Solve the following pairs of equations
(i) `x + y = 3.3, " " (0.6)/(3x - 2y) = -1, 3x - 2y != 0`
(ii) ` (x)/(3) + (y)/(4) = 4, " " (5x)/(6) - (y)/(8) = 4`
(iii) `4x + (6)/(y) = 15, " " 6x - (8)/(y) = 14, y != 0`
(iv) `(1)/(2x) - (1)/(y) = -1, " " (1)/(x) + (1)/(2y) = 8, x, y != 0`
(v) `43x + 67y = -24, " " 67x + 43y = 24`
(vi) `(x)/(a) + (y)/(b) = a + b , " " (x)/(a^(2)) + (y)/(b^(2)) = 2, a, b != 0`
(vii) `(2xy)/(x+y) = (3)/(2), " " (xy)/(2x-y) = (-3)/(10), x + y != 0, 2x - y != 0`

Explanation

(i) Given pair of linear equations are is
`" " x+y=3.3 " " ...(i)`
and `{:(ul(" "0.6" ")),(3x-2y):}=-1`
`rArr 0.6=-3x+2y`
`rArr " " 3x-2y=-0.6 " " ...(ii)`
Now, multiplying Eq. (i) by 2 and then adding with Eq. (ii), we get
`rArr " " 2x+2y=6.6`
`rArr " " 3x-2y=-0.6`
`" " 5x=6 rArr x=(6)/(5)=1.2`
Now, put the value of x in Eq. (i) , we get
`" " 1.2+y=3.3`
`rArr " " y=3.3 -1.2`
`rArr " " y=2.1`
Hence, the required values of x and y are 1.2 and 2.1, respectively.
(ii) Given, pair of linear equations is
`" " (x)/(3)+(y)/(4)=4`
On multiplying both sides by LCM (3, 4)=12, we get
4x+3y=48 `" " ...(i)`
and `(5x)/(6)-(y)/(8)=4`
On multiplying both sides by LCM (6, 8)=24, we get
`20x-3y=96 " " ...(ii)`
Now, adding Eqs. (i) and (ii), we get
`" " 24x=144`
`rArr " " x=6`
Now, put the value of x in Eq. (i) , we get
`4xx6+3y=48`
`rArr " " 3y=48-24`
`rArr " " 3y=24 rArr y=8`
Hence, the required values of x and y are 6 and 8, respectively.
(iii) Given, pair of linear equations are
` 4x+(6)/(y)=15 " "...(i)`
and `" " 6x-(8)/(y)=14, y != 0 " " ...(ii)`
Let `u=(1)/(y)`, then above equation becomes
`4x+6u=15 " " ... (iii)`
and `" " 6x-8u=14 " " ...(iv)`
On multiplying Eq. (iii) by 8 and Eq. (iv) by 6 and then adding both of them, we get
`32x+48u=120`
`36x-48u=84 rArr 68x=204`
`rArr " " x=3`
Now, put the value of x in Eq. (iii), we get
`" " 4xx3+ 6u=15`
`rArr " " 6u=15-12 rArr 6u=3`
`rArr " " u=(1)/(2)rArr(1)/(y)=(1)/(2) " " `[`:'u=(1)/(y)`]
`rArr " " y=2`
Hence, the required values of x and y are 3 and 2, respectively.
(iv) Given pair of linear equations is
`(1)/(2x)-(1)/(y)=-1 " " ...(i)`
and `" " (1)/(x)+(1)/(2y)=8,x,y!=0 " " ...(ii)`
Let ` u=(1)/(x)` and `v=(1)/(y)`, then the above equations becomes
`(u)/(2)-v=-1`
`rArr " " u-2v=-2 " " ...(iii)`
and `" " u+(v)/(2)=8`
`rArr " " 2u+v=16 " " ...(iv)`
On, multiplying Eq. (iv) by 2 and then adding with Eq. (iii), we get
`{:(4u+2v=32),(ul(u-2v=-2)),(" " 5u=30):}`
`rArr " " u=6`
Now, put the value of u in Eq. (iv), we get
`2xx6+v=16`
`rArr " " v=16-12=4`
`rArr " " v=4`
`:. " " x=(1)/(u)=(1)/(6)` and `y=(1)/(v)=(1)/(4)`
Hence, the required values of x and y are `(1)/(6)` and `(1)/(4)`, respectively.
(v) Given pair of linear equations is
`43x+67y=-24 " " ...(i)`
and `" " 67x+43y=24 " " ...(ii)`
On multiplying Eq. (i) by 43 and Eq. (ii) by 67 and then subtracting both of them, we get
`{:((67)^(2)x+43xx67y=24xx67),(ul(underset(-)((43))^(2)x+underset(-)43xx67y=-underset(+)24xx43)),({(67)^(2)-(43)^(2)}x=24(67+43)):}`
`rArr " " (67+43)(67-43)x=24xx110 " " `[`:' (a^(2)-b^(2))=(a-b)(a+b)`]

