Question

2.
6) For which values ofa and b does the following pair of linear equations have an
infinite number of solutions?
2x + 3y - 7
(a - b) x + (a+b) y 3a + b 2​

Answer 1

Answer:

hi

Step-by-step explanation:

i have answered in that pic

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The following pair of linear equations have an infinite number of solution.

2x + 3y - 7

(a-b)x + (a+b)y = 3a + b - 2

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The value of a and b.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of the infinite many solution :

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

Therefore;

$\bullet\sf{2x+3y-7=0........................(1)}\\\bullet\sf{(a-b)x+(a+b)y-(3a+b-2)=0......................(2)}$

As we know that given equation compared with;

$\bullet\sf{a_1x+by_1+c_1=0}\\\bullet\sf{a_2x+by_2+c_2=0}$

$\sf{\dfrac{a_1}{a_2} =\dfrac{2}{a-b} \:\:,\:\dfrac{b_1}{b_2} =\dfrac{3}{a+b} \:\:,\:\:\dfrac{c_1}{c_2} =\dfrac{-7}{-3a+b-2} }$

Now;

$\longrightarrow\bf{\dfrac{a_1}{a_2} =\dfrac{c_1}{c_2} }\\\\\\\longrightarrow\sf{\dfrac{2}{a-b} =\dfrac{7}{3a+b-2} }\\\\\\\longrightarrow\sf{2(3a+b-2)=7(a-b)}\\\\\\\longrightarrow\sf{6a+2b-4=7a-7b}\\\\\\\longrightarrow\sf{6a-7a+2b+7b=4}\\\\\\\longrightarrow\bf{-a+9b=4.................(3)}$

&

$\longrightarrow\bf{\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2} }\\\\\\\longrightarrow\sf{\dfrac{2}{a-b} =\dfrac{3}{a+b }}\\\\\\\longrightarrow\sf{2(a+b)=3(a-b)}\\\\\\\longrightarrow\sf{2a+2b=3a-3b}\\\\\\\longrightarrow\sf{2a-3a+2b+3b=0}\\\\\\\longrightarrow\bf{-a+5b=0........................(4)}$

$\bf{\underline{\underline{\pink{\bf{Using\:Substitution\:Method\::}}}}}$

From equation (3),we get;

$\longrightarrow\sf{-a+9b=4}\\\\\longrightarrow\sf{9b=4+a}\\\\\longrightarrow\sf{b=\dfrac{4+a}{9} ........................(5)}$

Putting the value of b in equation (4),we get;

$\longrightarrow\sf{-a+5\bigg(\dfrac{4+a}{9} \bigg)=0}\\\\\\\longrightarrow\sf{-a+\dfrac{20+5a}{9} =0}\\\\\\\longrightarrow\sf{-9a+20+5a=0}\\\\\\\longrightarrow\sf{-4a=-20}\\\\\\\longrightarrow\sf{a=\cancel{\dfrac{-20}{-4}}}\\\\\\\longrightarrow\sf{\red{a=5}}$

Putting the value of a in equation (5),we get;

$\longrightarrow\sf{b=\dfrac{4+5}{9} }\\\\\\\longrightarrow\sf{b=\cancel{\dfrac{9}{9} }}\\\\\\\longrightarrow\sf{\red{b=1}}$

Thus;

The value of a is 5 and b is 1 .