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 = 0\\&\left( {a\!-\!b} \right)x \!+\! \left( {a + b} \right)y \!- \!\left( {3a + b-2} \right)\! = \!0\end{align}

\begin{align} &\frac{{{a_1}}}{{{a_2}}} \!=\! \frac{2}{{a \!-\! b}}, \\&\frac{{{b_1}}}{{{b_2}}} \!=\! \frac{3}{{a\! + \! b}},\\& \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:

(ii)

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

\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \frac{3}{{2k - 1}}, \quad \frac{{{b_1}}}{{{b_2}}}= \frac{1}{{k - 1}},\\\,\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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