Find the value of k for which x 3y 6 2x ky 12 has unique solution

The given system of equations can be written as(k – 3) x + 3y - k = 0kx + ky - 12 = 0This system is of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`where, `a_1 = k, b_1= 3, c_1= -k and a_2 = k, b_2 = k, c_2= -12`For the given system of equations to have a unique solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒ (k−3)/k = 3/k = (−k)/(−12)`

`⇒ k – 3 = 3 and k^2 = 36`

⇒ k = 6 and k = ±6⇒ k = 6

Hence, k = 6.

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Find the values of a and b for which the system of linear equations has an infinite number of solutions:
(a – 1) x + 3y = 2, 6x + (1 – 2b)y = 6

The given system of equations can be written as(a – 1) x + 3y = 2⇒(a – 1) x + 3y – 2 = 0           ….(i)and 6x + (1 – 2b)y = 6⇒6x + (1 – 2b)y – 6 = 0         ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`where, `a_1 = (a – 1), b_1= 3, c_1= -2 and a_2 = 6, b_2 = (1 – 2b), c_2= -6`For an infinite number of solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒ (a−1)/6 = 3/((1−2b)) = (−2)/(−6)``⇒ (a−1)/6 = 3/((1−2b)) = 1/3``⇒ (a−1)/6 = 1/3 and 3/((1−2b)) = 1/3`⇒ 3a – 3 = 6 and 9 = 1 – 2b⇒ 3a = 9 and 2b = -8⇒ a = 3 and b = -4

∴ a = 3 and b = -4

Concept: Pair of Linear Equations in Two Variables

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The given system of equations can be written as(2a – 1) x + 3y = 5⇒(2a – 1) x + 3y – 5 = 0               ….(i)and 3x + (b – 1)y = 2⇒3x + (b – 1)y – 2 = 0                        ….(ii)

These equations are of the following form:

`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = (2a – 1), b_1= 3, c_1= -5 and a_2 = 3, b_2 = (b – 1), c_2= -2`For an infinite number of solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒ ((2a−1))/3 = 3/((b−1)) = (−5)/(−2)``⇒ ((2a−1))/6 = 3/((b−1)) = 5/2``⇒ ((2a−1))/6 = 5/2 and 3/((b−1)) = 5/2`⇒ 2(2a – 1) = 15 and 6 = 5(b – 1)⇒ 4a – 2 = 15 and 6 = 5b – 5⇒ 4a = 17 and 5b = 11

`∴ a = 17/4 and b = 11/5`

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The given system of equations can be written as2x - 3y = 7⇒2x - 3y - 7 = 0                        ….(i)and (a + b)x - (a + b – 3)y = 4a + b⇒(a + b)x - (a + b – 3)y - 4a + b = 0               ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`Here,` a_1 = 2, b_1= -3, c_1= -7 and a_2 = (a + b), b_2= -(a + b - 3), c_2= -(4a + b)`For an infinite number of solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``2/(a+b) = (−3)/(−(a+b−3)) = (−7)/(−(4a+b))``⇒ 2/(a+b) = 3/((a+b−3)) = 7/((4a+b))``⇒ 2/(a+b) = 7/((4a+b)) and 3/((a+b−3)) = 7/((4a+b))`⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b - 3)⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b - 21⇒ 4a = 17 and 5b = 11∴ a = 5b                              …….(iii)and 5a = 4b – 21               ……(iv)

On substituting a = 5b in (iv), we get:

25b = 4b – 21⇒21b = -21⇒b = -1On substituting b = -1 in (iii), we get:a = 5(-1) = -5

∴a = -5 and b = -1.

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The given system of equations can be written as2x + 3y = 7⇒2x + 3y - 7 = 0 ….(i)and (a + b + 1)x - (a + 2b + 2)y = 4(a + b) + 1(a + b + 1)x - (a + 2b + 2)y – [4(a + b) + 1] = 0 ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 2, b_1= 3, c_1= -7 and a_2 = (a + b + 1), b_2 = (a + 2b + 2), c_2= – [4(a + b) + 1]`For an infinite number of solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``2/((a+b+1)) = 3/((a+2b+2)) = (−7)/(−[4(a+b)+1])``⇒ 2/((a+b+1)) = 3/((a+2b+2)) = 7/([4(a+b)+1])``⇒ 2/((a+b+1)) = 3/((a+2b+2))and 3/((a+2b+2)) = 7/([4(a+b)+1])`⇒ 2(a + 2b + 2) = 3(a + b + 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14⇒ a – b - 1=0 and 12a + 12b + 3 = 7a + 14b + 14⇒ a - b = 1 and 5a – 2b = 11a = (b + 1)                  …….(iii)5a - 2b = 11                  ……(iv)On substituting a = (b + 1) in (iv), we get:5(b + 1) – 2b = 11⇒5b + 5 – 2b = 11⇒ 3b = 6⇒ b = 2On substituting b = 2 in (iii), we get:

a = 3

∴a = 3 and b = 2.

