For what value of k, the linear equation 3x + ky = 12 has equal values of x and y for its solution.

The given system of equations:x + 2y = 5⇒ x + 2y - 5 = 0            ….(i)3x + ky + 15 = 0             …(ii) These equations are of the forms:`a_1x+b_1y+c_1 = 0 and a_2x+b_2y+c_2 = 0`where, `a_1 = 1, b_1= 2, c_1= -5 and a_2 = 3, b_2 = k, c_2 = 15`(i) For a unique solution, we must have:

∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/3 ≠ 2/k ⇒ k ≠ `6`

Thus for all real values of k other than 6, the given system of equations will have a unique solution.(ii) For the given system of equations to have no solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ 1/3 = 2/k≠ (−5)/15``⇒ 1/3 = 2/k and 2/k≠ (−5)/15`⇒k = 6, k ≠ -6

Hence, the required value of k is 6.

Question 30 Pair of Linear Equations in Two Variables - Exercise 3.3

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Answer:

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a1 = k, b1 = 3, c1 = -(k – 3)

a2 = 12, b2 = k, c2 = – k

Then,

a1 /a2 = k/12

b1 /b2 = 3/k

c1 /c2 = (k-3)/k

For no solution of the pair of linear equations,

a1/a2 = b1/b2≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k2 = 36

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k2 – 6k ≠ 0

so, k ≠ 0,6

Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.

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Text Solution

Solution : The given linear equation is 2x+cy=8. <br> Now, by condition, x and y-coordinate of given linear equation are same, i.e., x=y. <br> Put y=x in Eq. (i), we get <br> 2x+cx=8 <br> `implies " " cx=8-2x` <br> `implies " " c=(8-2x)/(x),x ne 0` <br> Hence, the required value of c is `(8-2x)/(x)`.

Option 1 : \(\dfrac{4}{3}\)

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60 Questions 60 Marks 60 Mins

Given:

Linear Equation:

3x + 4y = 12

x + ky = 4

Concept:

For the equation of the following form

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

To get Infinite Solution, we must have the following criteria,

\((\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2})\)

Calculation:

Comparing the equations, we get

a1 = 3

a2 = 1

b1 = 4

b2 = k

c1 = - 12

c2 = - 4

Substituting the values in \((\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2})\), we have,

\((\Rightarrow\dfrac{3}{1} = \dfrac{4}{k} = \dfrac{-12}{-4})\)

\((\Rightarrow\dfrac{3}{1} = \dfrac{4}{k} = \dfrac{-12}{-4})\)

= 3k = 4

⇒ Hence, k = 4/3

For a system of linear equation in two variables

Unique Solution: \((\dfrac{a_1}{a_2} ≠ \dfrac{b_1}{b_2} )\)

No Solution: \((\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}≠ \dfrac{c_1}{c_2})\)

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