# Linear Equations

## Exercise 3.6 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