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Question

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


A

a unique solution

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B

infinitely many solutions

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C

no solution

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D

2 solutions

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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.


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