Question

(i) For which values of $a$ and $b$, does the following pair of linear equations have an infinite number of solutions?
$2x + 3y = 7$ ; $\left( {a - b} \right)x + \left( {a + b} \right)y = 3a + b - 2$
(i) For which value of $k$ the following pair of linear equations have no solution?
$3x + y = 1$ ; $\left( {2k - 1} \right)x + \left( {k - 1} \right)y = 2k + 1$

Answer 1

Hint: This problem deals with solving the system of pairs of linear equations with an infinite number of solutions and no solutions. If the pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0$, ${a_2}x + {b_2}y + {c_2} = 0$ have infinite number of solutions, then it is given by:
$ \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
If the linear equations have no solutions, it is given by:
$ \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$

Complete step-by-step answer:
Given two equations $2x + 3y = 7$ and $\left( {a - b} \right)x + \left( {a + b} \right)y = 3a + b - 2$
Considering the first equation $2x + 3y = 7$, as given below:
$ \Rightarrow 2x + 3y = 7$
$ \Rightarrow 2x + 3y - 7 = 0$
Now considering the second equation $\left( {a - b} \right)x + \left( {a + b} \right)y = 3a + b - 2$, as given below:
$ \Rightarrow \left( {a - b} \right)x + \left( {a + b} \right)y = 3a + b - 2$
$ \Rightarrow \left( {a - b} \right)x + \left( {a + b} \right)y - \left( {3a + b - 2} \right) = 0$
Now given that these two linear equations have an infinite number of solutions, hence equating ratios of the like term coefficients, as given below:
$ \Rightarrow \dfrac{2}{{\left( {a - b} \right)}} = \dfrac{3}{{\left( {a + b} \right)}} = \dfrac{{ - 7}}{{ - \left( {3a + b - 2} \right)}}$
First considering the first equation of ratios, as given below:
$ \Rightarrow \dfrac{2}{{\left( {a - b} \right)}} = \dfrac{3}{{\left( {a + b} \right)}}$
$ \Rightarrow 2\left( {a + b} \right) = 3\left( {a - b} \right)$
Simplifying the equation and by grouping the like terms and unlike terms together, gives:
$ \Rightarrow 2a + 2b = 3a - 3b$
$ \Rightarrow a = 5b$
Now equating the first and the last ratio, as given below:
$ \Rightarrow \dfrac{2}{{\left( {a - b} \right)}} = \dfrac{{ - 7}}{{ - \left( {3a + b - 2} \right)}}$
$ \Rightarrow \dfrac{2}{{\left( {a - b} \right)}} = \dfrac{7}{{\left( {3a + b - 2} \right)}}$
Simplifying the above equation, as given below:
$ \Rightarrow 2\left( {3a + b - 2} \right) = 7\left( {a - b} \right)$
$ \Rightarrow 6a + 2b - 4 = 7a - 7b$
$ \Rightarrow a = 9b - 4$
Now substituting $a = 5b$, in the above obtained equation :
$ \Rightarrow 5b = 9b - 4$
$ \Rightarrow 4b = 4$
$\therefore b = 1$
Now finding the value of $a$, as given below:
$ \Rightarrow a = 5b$
$ \Rightarrow a = 5\left( 1 \right)$
$\therefore a = 5$
Hence the values of $a$ and $b$ are 5 and 1 respectively.
Given two equations $3x + y = 1$ and $\left( {2k - 1} \right)x + \left( {k - 1} \right)y = 2k + 1$, we have to find the value of $k$.
Consider the first equation $3x + y = 1$, as given below:
$ \Rightarrow 3x + y = 1$
$ \Rightarrow 3x + y - 1 = 0$
Now consider the second equation $\left( {2k - 1} \right)x + \left( {k - 1} \right)y = 2k + 1$, as given below:
$ \Rightarrow \left( {2k - 1} \right)x + \left( {k - 1} \right)y = 2k + 1$
$ \Rightarrow \left( {2k - 1} \right)x + \left( {k - 1} \right)y - \left( {2k + 1} \right) = 0$
Given that these two linear equations have no solution, then the ratios of $x$ and $y$ coefficients are equal, but these ratios are not equal to the ratios of the constants.
$ \Rightarrow \dfrac{3}{{\left( {2k - 1} \right)}} = \dfrac{1}{{\left( {k - 1} \right)}} \ne \dfrac{{ - 1}}{{ - \left( {2k + 1} \right)}}$
Now equating the first two ratios, as shown below:
$ \Rightarrow \dfrac{3}{{\left( {2k - 1} \right)}} = \dfrac{1}{{\left( {k - 1} \right)}}$
On further simplification of the above equation:
$ \Rightarrow 3\left( {k - 1} \right) = \left( {2k - 1} \right)$
$ \Rightarrow 3k - 3 = 2k - 1$
$\therefore k = 2$
Hence the value of $k$ is equal to 2.

Final Answer: The values of a and b are 5 and 1 respectively, whereas the value of k is 2.

Note:
Please note that while solving this problem when the linear pair of equations have infinite number of solutions, then that means that the lines are the exact same line, hence for which the ratios of the coefficients are equated. But whereas for the lines which have no solutions, these lines are parallel.