Algebra

Linear Equation

Medium

For some constant real numbers $p, k$ and $a$ consider the following system of linear equations in $x$ and $y$ :

$px - 4y = 2$

$3x + ky = a$

A necessary condition for the system to have no solution for $(x, y)$ is

$ap - 6 = 0$

$2$

$2a + k\neq 0$

$kp + 12 \neq 0$

✅ Correct Option: 3

If we are given:

I) $a₁x + b₁y + c₁ = 0$

II) $a₂x + b₂y + c₂ = 0$

We have the following conditions:

Case	Condition	Type of Solution
Unique Solution	$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$	Intersecting lines
Infinite Solutions	$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$	Coincident lines
No Solution	$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$	Parallel lines

Rewrite the equations (in the question) in the general format of $ax + by + c =0$ :

I) $px-4y-2 =0$

I) $3x+ky-a =0$

Let's identify our coefficients:

$a_1 = p$ , $b_1 = -4$ , $c_1 = 2$

$a_2 = 3$ , $b_2 = k$ , $c_2 = a$

For no solution:

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

$\dfrac{p}{3} = \dfrac{-4}{k} \neq \dfrac{2}{a}$

This gives us two separate conditions to work with.

Case 1:

$\dfrac{p}{3} = \dfrac{-4}{k}$

$pk = 3 \times (-4)$

$pk = -12$

Case 2:

$\dfrac{-4}{k} \neq \dfrac{2}{a}$

$-4a \neq 2k$

$-2a \neq k$

$2a + k \neq 0$

For the system to have no solution, we need both conditions:

$pk = -12$ (makes the lines parallel)

$2a + k \neq 0$ (ensures they're not the same line)

Key Insight: The first condition creates parallel lines, while the second condition prevents them from being identical (slope differs). Together, they guarantee no intersection point exists!

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For some constant real numbers p,kp, kp,k and aaa consider the following system of linear equations in xxx and yyy: px−4y=2px - 4y = 2px−4y=2 3x+ky=a3x + ky = a3x+ky=a A necessary condition for the system to have no solution for (x,y)(x, y)(x,y) is

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CAT Past Year Questions (Topic-Wise)

For some constant real numbers $p, k$ and $a$ consider the following system of linear equations in $x$ and $y$ :

$px - 4y = 2$

$3x + ky = a$

A necessary condition for the system to have no solution for $(x, y)$ is