Answer: We have
5/x + 6y = 13 and
3/x + 4y = 7
Lets simplify these equations. Assuming 1/x = z, we can rewrite them,
5/x + 6y = 13
⇒ 5z + 6y = 13 …(i)
3/x + 4y = 7
⇒ 3z + 4y = 7 …(ii)
To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.
Lets multiply eq.(i) by 3 and eq.(ii) by 5, so that variable z in both the equations have same coefficient.
Recalling equations (i) & (ii),
5z + 6y = 13 [×3
3z + 4y = 7 [×5
⇒ - 2y = 4
⇒ y = - 2
Substitute y = - 2 in eq.(i)/eq.(ii), as per convenience of solving.
Thus, substituting in eq.(ii), we get
3z + 4( - 2) = 7
⇒ 3z – 8 = 7
⇒ 3z = 7 + 8
⇒ 3z = 15
⇒ z = 5
Thus, z = 5 and y = - 2
As z = 1/x,
⇒ 5 = 1/x
⇒ x = 1/5
Hence, we have x = 1/5 and y = - 2
