Class 10 Mathematics

Linear Equations

Exercise 3.6 (NCERT Book) Part 2

Solve the following pairs of equations by reducing them to a pair of linear equations:

(d) `(5)/(x-1)+(1)/(y-2)=2` and `(6)/(x-1)-(3)/(y-2)=1`

Solution: First equation:

`(5)/(x-1)+(1)/(y-2)=2`

Or, `(5(y-2)+x-1)/((x-1)(y-2))=2`

Or, `(5y-10+x-1)/(xy-2x-y+2)=2`

Or, `x+5y-11=2xy-4x-2y+4`
Or, `x+5y-11+4x+2y-4=2xy`
Or, `5x+7y-15=2xy`


Second equation:
`(6)/(x-1)-(3)/(y-2)=1`

Or, `(6y-12-3x+3)/(xy-2x-y+2)=1`

Or, `6y-3x-9=xy=2x-y+2`
Or, `6y-3x-9+2x+y-2=xy`
Or, `7y-x-11=xy`

Multiply the second equation by 2 and subtract first equation from resultant:

`14y-2x-22-(7y+5x-15)=2xy-2xy`
Or, `14y-2x-22-7y-5x+15=0`
Or, `7y-7x-7=0`
Or, `y-x-1=0`
Or, `y=x+1`

Substituting the value of x in first equation, we get;

`5x + 7y – 15 = 2xy`
Or, `5x + 7x + 7 – 15 – 2x(x + 1)`
Or, `12x – 8 = 2x^2 + 2x`
Or, `10x – 8 = 2x^2`

Or, `x^2 = 5x – 4`

Similarly, substituting the value of y in second equation, we get;

`7y – x – 11 = xy`
Or, `7x + 7 – x – 11 = x^2 + 1`

Or, `6x – 5 = x^2`

From above two equations, it is clear;

`5x – 4 = 6x – 5`
Or, `5x + 1 = 6x`
Or, `x = 1`
Hence, `y = x + 1 = 2`
Hence, `x = 1` and `y = 2`


(e) `(7x-2y)/(xy)=5` and `(8x+7y)/(xy)=15`

Solution: First equation:

`(7x-2y)/(xy)=5`

Or, `7x-2y=5xy`

Second equation:
`(8x+7y)/(xy)=15`

Or, `8x+7y=15xy`

Multiplying the first equation by 3 and subtracting the resultant from second equation, we get;

`8x + 7y – 21x + 6y = 15xy – 15xy`
Or, `- 3x + 13y = 0`
Or, `y – x = 0`

Or, `x = y`

Substituting the value of x in first equation, we get;

`7x – 2y = 5xy`
Or, `7y – 2y = 5y^2`
Or, `5y = 5y^2`
Or, `y = 1`
Hence, `x = 1` and `y = 1`

(f) `6x + 3y = 6xy` and `2x + 4y = 5xy`

Solution: Multiplying first equation by 5 and second equation by 6 and subtracting them, we get;

`30x + 15y – 12x – 24y = 30xy – 30xy`
Or, `18x – 9y = 0`
Or, `2x – y = 0`
Or, `2x = y`

Substituting the value of y in first equation, we get;

`6x + 3y = 6xy`
Or, `6x + 6x = 12x^2`
Or, `12x = 12x^2`
Or, `x = 1`

Substituting the value of x in second equation, we get;

`2 + 4y = 5y`
Or, `y = 2`
Hence, `x = 1` and `y = 2`


(g) `(1)/(x+y)+(2)/(x-y)=2` and `(15)/(x+y)-(5)/(x-y)=-2`

Solution: First equation:

`(1)/(x+y)+(2)/(x-y)=2`

Or, `(10x-10y+2x+2y)/((x+y)(x-y))=4`

Or, `12x-8y=4(x+y)(x-y)`
Or, `3x-2y=x^2-y^2`

Second equation:
`(15)/(x+y)-(5)/(x-y)=-2`

Or, `(15x-15y-5x-5y)/((x+y)(x-y))=-2`

Or, `10x-290y=-2(x^2-y^2)`
Or, `-5x+10y=x^2-y^2`

From first and second equations, it is clear;

`3x-2y=10y-5x`
Or, `8x=12y`
Or, `x=(3)/(2)y`

Substituting the value of x in first equation, we get;

`3x-2y=x^2-y^2`
Or, `3xx(3)/(2)y-2y=(9)/(4)y^2-y^2`

Or, `(9y-4y)/(2)=(9y^2-4y^2)/(4)`

Or, `5y=(5y^2)/(2)`

Or, `10y=5y^2`
Or, `2y=y^2`
Or, `y=2`
Hence, `x=2xx(3)/(2)=3`

(h) `(1)/(3x+y)+(1)/(3x-y)=3/4` and `(1)/(2(3x+y))-(1)/(2(3x-y))=-1/8`

Solution: First equation:

`(1)/(3x+y)+(1)/(3x-y)=3/4`

Or, `(3x-y+3x+y)/(9x^2-y^2)=3/4`

Or, `6x\xx4=3(9x^2-y^2)`
Or, `8x=9x^2=y^2`

Second equation:
`(1)/(2(3x+y))-(1)/(2(3x-y))=-1/8`

Or, `(1)/ (3x+y)-(1)/(3x-y)=-2/8=-1/4`

Or, `(3x-y-3x-y)/(9x^2-y^2)=-1/4`

Or, `-2yxx4=-(9x^2-y^2)`
Or, `8y=9x^2-y^2`

From first and second equation, it is clear;

`8x = 8y`
Or, `x = y`

Substituting the value of x in first equation, we get;

`8x = 9x^2 – y^2`
Or, `8y = 9y^2 – y^2 = 8y^2`
Or, `8y = 8y^2`
Or, `y = 1`

Substituting the value of y in second equation, we get;

`8 = 9x^2 – 1`
Or, `9 = 9x^2`
Or, `x = 1`
Hence, `x = 1` and `y = 1`


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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 1)

Exercise:3.6 (part 3)