1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Solve: 5x−1+1y−2=2 6x−1−3y−2=1 (where x≠1,y≠2) The value of x and y are.

A
x=4, y=5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x=5, y=4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x=8, y=5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A x=4, y=5If we substitute 1x−1 as p and 1y−2 as q in the given equations. (As x≠1,y≠2) We get the equations as 5p+q=2 6p−3q=1 Now, we can solve the pair of equation by method of elimination. 15p+3q=6 6p−3q=1 On adding both the above equations, we get p=13 Now by substituting the value in one of the equation we find q=13 As we have assumed p=1x−1 Therefore, p=1x−1=13 x=4 SImilarly we assumed q=1y−2 Hence 1y−2=13 y−2=3 Thus x=4 and y=5

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Solving Linear Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program