Question

the pair of equations x +2y -5 =0 and -4x -8y +20=0 have a) unique solution b) exactly two solutions c) infinitely many solutions d) no solutions​

Answer 1

Step-by-step explanation:

x+2y-5=0

-4x-8y+20=0

solve euations

4x+8y-20=0

-4x-8y+20=0

the ans is no solution

option D is your answer

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

The pair of equations x + 2y - 5 =0 and - 4x - 8y + 20 = 0 have

a) unique solution

b) exactly two solutions

c) infinitely many solutions

d) no solutions

CONCEPT TO BE IMPLEMENTED

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

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

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

EVALUATION

Here the given system of equations are

x + 2y - 5 = 0 and - 4x - 8y + 20 = 0

Comparing with the equation

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = 1 , b₁ = 2 , c₁ = - 5 and a₂ = - 4 , b₂ = - 8 , c₂ = 20

Now

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{1}{ - 4} = - \frac{1}{4}}$

$\displaystyle \sf{ \frac{b_1}{b_2} = \: \frac{2}{ - 8}} = - \frac{1}{4}$

$\displaystyle \sf{ \frac{c_1}{c_2} = \frac{ - 5}{20} = - \frac{1}{4} }$

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

So the given system of equations have Infinite number of solutions

FINAL ANSWER

Hence the correct option is

c) infinitely many solutions

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