wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

The system of pair of equations 4x−3y+12=0 and 2x+3y−15=0 has


A

a unique solution

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

infinitely many solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

no solution

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

2 solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

a unique solution


Given equations: 4x3y+12=0 and 2x+3y15=0

On comparing with the a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1=4, b1=3, c1=12
a2=2, b2=3, c2=15

Here,

a1a2=42=2 and b1b2=33=1

a1a2b1b2

We know that when a pair of linear equations in two variables a1x+b1y+c1=0 and a2x+b2y+c2=0, have

a1a2b1b2

then the lines representing these equations will intersect and have a unique solution.

Therefore, the given pair of equations has a unique solution.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Graphical Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon