Question

In the following system of equation determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it.
$-3x+4y=5$

$\frac{9}{2}x-6y+\frac{15}{2}=0$.

Answer 1

The given system of equations may be written as
$-3x+4y-5=0$

$\frac{9}{2}x-6y+\frac{15}{2}=0$

The given system of equations is of the form
$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where, $a_1=-3,b_1=4,c_1=-5$ and $a_2=\frac{9}{2},b_2=-6,c_2=\frac{15}{2}$

We have, $\frac{a_1}{a_2}=\frac{-3}{9/2}=\frac{-2}{3},\frac{b_1}{b_2}=\frac{4}{-6}=\frac{-2}{3}$ and $\frac{c_1}{c_2}=\frac{-5}{15/2}=\frac{-2}{3}$

Clearly, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, the given system of equations has infinitely many solutions.