Question

For which values of a and b will the following
pair of
linear equations have in infinetely many
salutions ? x+2y = I ) and (a-b)x+(a+b)y =a+b-2​

Answer 1

Answer:

Answer: The required values are a = 3 and b = 1.

Step-by-step explanation: We are given to find the values of a and b for which the following pair of linear equations will have infinitely many solutions :

\begin{lgathered}x+2y=1,\\\\(a-b)x+(a+b)y=a+b-2.\end{lgathered}

x+2y=1,

(a−b)x+(a+b)y=a+b−2.

We know that

for a system of linear equations to have infinitely many solutions, the coefficients of the unknown variables (x and y) and the constant terms must be in proportion.

So, for the given system, we must have

\dfrac{1}{a-b}=\dfrac{2}{a+b}=\dfrac{1}{a+b-2}.

a−b

1

=

a+b

2

=

a+b−2

1

.

We have from above that

\begin{lgathered}\dfrac{1}{a-b}=\dfrac{2}{a+b}\\\\\Rightarrow a+b=2(a-b)\\\\\Rightarrow a+b=2a-2b\\\\\Rightarrow 2a-a=b+2b\\\\\Rightarrow a=3b~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\end{lgathered}

a−b

1

=

a+b

2

⇒a+b=2(a−b)

⇒a+b=2a−2b

⇒2a−a=b+2b

⇒a=3b (i)

and

Again, substituting the value of b in equation (i), we get

a=3\times1=3.a=3×1=3.

Thus, the required values are a = 3 and b = 1.