Question

The pair of linear equations x – 2y = 5 and 2x – 4y = 1 have :
(a) Many Solutions
(b) No Solution
(c) One Solution
(d)Two Solution

Answer 1

Answer:

No solution

Step-by-step explanation:

is your answer

Answer 2

The pair of linear equations x – 2y = 5 and 2x – 4y = 1 have No Solution

Given :

The pair of equations x – 2y = 5 and 2x – 4y = 1

To find :

The pair of linear equations x – 2y = 5 and 2x – 4y = 1 have :

(a) Many Solutions

(b) No Solution

(c) One Solution

(d)Two Solution

Concept :

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y = c_1 \: and \: \: a_2x+b_2y = c_2}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

No solution

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

Solution :

Step 1 of 2 :

Write down the given pair of equations

Here the given pair of linear equations are

x – 2y = 5 - - - - - (1)

2x – 4y = 1 - - - - - (2)

Step 2 of 2 :

Find the number of solutions

x – 2y = 5 - - - - - (1)

2x – 4y = 1 - - - - - (2)

Comparing with the equations

a₁x + b₁y = c₁ and a₂x + b₂y = c₂ we get

a₁ = 1 , b₁ = - 2 , c₁ = 5 and a₂ = 2 , b₂ = - 4 , c₂ = 1

Now we have ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{1}{2}$

$\displaystyle \sf \frac{b_1}{b_2} = \frac{-2}{ - 4} = \frac{1}{2}$

$\displaystyle \sf \frac{c_1}{c_2} = \frac{5}{ 1} = 5$

Thus we get ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{b_1}{b_2} = \ne \frac{c_1}{c_2}$

So the pair of linear equations x – 2y = 5 and 2x – 4y = 1 have no Solution

Hence the correct option is (b) No Solution

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