Question

find the values of a and b which for simultanious equation x + 2y = 1 and ( a - b)x + ( a + b) y = a+b - 2 have infinitly many solutions​

Answer 1

Given

Two equations

• x+2y-1=0 ------(i)
• (a-b)x+(a+b)y-(a+b)+2=0 ------- (ii)

To Find

• We have to find the value of a and b in case the given equations have infinitely many solutions

Solution :-

As we know that

For infinitely many solutions

$\bullet\boxed{\sf\ \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}}$

take two equations

$\sf\ \ a_1x+b_1y+c_1= 0----- (iii)\\ \\ \sf\ \ a_2x+b_2y+c_2=0-----(iv)$

On comparing eq.(i) with (iii) & eq.(ii) with (iv)

We get :-

$\bullet\sf\ \ a_1= 1 \ \ \ ;\ \bullet\sf\ \ a_2=(a-b)\\ \\ \bullet\sf\ b_1=2\ \ \ ;\ \bullet\sf\ \ b_2=(a+b)\\ \\ \bullet\sf\ c_1=(-1)\ \ \ ;\ \bullet\sf\ \ c_2=2-(a+b)$

Now substitute these values -

$:\implies\sf\ \dfrac{1}{(a-b)}= \dfrac{2}{(a+b)}=\dfrac{(-1)}{2-(a+b)}\\ \\ \\ :\implies\sf\ \ On\ comparing \ 1st\ and\ 2nd \ \\ \\ \\ :\implies\sf\ \dfrac{1}{(a-b)}=\dfrac{2}{(a+b)} \\ \\ \\ :\implies\sf\ a+b= 2(a-b)\\ \\ \\ :\implies\sf\ a+b=2a-2b\\ \\ \\ :\implies\sf\ b+2b= 2a-a\\ \\ \\ :\implies\underline{\boxed{\sf\ \ 3b=a}}------ (v)$

Now on comparing 2nd and 3rd

$:\implies\sf\ \dfrac{2}{(a+b)}=\dfrac{-1}{2-(a+b)}\\ \\ \\ :\implies\sf\ 2\big\{2-(a+b)\big\}= -(a+b)\\ \\ \\ :\implies\sf\ 4-2a-2b= -a-b\\ \\ \\ :\implies\sf\ \ 4= (-a+2a)+(-b+2b)\\ \\ \\ :\implies\sf\ \ 4= a+b\\ \\ \sf\ Put\ the\ value\ of\ (a)\\ \\ \\ :\implies\sf\ 4=3b+b\\ \\ \\ :\implies\sf\ \ \not{4}= \not{4}b\\ \\ \\ :\implies\underline{\boxed{\sf{\purple{\ \ b= 1}}}}$

• Put the value of b in eq.(v) to find the value of a

$:\implies\sf\ 3b=a\\ \\ \\ :\implies\sf\ 3(1)= a\\ \\ \\ :\implies\underline{\boxed{\sf{\blue{\ \ a= 3}}}}$

$\underline{\bigstar\sf The\ value\ of \ a= 3\ and\ b= 1}$

Answer 2

Answer:

Given :-

Two equation :

• x + 2y = 1
• (a - b)x + (a + b)y = (a + b) - 2

To Find :-

• What is the value of a and b.

Solution :-

Given two equation :

$\mapsto$ $\sf x + 2y = 1$$\sf \\ \implies \bold{x + 2y - 1 =\: 0\: -----\: Equation\: no\: (1)}$

$\mapsto$ $\sf (a - b)x + (a + b)y = (a + b) - 2$

$\sf \\ \implies \bold{(a - b)x + (a + b)y - (a + b + 2) =\: 0\: -----\: Equation\: no\: (2)}$

Now, as we know that,

${\red{\boxed{\large{\bold{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} =\: \dfrac{c_1}{c_2}}}}}}$

From this formula we get,

• $\sf a_1 =\: 1$
• $\sf b_1 =\: 2$
• $\sf c_1 =\: - 1$
• $\sf a_2 =\: (a - b)$
• $\sf b_2 =\: (a + b)$
• $\sf c_2 =\: - (a + b - 2)$

Then we get,

$\implies$ $\sf \dfrac{1}{a - b} =\: \dfrac{2}{a + b} =\: \dfrac{- 1}{- (a + b - 2)}$

Now, by taking first two parts we get,

⇒ $\sf \dfrac{1}{a - b} =\: \dfrac{2}{a + b}$

By doing cross multiplication we get,

⇒ $\sf 2(a - b) =\: a + b$

⇒ $\sf 2a - 2b =\: a + b$

⇒ $\sf 2a - a =\: 2b + b$

$\sf \implies \bold{\pink{a =\: 3b -----\: Equation\: no\: (3)}}$

Again, by taking last two parts we get,

↦ $\sf\dfrac{2}{a + b} =\: \dfrac{{\cancel{-}} 1}{{\cancel{-}} (a + b - 2)}$

By doing cross multiplication we get,

↦ $\sf 2(a + b - 2) =\: a + b$

↦ $\sf 2a + 2b - 4 =\: a + b$

↦ $\sf 2a - a + 2b - b =\: 4$

$\sf \implies \bold{\pink{a + b =\: 4\: -----\: Equation\: no\: (4)}}$

Now, by putting the value of equation no (3) in the equation no (4) we get,

↛ $\sf 3b + b =\: 4$

↛ $\sf 4b =\: 4$

↛ $\sf b =\: \dfrac{\cancel{4}}{\cancel{4}}$

➠ $\sf\bold{\purple{b =\: 1}}$

Again, by putting the value of b in the equation no (3) we get,

↛ $\sf a = 3b$

↛ $\sf a = 3(1)$

↛ $\sf a =\: 3 \times 1$

➠ $\sf\bold{\purple{a =\: 3}}$

$\therefore$ The value of a is 3 and the value of b is 1 .