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Solution :

(i) Given pair of linear equations are is <br> `" " x+y=3.3 " " ...(i)` <br> and `{:(ul(" "0.6" ")),(3x-2y):}=-1` <br> `rArr 0.6=-3x+2y` <br> `rArr " " 3x-2y=-0.6 " " ...(ii)` <br> Now, multiplying Eq. (i) by 2 and then adding with Eq. (ii), we get <br> `rArr " " 2x+2y=6.6` <br> `rArr " " 3x-2y=-0.6` <br> `" " 5x=6 rArr x=(6)/(5)=1.2` <br> Now, put the value of x in Eq. (i) , we get <br> `" " 1.2+y=3.3` <br> `rArr " " y=3.3 -1.2` <br> `rArr " " y=2.1` <br> Hence, the required values of x and y are 1.2 and 2.1, respectively. <br> (ii) Given, pair of linear equations is <br> `" " (x)/(3)+(y)/(4)=4` <br> On multiplying both sides by LCM (3, 4)=12, we get <br> 4x+3y=48 `" " ...(i)` <br> and `(5x)/(6)-(y)/(8)=4` <br> On multiplying both sides by LCM (6, 8)=24, we get <br> `20x-3y=96 " " ...(ii)` <br> Now, adding Eqs. (i) and (ii), we get <br> `" " 24x=144` <br> `rArr " " x=6` <br> Now, put the value of x in Eq. (i) , we get <br> `4xx6+3y=48` <br> `rArr " " 3y=48-24` <br> `rArr " " 3y=24 rArr y=8` <br> Hence, the required values of x and y are 6 and 8, respectively. <br> (iii) Given, pair of linear equations are <br> ` 4x+(6)/(y)=15 " "...(i)` <br> and `" " 6x-(8)/(y)=14, y != 0 " " ...(ii)` <br> Let `u=(1)/(y)`, then above equation becomes <br> `4x+6u=15 " " ... (iii)` <br> and `" " 6x-8u=14 " " ...(iv)` <br> On multiplying Eq. (iii) by 8 and Eq. (iv) by 6 and then adding both of them, we get <br> `32x+48u=120` <br> `36x-48u=84 rArr 68x=204` <br> `rArr " " x=3` <br> Now, put the value of x in Eq. (iii), we get <br> `" " 4xx3+ 6u=15` <br> `rArr " " 6u=15-12 rArr 6u=3` <br> `rArr " " u=(1)/(2)rArr(1)/(y)=(1)/(2) " " `[`:'u=(1)/(y)`] <br> `rArr " " y=2` <br> Hence, the required values of x and y are 3 and 2, respectively. <br> (iv) Given pair of linear equations is <br> `(1)/(2x)-(1)/(y)=-1 " " ...(i)` <br> and `" " (1)/(x)+(1)/(2y)=8,x,y!=0 " " ...(ii)` <br> Let ` u=(1)/(x)` and `v=(1)/(y)`, then the above equations becomes <br> `(u)/(2)-v=-1` <br> `rArr " " u-2v=-2 " " ...(iii)` <br> and `" " u+(v)/(2)=8` <br> `rArr " " 2u+v=16 " " ...(iv)` <br> On, multiplying Eq. (iv) by 2 and then adding with Eq. (iii), we get <br> `{:(4u+2v=32),(ul(u-2v=-2)),(" " 5u=30):}` <br> `rArr " " u=6` <br> Now, put the value of u in Eq. (iv), we get <br> `2xx6+v=16` <br> `rArr " " v=16-12=4` <br> `rArr " " v=4` <br> `:. " " x=(1)/(u)=(1)/(6)` and `y=(1)/(v)=(1)/(4)` <br> Hence, the required values of x and y are `(1)/(6)` and `(1)/(4)`, respectively. <br> (v) Given pair of linear equations is <br> `43x+67y=-24 " " ...(i)` <br> and `" " 67x+43y=24 " " ...(ii)` <br> On multiplying Eq. (i) by 43 and Eq. (ii) by 67 and then subtracting both of them, we get <br> `{:((67)^(2)x+43xx67y=24xx67),(ul(underset(-)((43))^(2)x+underset(-)43xx67y=-underset(+)24xx43)),({(67)^(2)-(43)^(2)}x=24(67+43)):}` <br> `rArr " " (67+43)(67-43)x=24xx110 " " `[`:' (a^(2)-b^(2))=(a-b)(a+b)`] <br> <br> `rArr " " 110xx24x=24xx110` <br> `rArr " " x=1` <br> Now, put the value of x in Eq. (i), we get <br> `43xx1+67y=-24` <br> `rArr " " 67y=-24-43` <br> `rArr " " 67y=-67`<br> `rArr " " y=-1` <br> Hence, the required values of x and y and 1 and -1, respectively. <br> (vi) Given pair of linear equations is <br> `(x)/(a)+(y)/(b)=a+b " " ...(i)` <br> and `" " (x)/(a^(2))+(y)/(b^(2))=2, a, b!=0 " " ...(ii)` <br> On multiplying Eq. (i) by `(1)/(2)` and then subtracting from Er. (ii) , we get <br> `{:((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)):}` <br> `rArr " " y((a-b)/(ab^(2)))=1-(b)/(a)=((a-b)/(a))` <br> `rArr " " y=(ab^(2))/(a)rArry=b^(2)` <br> Now, put the value of y in Eq. (ii), we get <br> `(x)/(a^(2))+(b^(2))/(b^(2))=2` <br> `rArr " " (x)/(a^(2))=2-1=1` <br> `rArr " " x=a^(2)` <br> Hence, the required values ofx and y are `a^(2) ` and `b^(2)`, respectively. <br> (vii) Given pair of equations is <br> `(2xy)/(x+y)=(3)/(2)`, where `x+y !=0` <br> `rArr " " (x+y)/(2xy)=(2)/(3)` <br> `rArr " " (x)/(xy)+(y)/(xy)=(4)/(3)` <br> `rArr " " (1)/(y)+(1)/(x)=(4)/(3) " " ...(i)` <br> and `" " (xy)/(2x-y)=(-3)/(10)`, where `2x-y!=0` <br> `rArr " " (2x-y)/(xy)=(-10)/(3)` <br> `rArr " " (2x)/(xy)-(y)/(xy)=(-10)/(3)` <br> `rArr " " (2)/(y)-(1)/(x)=(-10)/(3)` <br> Now, put `(1)/(x)=u` and `(1)/(y)=v`, then the pair of equations becomes <br> `v+u=(4)/(3) " " ...(iii)` <br> and `" " 2v-u=(-10)/(3) " " ...(iv)` <br> On adding both equations, we get <br> `3v=(4)/(3)-(10)/(3)=(-6)/(3)` <br> `rArr " " 3v=-2` <br> `rArr" "v=(-2)/(3)` <br> Now, put the value of v in Eq. (iii), we get <br> `(-2)/(3)+u=(4)/(3)` <br> `rArr " " u=(4)/(3)+(2)/(3)=(6)/(3)=2` <br> `rArr " " x=(1)/(u)=(1)/(2)` <br> and `" " y=(1)/(v)=(1)/((-2//3))=(-3)/(2)` <br> Hence, the required values of x and `(1)/(2)`and `(-3)/(2)`, respectively.