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The given system of equations can be written as2x + 3y - 7 = 0                                               ….(i)(a + b)x + (2a - b)y – 21 = 0                        ….(ii)This system is of the form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 2, b_1= 3, c_1= -7 and a_2 = a + b, b_2 = 2a - b, c_2= – 21`For the given system of linear equations to have an infinite number of solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒2/(a+b) = 3/(2a−b) = (−7)/(−21)``⇒ 2/(a+b) = (−7)/(−21) = 1/3 and 3/(2a−b) = (−7)/(−21) = 1/3`⇒ a + b = 6 and 2a –b = 9Adding a + b = 6 and 2a – b = 9 ,we get3a = 15 ⇒ a = `15/3` = 3Now substituting a = 5 in a + b = 6, we have5 + b = 6 ⇒ b = 6 – 5 = 1

Hence, a = 5 and b = 1.

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The given system of equations can be written as2x + 3y - 7 = 0                         ….(i)2ax + (a + b)y – 28 = 0             ….(ii)This system is of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`where, `a_1 = 2, b_1= 3, c_1= -7 and a_2 = 2a, b_2 = a + b, c_2= – 28`

For the given system of linear equations to have an infinite number of solutions, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒2/(2a) = 3/(a+b) = (−7)/(−28)``⇒ 2/(2a) =( −7)/(−28 )= 1/4 and 3/(a+b) = (−7)/(−28) = 1/4`⇒ a = 4 and a + b = 12Substituting a = 4 in a + b = 12, we get4 + b = 12 ⇒ b = 12 – 4 = 8

Hence, a = 4 and b = 8.

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The given system of equations:8x + 5y = 98x + 5y - 9 = 0                   ….(i)kx + 10y = 15kx + 10y - 15 = 0                  ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 8, b_1= 5, c_1= -9 and a_2 = k, b_2 = 10, c_2= – 15`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e.,  8/k = 5/10 ≠( −9)/(−15)``i.e. , 8/k = 1/2 ≠ 3/5``8/k = 1/2 and 8/k ≠ 3/5``⇒ k = 16 and k ≠ 40/3`

Hence, the given system of equations has no solutions when k is equal to 16.

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The given system of equations:kx + 3y = 3kx + 3y - 3 = 0                    ….(i)12x + ky = 612x + ky - 6 = 0                    ….(ii)

These equations are of the following form:

`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = k, b_1= 3, c_1= -3 and a_2 = 12, b_2 = k, c_2= –6`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``i .e.,  k/12 = 3/k ≠ (−3)/(−6)``k/12 = 3/k and 3/k ≠ 1/2``⇒ k^2 = 36 and k ≠ 6``⇒ k = ±6 and k ≠ 6`

Hence, the given system of equations has no solution when k is equal to -6.

Page 10

The given system of equations:3x - y - 5 = 0                      ….(i)And, 6x - 2y + k = 0             ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where,`a_1 = 3, b_1= -1, c_1= -5 and a_2 = 6, b_2= -2, c_2 = k`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e., 3/6 = (−1)/(−2) ≠ −5/k``⇒(−1)/(−2) ≠ (−5)/k ⇒ k ≠ -10`

Hence, equations (i) and (ii) will have no solution if k ≠ -10.

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The given system of equations can be written askx + 3y + 3 - k = 0                       ….(i)12x + ky - k = 0                           ….(ii)This system of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`

where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ k/12 = 3/k ≠ (3−k)/(−k)``⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)``⇒ k^2 = 36 and -3 ≠ 3 - k`⇒ k = ±6 and k ≠ 6⇒k = -6

Hence, k = -6.

Page 12

The given system of equations:5x - 3y = 0                 ….(i)2x + ky = 0                 ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 5, b_1= -3, c_1 = 0 and a_2 = 2, b_2 = k, c_2 = 0`For a non-zero solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2)``⇒ 5/2 = (−3)/k``⇒5k = -6 ⇒ k = (−6)/5`

Hence, the required value of k is `(−6)/5`.