For what value of C the pair of equations CX Y 2 and 6x 2y 3 will have infinitely many solutions?

Question 5 Pair of Linear Equations in Two Variables - Exercise 3.1

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Answer:

For infinitely many solutions,

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ \Rightarrow \frac{c}{6}=\frac{-1}{-2}=\frac{2}{3} \\ \text { Ratio } I \ {Ratio} II \ {Ratio} III

From ratios I and II, 2c = 6 ⇒ c = 3

From ratios I and III, 3c = 12 ⇒ c = 4

As from the ratios, values of c are not common. So, there is no value of c for which lines have infinitely many solutions.

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Text Solution

2-3-12no value

Answer : D

Solution : Condition for infinitely many solutions <br> `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) =(c_(1))/(c_(2)) " "`...(i) <br> The given lines are `cx-y = 2 ` and `6x - 2y = 3` <br> Here, `" " a_(1) = c, b_(1) = -1, c_(1) = -2` <br> and `" " a_(2) = 6, b_(2) = -2, c_(2) = -3` <br> From Eq. (i), `" " (c)/(6) =(-1)/(-2) = (-2)/(-3)` <br> Here, `" " (c)/(6)=(1)/(2)` and `(c)/(6) =(2)/(3)` <br> `rArr " " c = 3` and `c = 4` <br> Since, c has different values. <br> Hence, for no value of c the pair of equations will have infinitely many solutions .

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$\Rightarrow \frac{c}{6}=\frac{1}{2}and\frac{c}{6}=\frac{2}{3}\Rightarrow c=3andc=4$

c has two different values when two different relations are considered.

Hence, for no value of c the pair of equations will have infinitely many solutions.