Question

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

Answer 1

In these types of questions we have to compare the coefficients of both the given equations and by using condition for equations to have infinite number of solution,we will get the value for $a$ and $b$

Complete step-by-step answer:

1. It is given that the equations are $2x + 3y = 7.......(1)$
   $(a - b)x + (a + b)y = 3a + b - 2.......(2)$
   We have to find the values for $a$ and $b$.
   For the equation to have an infinite number of solutions, the ratios of the coefficients of $x$,$y$ and constant terms in the two equations should be equal.
   Comparing eqn. $(1)$ coefficients with ${a_1}x + {b_1}y + {c_1} = 0$
   $\therefore {a_1} = 2,{b_1} = 3,{c_1} = - 7$
   And similarly, comparing eqn. $(2)$ the coefficient with ${a_2}x + {b_2}y + {c_2} = 0$
   $\therefore {a_2} = (a - b),{b_2} = (a + b),{c_2} = (3a + b - 2)$
   It is given that the equation has infinite number of solutions
   So, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
   ∴ We can write $\dfrac{2}{{a - b}} = \dfrac{3}{{a + b}} = \dfrac{7}{{3a + b - 2}}....\left( 3 \right)$
   We are taking two terms in $\left( 3 \right)$ to solve at a time for which solving the values for $a$ and $b$
   $\dfrac{2}{{a - b}} = \dfrac{3}{{a + b}}$
   Let us take a cross multiply for both sides,
   $2(a + b) = 3(a - b)$
   On multiplying the terms,
   $2a + 2b = 3a - 3b$
   By collecting same terms on each side and simplify we get,
   $a = 5b...(4)$
   On solving these two terms in equation $\left( 3 \right)$.
   We can write it as $\dfrac{2}{{a - b}} = \dfrac{7}{{3a + b - 2}}$
   Let us cross multiply both the fractions
   $2(3a + b - 2) = 7(a - b)$
   On multiplying the terms, we get
   $6a + 2b - 4 = 7a - 7b$
   Collecting the same term on each sides and simplify we get,
   $a - 9b + 4 = 0....(5)$
   Now, substituting the value of $a = 5b$in the equation $(5)$ we get,
   $5b - 9b + 4 = 0$
   Subtracting the same values
   $- 4b + 4 = 0$
   Taking negative term to the right side of the expression
   $b = \dfrac{4}{4}$
   $b = 1$
   Substituting the value of $b = 1$ in the equation $(4)$ we get,
   $a = 5b = 5(1) = 5$
   Hence, for $a = 5$ and $b = 1$ the given set of equations have infinitely many solutions.

Note: For any two points, exactly one line can draw through both the points. It concludes that if two lines intersect in at least two points, then they must be the same line.Students should remember the condition that for equations to have infinite number of solutions i.e $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ for solving these types of problems.

Hint: In these types of questions we have to compare the coefficients of both the equations and by using condition for equations to have no solution,we will get the value for $k$.

Complete step-by-step answer:
It is given that the equations are
$3x + y = 1...(1)$
$(2k - 1)x + (k - 1)y = 2k + 1...(2)$
We have to find the value for $k$
Comparing eqn. $(1)$ with coefficient ${a_1}x + {b_1}y + {c_1} = 0$
$\therefore {a_1} = 3,{b_1} = 1,{c_1} = - 1$
Similarly, in eqn. $(2)$ with coefficient ${a_2}x + {b_2}y + {c_2} = 0$
$\therefore {a_2} = (2k - 1),{b_2} = (k - 1),{c_2} = - (2k + 1)$
For equations to have no solution it should satisfy the condition i.e $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
Here we have to find the value for $k$, so we take the first two terms, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}}$
Now, we can write it as $\dfrac{3}{{(2k - 1)}} = \dfrac{1}{{(k - 1)}}$
Let us cross multiply both the fractions
$3(k - 1) = 1(2k - 1)$
Open the brackets by multiplying the terms
$3k - 3 = 2k - 1$
By collecting same terms on each side we get,
$3k - 2k = 3 - 1$
On subtracting we get the value of $k$
$k = 2$
Therefore for the value of $k = 2$ the required given set of equations have no solutions.

Note: If there are no points common to both lines, hence there is no solution for the system.When there is no solution the equations are called inconsistent.Students should remember the condition that for equations to have no solution i.e $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ for solving these types of problems.