Question

Plz ans it with full explanation​

Answer 1

Answer:

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Step-by-step explanation:

so given simultaneous linear equations are

x+2y=1

(a-b)x+(a+b)y=a+b-2

so comparing these equations with

a1x+b1y=c1 and a2x+b2y=c2

we get

a1=1 a2=(a-b)

b1=2 b2=(a+b)

c1=1 c2=a+b-2

so now as here both equations have infinitely many solutions

we know that

Whenever any two equations have infinitely many solutions

it can be represented as

a1/a2=b1/b2=c1/c2

1/(a-b)=2/(a+b)=1/(a+b-2)

considering thus then first

1/(a-b)=2/(a+b)

ie a+b=2(a-b)

ie a+b=2a-2b

so a-3b=0 (1)

now considering

2/(a+b)=1/(a+b-2)

2(a+b-2)=a+b

ie 2a+2b-4=a+b

so a+b=4 (2)

multiplying (2) by 3

we get

3a+3b=12 (3)

adding (1) and (3)

we get

4a=12

a=3

substitute value of a in any of two equations

we get

b=1

hence (a,b)=(3,1) is solution of above second linear equation

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Find the values of a and b for which the simultaneous linear equations x + 2y = 1 and (a - b)x + (a + b)y = a + b - 2 have infinitely many solutions.

$\large\underline{\sf{Solution-}}$

Given pair of linear equation is

$\rm :\longmapsto\:x + 2y = 1 - - - (1)$

and

$\rm :\longmapsto\:(a - b)x + (a + b)y = a + b - 2 - - - (2)$

Now,

Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get  

• a₁ = 1

• b₁ = 2

• c₁ = 1

• a₂ = a - b

• b₂ = a + b

• c₂ = a + b - 2

Now, we know that,

System of two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have infinitely many solutions iff  

$\boxed{ \bf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b} = \dfrac{1}{a + b - 2}$

On taking first and second member, we get

$\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b}$

$\rm :\longmapsto\:a + b = 2(a - b)$

$\rm :\longmapsto\:a + b = 2a - 2b$

$\rm :\longmapsto\:a - 2a = - b- 2b$

$\rm :\longmapsto\: - a = - 3b$

$\bf :\longmapsto\: a = 3b - - - - - (1)$

Now, On taking first and third member, we get

$\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{1}{a + b - 2}$

$\rm :\longmapsto\:a + b - 2 = a - b$

$\rm :\longmapsto \: b - 2 = - b$

$\rm :\longmapsto \: b + b = 2$

$\rm :\longmapsto \: 2b= 2$

$\bf :\longmapsto \: b= 1$

On substituting the value of b in equation (1), we get

$\bf :\longmapsto\:a = 3 \times 1 = 3$

$\red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{a \: = \: 3} \\ \\ &\sf{b \: = \: 1} \end{cases}\end{gathered}\end{gathered}}$

Additional Information :-

1. System of two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have unique solution iff  

$\boxed{ \bf{ \: \frac{a_1}{a_2} \: \ne \: \frac{b_1}{b_2} }}$

2. System of two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have no solutions iff  

$\boxed{ \bf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}}}$