For what value of will the following pair of linear equations have no solution 2x 3y 9?

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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.

One of the most important and exam oriented question from linear equations in two variables in which we have given two equations 2x+3y=9, 6x+(k-2)y=3k-2 and we have to find the value of k for which the given equations have no solutions.

Kindly solve the above equations by using the properties of linear equations

RS Aggarwal, class 10, chapter 3F, question no 4

4

For what value of k will the following pair of linear equations have no solution?

2x + 3y = 9, 6x + (k – 2) y = (3k – 2) .

There are two equations given in the question:

2x + 3y – 9 = 0 (i)

And, 6x + (k – 2)y – (3k – 2) = 0 (ii)

These given equations are in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 where,

a1 = 2, b1 = 3 and c1 = – 9

Also, a2 = 6, b2 = (k – 2) and c2 = – (3k – 2)

Now, for the given pair of linear equations having no solution we must have:

k = 11,

k = 11, 3 (3k – 2)

∴ k = 11 and 1

Hence, the value of k is 11

Find the value of k for which the system of linear equations has an infinite number of solutions.
2x + 3y=9,
6x + (k – 2)y =(3k – 2

The given pair of linear equations are
2x + 3y – 9 = 0                            ……(i)
6x + (k – 2)y – (3k – 2) = 0           ……(ii)
Which is of the form `a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0`, where
`a_1 = 2, b_1 = 3, c_1 = -9, a_2 = 6, b_2 = k – 2 and c_2 = -(3k – 2)`
For the given pair of linear equations to have infinitely many solutions, we must have
`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`
`⇒ 2/6 = 3/(k−2) ≠ (−9)/(−(3k−2))`
`⇒ 2/6 = 3/(k−2) , 3/(k−2) ≠ (−9)/(−(3k−2))`
`⇒ k = 11, 3/(k−2) ≠ 9/((3k−2))`
⇒ k = 11, 3(3k – 2) ≠ 9(k – 2)
⇒ k = 11, 1 ≠ 3 (true)
Hence, k = 11.

Solution : For no solution, we have ` (a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)`.
` therefore (2)/(6) = (3)/(k - 2 ) and (2)/(6) ne = (9)/(3k - 2) rArr ( 3)/(k - 2 ) = (1)/(3) and (9)/(3k - 2) ne (1)/(3)`
` rArr k - 2 = 9 and 3k - 2 ne 27 rArr k = 11 and k ne (29)/(3)`.
Hence, ` k = 11 `

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For what value of k will the following pair of linear equation have no solution? 2x+3y=9,6x+k 2y=3k 2.

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2x+3y=96x+(k−2)y=(3k−2)Equation can be written as 2x+3y−9=06x+(k−2)y−(3k−2)=0 Compare witha1x+by1+c1=0a2x+by2+c2=0a1=2b1=3c1=−9a2=6b2=k−2c2=−(3k−2) And for no solutiona1a2=b1b2≠c1c2Takea1a2=b1b226=3k−22k−4=182k=22k=11

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