Maths M.C.Q (1 Marks)

Given:
$x=-y \text { and } y>0$
Now, we have:
$x^2 y$
On substituting $x=-y$, we get:
$(-y)^2 y=y^3>0(\therefore y>0)$
This is true.
$x+y$
On substituting $x=-y$, we get:
$(-y)+y=0$
This is also true.
$xy$
On substituting $x=-y$, we get:
$(-y) y=-y^2<0(\therefore y>0)$
This is again true.
$\frac{1}{x}-\frac{1}{y}=0$
$\Rightarrow \frac{y-x}{x y}=0$
On substituting $x=-y$, we get:
$\frac{y-(-y)}{(-y) y}=0$
$\Rightarrow \frac{2 y}{-y^2}=0$
$\Rightarrow 2 y=0$
$\Rightarrow y=0$
Hence, from the above equation, we get $y=0$, which is wrong.

$3x + 2ky = 2$ and $2x + 5y + 1 = 0$
$3x + 2ky - 2 = 0$ and $2x + 5y + 1 = 0$
If the lines are parallel, it means the system has an infinite number of solutions.
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}.$
So, $\frac{3}{2}=\frac{2\text{k}}{5}=\frac{-2}{1}$
$\Rightarrow\frac{3}{2}=\frac{2\text{k}}{5}$
$\Rightarrow\text{k}=\frac{15}{4}$

The given system equations can be written as follows:
$2 x-3 y-7=0 \text { and }(a+b) x-(a+b-3) y-(4 a+b)=0$
The given equations are of the following form:
$a_1 x+b_1 y+c_1=0 \text { and } a_2 x+b_2 y+c_2=0$
Here, $a_1=2, b_1=-3 c_1=-7$ and $a_2=(a+b), b_2=-(a+b-3)$ and $c_2=-(4 a+b)$
$\therefore \frac{a_1}{a_2}=\frac{2}{(a+b)}, \frac{b_1}{b_2}=\frac{-3}{-(a+b-3)}=\frac{3}{-(a+b-3)} \text { and } \frac{c_1}{c_2}=\frac{-7}{-(4 a+b)}=\frac{7}{(4 a+b)}$
For an infinite number of solution, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore \frac{3}{(a+b)}=\frac{7}{(a+b-3)}=\frac{7}{(4 a+b)}$
Now, we have:
$\frac{2}{(a+b)}=\frac{3}{(a+b-3)}$
$\Rightarrow a-6=3 a+3 b$
$\Rightarrow a+b+6=0 \ldots$
Again, we have:
$\frac{3}{(a+b-3)}=\frac{7}{(4 a+b)}$
$\Rightarrow 12 a+3 b=7 a+7 b-21$
$\Rightarrow 5 a-7 b+21=0 \ldots \text { (ii) }$
On multiplying (i) by 4 , we get:
$4 a+4 b+24=0$
On substituting $a=-5$ in, we get:
$\Rightarrow-5+b+6=0$
$\Rightarrow b=-1$
$\therefore a=-5 \text { and } b=-1$

$6x - 2y + 9 = 0$ and $3x - y + 12 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{3}=2$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-1}=2$
$\frac{\text{c}_1}{\text{c}_2}=\frac{9}{12}=\frac{3}{4}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
We know that,
If in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
We have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2},$ then the given system has no solution.
So, the lines are parallel.

$5x - 15y = 8$ and $3\text{x}-9\text{y}=\frac{24}{5}$
$5x - 15y - 8 = 0$ and $15x - 45y - 24 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{5}{15}=\frac{1}{3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-15}{-45}=\frac{1}{3}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-24}=\frac{1}{3}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{b}_1}{\text{b}_2}.$
We know that,
If in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
We have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{b}_1}{\text{b}_2}.$ then the system has a unique solution.
So, the pair of lines are coincident.

$2x + 3y = 5$ and $4x + 6y = 15$
$\Rightarrow 2x + 3y - 5 = 0$ and $4x + 6y - 15 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-5}{-15}=\frac{1}{3}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
We know that,
The system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, the pair of equations has no solution.

$2x + 3y + 9 = 0$ and $x - 2y + 8 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{1}=2$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{-12}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{9}{-8}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
We know that,
If in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
We have $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ then the system has a unique solution.
So, the pair of lines are intersecting exactly at one point.

A system of equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$ is said to be inconsistent if it has no solution.
If a pair of linear equation are parallel,
It has no solutions.
If a pair of linear equation are coincident,
If has infinite number of solutions.
If a pair of linear equations are intersecting,
It has a unique solution.
So, the pair of linear will be parallel.

$x-2 y=3 \text { and } 3 x+k y=1$
We know that, the system of linear equations $a_1 x+b_1 x+c_1=0, a_2 x+b_2 y+c_2=0$
has a unique solution if $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$\text { So, } \frac{1}{3} \neq \frac{-2}{k}$
$\Rightarrow k \neq-6$

$kx - y = 2$ and $6x - 2y = 3$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$
has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{\text{k}}{6}\neq\frac{-1}{-2}$
$\Rightarrow\text{k}\neq3.$

A system of equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$ is said to be consistent if it has at least one solution.
If a pair of linear equations are parallel,
It has no solutions.
If a pair of linear equations are coincident,
It has infinite number of solutions.
If a pair of linear equation are intersecting,
It has a unique solution.
So, the pair of linear can be intersecting or coincident.

$kx - 2y = 3$ and $3x + y = 5$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$
has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{\text{k}}{3}=\frac{-2}{1}$
$\Rightarrow\text{k}\neq-6.$
Thus, $k$ can take any real values except $-6.$