Question

In the following system of equations, determine whether the system has a unique solution, no solution or infinitely many solution. In case if there is a unique solution, find it.
$2x+3y=7$
$6x+5y=11$

• A. $x=\frac{1}{4},y=\frac{5}{2}$
• B. $x=-\frac{1}{4},y=-\frac{5}{2}$
• C. $x=-\frac{1}{4},y=\frac{5}{2}$
• D. None of these

Answer 1

The correct option is C $x=-\frac{1}{4},y=\frac{5}{2}$

The given system of equations may be written as
$2x+3y-7=0$
$6x+5y-11=0$

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, $a_1=2,b_1=3,c_1=-7$ and $a_2=6,b_2=5,c_2=-11$

We have, $\frac{a_1}{a_2}=\frac{2}{6}=\frac{1}{3}$ and $\frac{b_1}{b_2}=\frac{3}{5}$

Clearly, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

So, the given system of equations has a unique solution.
To find the solution, we use the cross-multiplication method.

By cross-multiplication, we have
$\frac{x}{(3\times -11)-(5\times -7)}=\frac{-y}{(2\times -11)-(6\times -7)}=\frac{1}{(2\times 5)-(6\times 3)}$

$\Rightarrow \frac{x}{-33+35}=\frac{-y}{-22+42}=\frac{1}{10-18}$

$\Rightarrow \frac{x}{2}=\frac{y}{-20}=\frac{1}{-8}$

$\Rightarrow x=\frac{-2}{8}=-\frac{1}{4}$ and $y=\frac{-20}{-8}=\frac{5}{2}$

Hence, the given system of equations has a unique solution given by $x=-\frac{1}{4},y=\frac{5}{2}$.