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(i) For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?
$2x + 3y =7$
$(a – b)x + (a + b)y = 3a + b – 2$.
(ii) For which value of $k$ will the following pair of linear equations have no solution?
$(2k – 1)x + (k – 1)y = 2k + 1$.
To do:
We have to find
(i) The values of $a$, $b$ and whether the given linear equations have an infinite number of solutions.
(ii) The value of $k$ and whether the given pair of linear equations has no solution.
Solution:
(i) Given equations are: $2x+3y=7$ ..... $( i)$
$( a-b)x+( a+b)y=3a+b-2$ ...... $( ii)$
It has infinitely many solutions
This implies,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$
$\frac{2}{a-b}=\frac{3}{a+b}$
$\Rightarrow 2( a+b)=3( a-b)$
$\Rightarrow 2a+2b=3a-3b$
$\Rightarrow a-5b=0$
$\Rightarrow a=5b$ ..... $( iii)$
Now, $\frac{3}{a+b}=\frac{7}{3a+b-2}$
$\Rightarrow 7a+7b=9a+3b-6$
$\Rightarrow 2a-4b=6$
$\Rightarrow a-2b=3$
$\Rightarrow 5b-2b=3$
$\Rightarrow 3b=3$
$\Rightarrow b=1$
$\therefore a=5\times1=5$ [$\because a=5b$ from $( iii)$]
Thus, $a=5$ and $b=1$.
$3x + y = 1$
$(2k – 1)x + (k – 1)y = 2k + 1$
The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.
Comparing the given system of equations with the standard form of equations, we have,
$a_1=3, b_1=1, c_1=1$ and $a_2=(2k-1), b_2=k-1, c_2=2k+1$
The condition for which the above system of equations has no solution is
$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} ≠ \frac{c_{1}}{c_{2}} \ $
Therefore,
$\frac{3}{2k-1}=\frac{1}{k-1}≠\frac{1}{2k+1}$
$3(k-1)=1(2k-1)$ and $1(2k+1)≠1(k-1)$
$3k-3=2k-1$ and $2k+1≠k-1$
$3k-2k=-1+3$ and $2k-k≠-1-1$
$k=2$ and $k≠-2$
The value of $k$ for which the given system of equations has no solution is $k=2$ and $k≠-2$.  
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