For what value of C do the 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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The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is no value.

Explanation:

The given equations of lines are cx – y = 2 and 6x – 2y = 3

⇒ cx – y – 2 = 0 and 6x – 2y – 3 = 0

Here, a1 = c

b1= –1

c1 = –2

And a2 = 6

b2 = –2

c2 = –3

Since, condition for infinitely many solutions is

`a_1/a_2 = b_1/b_2 = c_1/c_2`

⇒ `c/6 = (-1)/(-2) = (-2)/(-3)`

⇒ `c/6 = 1/2` and `c/6 = 2/3`

⇒ c = 3 and c = 4

Since, c has different values.

So, there exists no value of c for which given equations have infinitely many solutions.

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 .