The pair of equations $2 x+3 y=5$ and $4 x+6 y=15$ has (a) a unique solution (b) exactly two solutions (c) infinitely many solutions (d) no solution

Home » The pair of equations and has (a) a unique solution (b) exactly two solutions (c) infinitely many solutions (d) no solution

The pair of equations $2 x+3 y=5$ and $4 x+6 y=15$ has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

The pair of equations $2 x+3 y=5$ and $4 x+6 y=15$ has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

Answer: (d) no solution
Here, $a_{1}=3, b_{1}=2 k, c_{1}=-2, a_{2}=2, b_{2}=5$ and $c_{2}=1$
$\therefore \frac{a_{1}}{a_{4}}=\frac{3}{2}, \frac{b_{1}}{b_{L}}=\frac{2 k}{5}$ and $\frac{c_{1}}{c_{L}}=\frac{-2}{1}$
For parallel lines, we have:
$\frac{a_{1}}{a_{3}}=\frac{b_{1}}{b_{7}} \neq \frac{c_{1}}{c}$
Sol:
The given system of equations can be written as:
$2 x+3 y-5=0$ and $4 x+6 y-15=0$
The given equations are of the following form:
$a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$
Here, $a_{1}=2, \mathrm{~b}_{1}=3, \mathrm{c}_{1}=-5, \mathrm{a}_{2}=4, \mathrm{~b}_{2}=6$ and $\mathrm{c}_{2}=-15$
$\therefore \frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-15}=\frac{1}{3}$
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{z}}$
As a result, the given system has no solution.