `(4)/(x)+3y=14` and `(3)/(x)-4y=23`
Answer: First equation:
`(4)/(x)+3y=14`
Or, `(4+3xy)/(x)=14`
Or, `4+3xy=14x`
Or, `4+3xy-14x=0`
Second equation:
`(3)/(x)-4y=23`
Or, `(3-4xy)/(x)=23`
Or, `3-4xy=23x`
Or, `3-4xy-23x=0`
Multiply first equation by 4 and second equation by 3 and add the resultant equations;
`16 + 12xy – 56x + 9 - 12xy - 69x = 0`
Or, `25 – 125x = 0`
Or, `125x = 25`
Or, `x = 1/5`
Substituting the value of x in first equation, we get;
`4+3xy-14x=0`
Or, `4+3xx(1)/(5)y-14xx(1)/(5)=0`
Or, `(20+3y-14)/(5)=0`
Or, `6+3y=0`
Or, `3y=-6`
Or, `y=-(6)/(3)=-2`
Here, `x=1/5` and `y=-2`