Question

# Solve the following pair of equations. 5x−1+1y−2=2 6x−1−3y−2=1 Here, x ≠ 1 and y ≠ 2 x = 4, y = 5 x = 5, y = 3 x=13, y=13 x = 7, y = 11

Solution

## The correct option is A x = 4, y = 5  If we substitute   1x−1 as p and  1y−2 as q in the given equations, we get the equations as 5p+q=2   .......(1) 6q−3q=1   .......(2) Now, we can solve the pair of equations by method of elimination. On multiplying the first equation by 3 and then adding it to (2), we get 15p+3q=6   6p−3q=1 _____________         p=13 Now by substituting the value of p in equation (2) we get q=13 Now, p=1x−1 ⇒1x−1=13 ⇒x=4 Similarly we assumed q=1y−2 ⇒1y−2=13 ⇒y−2=3 ⇒y=5 ∴x=4 and y=5

Suggest corrections