(i) Given equations are 8x - 3y = 5xy
and 6x - 5y = -2xy
The given equations are not linear in the variable x and y. These can be reduced into linear equations.
If we put x = 0 in either of the equations, we get y = 0. Hence, x = 0 and y = 0 is a solution of these equations.
To find other solution, dividing each of th given equations by xy.
So, `(8)/(y) - (3)/(x) = 5` .... (1)
and `(6)/(y) - (5)/(x) = -2` ....(2)
Let `(1)/(y)` = a and `(1)/(x)` = b then from equations (1) and (2), we get
`{:(8a - 3b = 5),(6a - 5b =-2):}}"linear equations "{:(....(3)),( ....(4)):}`
Multiplying equation (3) by 5 and (4) by 3, we get
40a - 15b = 25 ....(5)
18a - 15b = - 6 ....(6)
Subtracting equation (6) from (5), we get
22a = 31
implies `a = (31)/(22)`
Putting `a = (31)/(22)` in equation (3), we get
`8 xx (31)/(22) - 3b = 5`
`(124)/(11) - 3b = 5 implies -3b = 5 - (124)/(11)`
implies `-3b = - (69)/(11) implies b = (23)/(11)`
Now, `(1)/(y) = a implies (1)/(y) = (31)/(22) implies y = (22)/(31)`
and `(1)/(x) = b implies (1)/(x) = (23)/(11) implies x = (11)/(23)`
Hence, the solution of the given two equations are `{:(x = 0),(y = 0):}}` and `{:(x = (11)/(23)),(y = (22)/(31)):}}`.
So, the given system of equations has two solutions.
(ii) We have,
2x + 3y = 5xy .... (1)
3x - 2y = xy .... (2)
Dividing equations (1) and (2) by xy, we get
`(2x)/(xy) + (3y)/(xy) = (5xy)/(xy) implies (2)/(y) + (3)/(x) = 5 " "...(3)`
and `(3x)/(xy) - (2y)/(xy) = (xy)/(xy) implies (3)/(y) - (2)/(x) = 1 " "....(4)`
Let `(1)/(x) = u` and `(1)/(y) = v`.
`therefore` Equations (3) and (4) become,
2v + 3u = 5 ....(5)
3v - 2u = 1 ....(6)
Now this is a pair of linear equations,multiplying equation (5) by 2 and equation (6) by 3, we get
4v + 6u = 10 ....(7)
9v - 6u = 3 ....(8)
On adding equations (7) and (8), we get
13v = 13 implies v = 1
Putting v = 1 in equation (5), we get
2(1) + 3u = 5 implies 3u = 3 implies u = 1
Now, u = 1 implies `(1)/(x) = 1` implies x = 1
and v = 1 implies `(1)/(y) = 1` implies y = 1
Hence, required solution is `{:(x = 1),(y = 1):}}`.
