Consistency of a Pair of Linear Equations

Video Explanation

Question

Find the values of \(k\) for which the system

\[ 2x + ky = 1, \qquad 3x – 5y = 7 \]

will have (i) a unique solution, (ii) no solution. Also, check whether there is a value of \(k\) for which the system has infinitely many solutions.

Solution

Step 1: Write in Standard Form

\[ 2x + ky – 1 = 0 \quad (1) \]

\[ 3x – 5y – 7 = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 2, \quad b_1 = k, \quad c_1 = -1 \]

\[ a_2 = 3, \quad b_2 = -5, \quad c_2 = -7 \]

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(i) Condition for a Unique Solution

A pair of linear equations has a unique solution if

\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

\[ \frac{2}{3} \neq \frac{k}{-5} \]

\[ k \neq -\frac{10}{3} \]

Hence, the system has a unique solution for all real values of \(k\) except \(k = -\dfrac{10}{3}\).

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(ii) Condition for No Solution

A pair of linear equations has no solution if

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

\[ \frac{2}{3} = \frac{k}{-5} \]

\[ k = -\frac{10}{3} \]

Now check the third ratio:

\[ \frac{c_1}{c_2} = \frac{-1}{-7} = \frac{1}{7} \]

Since

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, \]

the system is inconsistent.

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(iii) Possibility of Infinitely Many Solutions

A pair of linear equations has infinitely many solutions if

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

But here,

\[ \frac{2}{3} \neq \frac{1}{7} \]

Hence, there is no value of \(k\) for which the given system has infinitely many solutions.

Conclusion

(i) The system has a unique solution for:

\[ \boxed{k \neq -\dfrac{10}{3}} \]

(ii) The system has no solution for:

\[ \boxed{k = -\dfrac{10}{3}} \]

(iii) There is no value of \(k\) for which the system has infinitely many solutions.