Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if Option 1: k ≠ 2 Option 2: k, = 2 Option 3: k = 2 Option 4: k, ≠ 2

Answer: k, ≠ 2. Solution: We have two equations: kx + y = 3 ... (1) 4x + ky = 4 ... (2) The big question: When do these have exactly ONE solution? Think of each equation as a line on a graph. For exactly one solution, these lines must intersect at exactly one point (not be parallel, not be the same line). We look at the coefficient matrix formed by the numbers in front of x and y: bmatrix k & 1 \\ 4 & k bmatrix If this matrix has a non-zero determinant, we get exactly one solution. For any 2 x 2 matrix bmatrix a & b \\ c & d bmatrix, determinant = ad - bc Our determinant = k * k - 1 * 4 = k^2 - 4 For a unique solution, we need: k^2 - 4 ≠ 0 k^2 ≠ 4 k ≠ 2 and k ≠ -2 Let's verify why k = 2 and k = -2 don't work: When k = 2: Equation 1: 2x + y = 3 Equation 2: 4x + 2y = 4 Multiplying equation 1 by 2: 4x + 2y = 6 But equation 2 says: 4x + 2y = 4 Contradiction! No solution exists. When k = -2: Equation 1: -2x + y = 3 Equation 2: 4x - 2y = 4 From equation 1: y = 3 + 2x Substituting into equation 2: 4x - 2(3 + 2x) = 4 4x - 6 - 4x = 4 -6 = 4 which is impossible! Let's verify with k = 1 (since 1 ≠ ± 2): x + y = 3 4x + y = 4 Subtracting: 3x = 1, so x = 1/3 Then: y = 3 - 1/3 = 8/3 One unique solution exists! The equations have a unique solution if and only if k ≠ 2 and k ≠ -2. Quick memory trick: Calculate k^2 - 4 ≠ 0 to find when the system has exactly one solution.

Algebra

Linear Equation

Easy

Let $k$ be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution if and only if

$k ≠ 2$

$|k| = 2$

$k = 2$

$|k| ≠ 2$

✅ Correct Option: 4

We have two equations:

$kx + y = 3$ ... (1)

$4x + ky = 4$ ... (2)

The big question: When do these have exactly ONE solution?

Think of each equation as a line on a graph. For exactly one solution, these lines must intersect at exactly one point (not be parallel, not be the same line).

We look at the coefficient matrix formed by the numbers in front of $x$ and $y$ :

$\begin{bmatrix} k & 1 \\ 4 & k \end{bmatrix}$

If this matrix has a non-zero determinant, we get exactly one solution.

For any $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , determinant $= ad - bc$

Our determinant $= k \cdot k - 1 \cdot 4 = k^2 - 4$

For a unique solution, we need:

$k^2 - 4 \neq 0$

$k^2 \neq 4$

$k \neq 2$ and $k \neq -2$

Let's verify why $k = 2$ and $k = -2$ don't work:

When $k = 2$ :

Equation 1: $2x + y = 3$

Equation 2: $4x + 2y = 4$

Multiplying equation 1 by 2: $4x + 2y = 6$

But equation 2 says: $4x + 2y = 4$

Contradiction! No solution exists.

When $k = -2$ :

Equation 1: $-2x + y = 3$

Equation 2: $4x - 2y = 4$

From equation 1: $y = 3 + 2x$

Substituting into equation 2: $4x - 2(3 + 2x) = 4$

$4x - 6 - 4x = 4$

$-6 = 4$ which is impossible!

Let's verify with $k = 1$ (since $1 \neq \pm 2$ ):

$x + y = 3$

$4x + y = 4$

Subtracting: $3x = 1$ , so $x = \dfrac{1}{3}$

Then: $y = 3 - \dfrac{1}{3} = \dfrac{8}{3}$

One unique solution exists!

The equations have a unique solution if and only if $k \neq 2$ and $k \neq -2$ .

Quick memory trick: Calculate $k^2 - 4 \neq 0$ to find when the system has exactly one solution.

CAT 2020 Slot 1

A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of $5$ children. How many toffees were there in his stock initially?

2025 Slot 3

In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is

CAT 2020 Slot 2

Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and $10$ less sharpeners. If the cost of one sharpener is Rs. $2$ more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

CAT 2023 Slot 3

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up $40 \%$ of his stock. That day, he sells half of the mangoes, $96$ bananas and $40 \%$ of the apples. At the end of the day, he ends up selling $50 \%$ of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

CAT 2022 Slot 1

Let a and b be natural number. If $a^2 + ab + a = 14$ and $b^2 + ab + b = 28$ , then $(2a + b)$ equals

CAT 2023 Slot 3

For some real numbers $a$ and $b$ , the system of equations $x + y = 4$ and $(a + 5)x + (b^2 - 15)y = 8b$ has infinitely many solutions for $x$ and $y$ . Then, the maximum possible value of $ab$ is

Let kkk be a constant. The equations kx+y=3kx + y = 3kx+y=3 and 4x+ky=44x + ky = 44x+ky=4 have a unique solution if and only if

Solution

Related questions:

Keyboard Shortcuts

CAT Past Year Questions (Topic-Wise)

Let $k$ be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution if and only if