Question

A pair of linear equations which has a unique solution $x=2,y=-3$ is

• A. x + y = -1 ; 2x - 3y = -5
• B. 2x - y = 1 ; 3x + 2y = 0
• C. 2x + 5y = -11 ; 4x + 10y = -22
• D. x - 4y -14 = 0 ; 5x - y - 13 = 0

Answer 1

Answer: (D)

We know that, if pair of linear equations $a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$ have the unique solution then $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Now lets check the options,

A. $2x-y=1;3x+2y=0$

$\frac{a_1}{a_2}=\frac{2}{3}$

$\frac{b_1}{b_2}=\frac{-1}{2}$

Since, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, which has unique solution.

Now, lets substitute $x=2,y=-3$ in both equations.

$2x-y=1$

$\Rightarrow$ LHS $=2\left(2\right)-3=4-3=1=$ RHS

3x + 2y = 0

$\Rightarrow$ LHS $=3\left(2\right)+2(-3)=6-6=0$ RHS

Hence, $x=2,y=-3$ is an unique solution of $2x-y=1;3x+2y=0$.

B. $2x+5y=-11;4x+10y=-22$

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{5}{10}=\frac{1}{2}$

Since, $\frac{a_1}{a_2}=\frac{b_1}{b_2}$, which has no unique solution.

C. $x-4y-14=0;5x-y-13=0$

$\frac{a_1}{a_2}=\frac{1}{5}$

$\frac{b_1}{b_2}=\frac{-4}{-1}=4$

Since, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, which has unique solution.

Now, lets substitute $x=2,y=-3$ in both equations.

$x-4y-14=0$

LHS $=2-4(-3)-14=2+12-14=0=$ RHS

$5x-y-13=0$

LHS $=5\left(2\right)-(-3)-13=10+3-13=13-13=0=$ RHS

Hence, $x=2,y=-3$ is an unique solution of $x-4y-14=0;5x-y-13=0$.

D. $x+y=-1;2x-3y=-5$

$\frac{a_1}{a_2}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{1}{-3}$

Since, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, which has unique solution.

Now, lets substitute $x=2,y=-3$ in both equations.

$x+y=-1$

LHS $=2-3=-1=$ RHS

$2x-3y=-5$

LHS $=2\left(2\right)-3(-3)=4+9=13\ne$ RHS

Hence, $x=2,y=-3$ is not an unique solution of $x+y=-1;2x-3y=-5$.

Hence, Option A, Option C are correct.