Question

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

• A. a unique solution
• B. infinitely many solutions
• C. no solution
• D. 2 solutions

Answer 1

The correct option is A

a unique solution

Given equations: $4x-3y+12=0$ and $2x+3y-15=0$

On comparing with the $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$a_1=4,b_1=-3,c_1=12$
$a_2=2,b_2=3,c_2=-15$

Here,

$\frac{a_1}{a_2}=\frac{4}{2}=2$ and $\frac{b_1}{b_2}=\frac{-3}{3}=-1$

$\Rightarrow \frac{a_1}{a_2}$ ≠ $\frac{b_1}{b_2}$

We know that when a pair of linear equations in two variables $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, have

$\Rightarrow \frac{a_1}{a_2}$ ≠ $\frac{b_1}{b_2}$

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

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