[28.(1)/(5x)+(y)/(9)=5],[(1)/(3x)+(y)/(2)=14]

Solve the following pairs of linear equations (1)/(5x)+(y)/(9)=5 and (1)/(3x)+(y)/(2)=14

On the ellipse 4x^(2)+9y^(2)=1, the points at which the tangents are parallel to the line 8x=9y are ((2)/(5),(1)/(5))( b ) (-(2)/(5),(1)/(5))(-(2)/(5),-(1)/(5))( d) ((2)/(5),-(1)/(5))

The plane containing the two lines (x-3)/(1) = (y-2)/(4) = (z-1)/(5) and (x-2)/(1) = (y+3)/(-4) = (z+1)/(5) is 11x+my + nz = 28 where .............

(1-5x)(9x+1)=……

(x)/(x^(2)-5x+9)<=1

A point on the ellipse,4x^(2)+9y^(2)=36, where the normal is parallel to the line,4x-2y-5=0 is (A) ((8)/(5),-(9)/(5))(B)(-(9)/(5),(8)/(5))(C)((8)/(5),(9)/(5))(D)((9)/(5),(8)/(5))

If (2)/(1!9!)+(2)/(3!7!)+(1)/(5!5!)=(2)/(a!), then orthocentre of the triangle having sides x-y+1=0,x+y+3=0 and 2x+5y-2=0

28-31x-5x^(2)

Solve the following equations by Elimination method : 1/3x +y/9 =5 , 1/5x +y/2 =16 .

When x=5 and y=2 , the expression (xy)/(70)+(9)/(5(x+y))+(1)/(x+y)=?