The pair of linear equations mx + 2y + 3 = 0 and 3x + 6y + 2 = 0 intersect each other, if Option 1: m≠2 Option 2: m = 1 Option 3: m ≠ 3 Option 4: m ≠ 1

Answer: m ≠ 1. Solution: Two straight lines are given: Line 1: mx + 2y + 3 = 0 Line 2: 3x + 6y + 2 = 0 Two lines can either intersect, be parallel, or be coincident. For lines a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0: Lines are parallel when: a_1/a_2 = b_1/b_2 Lines intersect when: a_1/a_2 ≠ b_1/b_2 From the given equations: Line 1: a_1 = m, b_1 = 2, c_1 = 3 Line 2: a_2 = 3, b_2 = 6, c_2 = 2 The ratios are: a_1/a_2 = m/3 b_1/b_2 = 2/6 = 1/3 For the lines to intersect: a_1/a_2 ≠ b_1/b_2 m/3 ≠ 1/3 m ≠ 1 Therefore, the lines intersect each other when m ≠ 1.

Quantitative Reasoning

Linear Equation

Medium

The pair of linear equations mx + 2y + 3 = 0 and 3x + 6y + 2 = 0 intersect each other, if

m≠2

m = 1

m ≠ 3

m ≠ 1

✅ Correct Option: 4

Two straight lines are given:

Line 1: $mx + 2y + 3 = 0$

Line 2: $3x + 6y + 2 = 0$

Two lines can either intersect, be parallel, or be coincident.

For lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ :

Lines are parallel when: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$

Lines intersect when: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

From the given equations:

Line 1: $a_1 = m$ , $b_1 = 2$ , $c_1 = 3$

Line 2: $a_2 = 3$ , $b_2 = 6$ , $c_2 = 2$

The ratios are:

$\dfrac{a_1}{a_2} = \dfrac{m}{3}$

$\dfrac{b_1}{b_2} = \dfrac{2}{6} = \dfrac{1}{3}$

For the lines to intersect:

$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

$\dfrac{m}{3} \neq \dfrac{1}{3}$

$m \neq 1$

Therefore, the lines intersect each other when $m \neq 1$ .

2026: 20 May Shift 1

One says, "Give me a hundred rupees, friend! I shall then become twice as rich as you." The other replies, "If you give me ten rupees, I shall be six times as rich as you." What is the amount of their (respective) capital?

2026: 14 May Shift 2

Solve the linear equation for $x$ :

$3x - 7 = 2x + 5$

2025: 19 May Shift 1

If 2x + 3y = 26 and y - x = 2, then the value of x + y is:

2025: 20 May Shift 1

If $6x - 10y = 10$ and $\frac{x}{x+y} = \frac{5}{7}$ , then (x-y) is equal to:

2025: 3 June Shift 1

Which one of the following pairs of linear equations has/ have infinite many solutions?

(A) $9x + 3y + 12 = 0$ , and $18x + 6y + 24 = 0$

(B) $2x - 3y = 13$ , and $7x - 2y = 20$

(D) $3x + y - 3 = 0$ and $2x + \frac{2}{3}y = 2$

Choose the correct answer from the options given below:

2026: 20 May Shift 2

Which of the following pairs of linear equations is/are consistent?

A. x + y – 5 = 0, 3x + 2y – 10 = 0

B. 3x – 2y – 6 = 0, 6x – 4y – 12 = 0

C. x – y – 9 = 0, 3x – 3y – 27 = 0

D. x + y – 4 = 0 , 3x + 3y – 12 = 0

Choose the correct option from the given below:

The pair of linear equations mx + 2y + 3 = 0 and 3x + 6y + 2 = 0 intersect each other, if

Solution

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