Math, asked by seetharamb22, 6 months ago

1. Do the following pair of linear equations have no solution? Justify your answer.
(i) 2x + 4y = 3
12y + 6x = 6
(ii) x = 2y
y = 2x
(iii) 3x + y – 3 = 0
2x + 2/3y = 2

Answers

Answered by padmamaloth1986
5

Step-by-step explanation:

The Condition for no solution = a1/a2 = b1/b2 ≠ c1/c2 (parallel lines)

(i) Yes.

Given pair of equations are

2x+4y - 3 = 0 and 6x + 12y - 6 = 0

Comparing the equations with ax+ by +c = 0;

We get,

a1 = 2, b1 = 4, c1 = - 3;

a2 = 6, b2 = 12, c2 = - 6;

a1 /a2 = 2/6 = 1/3b1 /b2 = 4/12 = 1/3

c1 /c2 = - 3/ - 6 = ½

Here, a1/a2 = b1/b2 ≠ c1/c2, i.e parallel lines

Hence, the given pair of linear equations has no solution.

(ii) No.

Given pair of equations,

x = 2y or x - 2y = 0

y = 2x or 2x - y = 0;

Comparing the equations with ax+ by +c = 0;

We get,

a1 = 1, b1 = - 2, c1 = 0;

a2 = 2, b2 = - 1, c2 = 0;

a1 /a2 = ½

b1 /b2 = -2/-1 = 2

Here, a1/a2 ≠ b1/b2.

Hence, the given pair of linear equations has unique solution.

(iii) No.

Given pair of equations,

3x + y - 3 = 0

2x + 2/3 y = 2

Comparing the equations with ax+ by +c = 0;

We get,

a1 = 3, b1 = 1, c1 = - 3;

a2 = 2, b2 = 2/3, c2 = - 2;

a1 /a2 = 2/6 = 3/2

b1 /b2 = 4/12 = 3/2

c1 /c2 = - 3/-2 = 3/2

Here, a1/a2 = b1/b2 = c1/c2, i.e coincident lines

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2. If sin A = ½ , then the value of cot A is  

(A) √3 (B) 1/√3 (C) √3/2 (D) 1 brainly.in/question/26614602

3. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

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2. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have

(A) a unique solution (B) exactly two solutions

(C) infinitely many solutions (D) no solution

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Answered by sidhhobanoth03
0

Answer:

2 is the answer

Step-by-step explanation:

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