Ex.3.5 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

(i) For which values of \(a\) and \(b\) will the following pair of linear equations have an infinite number of solutions?

\[\begin{align}2x + 3y\, &= 7\\\left( {a-b} \right)x + \left( {a + b} \right)y &= 3\,a\, + \,b\, - \,2\end{align}\]

(ii) For which value of \(k\) will the following pair of linear equations have no solution?

\[\begin{align}3x + y &= 1\\\left( {2k-1} \right)x + \left( {k-1} \right)y &= 2k + 1\end{align}\]

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.5 | Question 2

Text Solution

Steps:

(i) \(\begin{align}&2x + 3y\, = 7\\&\left( {a-b} \right)x + \left( {a + b} \right)y = 3\,a\, + \,b\, - \,2\end{align}\)

\[\begin{align}2x + 3y - 7 &= 0\\\left( {a-b} \right)x + \left( {a + b} \right)y - \left( {3a + b-2} \right) &= 0\\ \frac{{{a_1}}}{{{a_2}}} = \frac{2}{{a - b}}, \quad\frac{{{b_1}}}{{{b_2}}} = \frac{3}{{a + b}},\quad \frac{{{c_1}}}{{{c_2}}} &= \frac{7}{{3a + b - 2}}\end{align}\]

For infinitely many solutions,

\[\begin{align}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}&=\frac{{{c}_{1}}}{{{c}_{2}}} \\ \frac{2}{a-b}&=\frac{7}{3a+b-2} \\ 6a+2b-4&=7a-7b \\ a-9b&=-4 \qquad\; \ldots (1) \\\\ \frac{2}{a-b}&=\frac{3}{a+b} \\ 2a+2b&=3a-3b \\ a-5b&=0\text{ }\,\qquad\;\;\ldots (2) \\\end{align}\]

Subtracting (1) from (2), we obtain

\[\begin{align}4b &= 4\\b &= 1\end{align}\]

Substituting \(b = 1\) in equation (2), we obtain

\[\begin{align}a-5\,\, \times \,\,1\,\, &= 0\\a &= 5\end{align}\]

Hence, \(a = 5\) and \(b = 1\) are the values for which the given equations give infinitely many solutions.

Steps:

\(\begin{align}\left( \rm{ii} \right) \quad &3x + y - 1 = 0\\&\left( {2k-1} \right)x + \left( {k-1} \right)y - 2k-1 = 0\end{align}\)

\[\begin{align}3x + y - 1& = 0\\\left( {2k-1} \right)x + \left( {k-1} \right)y - 2k-1 &= 0\\
\frac{{{a_1}}}{{{a_2}}} = \frac{3}{{2k - 1}}, \quad \frac{{{b_1}}}{{{b_2}}}= \frac{1}{{k - 1}},\quad{\rm{ }}\,\frac{{{c_1}}}{{{c_2}}} = \frac{{ - 1}}{{ - 2k - 1}}\,\, &= \,\,\frac{1}{{2k + 1}}\end{align}\]

For no solution,

\[\begin{align} \frac{{{a}_{1}}}{{{a}_{2}}}&=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}~ \\
 \frac{3}{2k-1}&=\frac{1}{k-1}\ne \frac{1}{2k+1} \\ \frac{3}{2k-1}&=\frac{1}{k-1} \\3k-3&=2k-1 \\k&=2\end{align}\]

Hence, for \(k = 2\) the given equation has no solution.

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