`rArr " " 110xx24x=24xx110`
`rArr " " x=1`
Now, put the value of x in Eq. (i), we get
`43xx1+67y=-24`
`rArr " " 67y=-24-43`
`rArr " " 67y=-67`
`rArr " " y=-1`
Hence, the required values of x and y and 1 and -1, respectively.
(vi) Given pair of linear equations is
`(x)/(a)+(y)/(b)=a+b " " ...(i)`
and `" " (x)/(a^(2))+(y)/(b^(2))=2, a, b!=0 " " ...(ii)`
On multiplying Eq. (i) by `(1)/(2)` and then subtracting from Er. (ii) , we get
`{:((x)/(a^(2))+(y)/(b^(2))=2),(ul((x)/underset(-)(a^(2))+(y)/underset(-)(ab)=underset(-)1+(b)/(a))),(y((1)/(b^(2))-(1)/(ab))=2-1-(b)/(a)):}`
`rArr " " y((a-b)/(ab^(2)))=1-(b)/(a)=((a-b)/(a))`
`rArr " " y=(ab^(2))/(a)rArry=b^(2)`
Now, put the value of y in Eq. (ii), we get
`(x)/(a^(2))+(b^(2))/(b^(2))=2`
`rArr " " (x)/(a^(2))=2-1=1`
`rArr " " x=a^(2)`
Hence, the required values ofx and y are `a^(2) ` and `b^(2)`, respectively.
(vii) Given pair of equations is
`(2xy)/(x+y)=(3)/(2)`, where `x+y !=0`
`rArr " " (x+y)/(2xy)=(2)/(3)`
`rArr " " (x)/(xy)+(y)/(xy)=(4)/(3)`
`rArr " " (1)/(y)+(1)/(x)=(4)/(3) " " ...(i)`
and `" " (xy)/(2x-y)=(-3)/(10)`, where `2x-y!=0`
`rArr " " (2x-y)/(xy)=(-10)/(3)`
`rArr " " (2x)/(xy)-(y)/(xy)=(-10)/(3)`
`rArr " " (2)/(y)-(1)/(x)=(-10)/(3)`
Now, put `(1)/(x)=u` and `(1)/(y)=v`, then the pair of equations becomes
`v+u=(4)/(3) " " ...(iii)`
and `" " 2v-u=(-10)/(3) " " ...(iv)`
On adding both equations, we get
`3v=(4)/(3)-(10)/(3)=(-6)/(3)`
`rArr " " 3v=-2`
`rArr" "v=(-2)/(3)`
Now, put the value of v in Eq. (iii), we get
`(-2)/(3)+u=(4)/(3)`
`rArr " " u=(4)/(3)+(2)/(3)=(6)/(3)=2`
`rArr " " x=(1)/(u)=(1)/(2)`
and `" " y=(1)/(v)=(1)/((-2//3))=(-3)/(2)`
Hence, the required values of x and `(1)/(2)`and `(-3)/(2)`, respectively.