Class 10 Mathematics

Linear Equations

Exercise 3.6 (NCERT) 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)`

Solution: 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`

linear equations exercise solution

Solution: 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`

Solution: 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`


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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 1)

Exercise:3.5 (part 2)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)