Question

find the value of k for which each of the following systems of linear equation has an infinite number of solutions:
2x -3y =7 ,
(k-1)x + (k+2)y = 3k


​CBSE class X

Answer 1

Answer:-

• 1st equation 2x -3y -7 = 0

• 2nd equation (k-1)x +(k+2) -3k = 0

where

• a₁ = 2 , b₁ = -3 & c₁ = -7

• a₂ = k-1 , b₂ = k+2 & c₂ = -3k

As we know that the condition for equation which has infinitely many solution .

• a₁/a₂ = b₁/b₂ = c₁/c₂

substitute the value we get

$:\implies$ 2/(k-1) = -3/(k+2) = -7/-3k

$:\implies$ 2/(k-1) = -3(k+2) = 7/3k

$:\implies$ 2(k+2) = -3(k-1) or -3(3k) = 7(k+2)

$:\implies$ 2k + 4 = -3k +3 or -9k = 7k +14

$:\implies$ 5k = -1 or -16k = 14

$:\implies$ k = -1/5 or k = -14/16

$:\implies$ k = -1/5 or k = -7/8

• Hence, the value of k = -1/5 & -7/8 .

Answer 2

★ Step by step explanation :

• Here , we know that each of the following systems of linear equation has an infinite number of solutions.

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Therefore,

$\qquad:\implies\:\sf \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

★ Values that we have :

• 1st eqn = 2x - 3y - 7 = 0
• 2nd eqn = (k - 1)x + (k + 2)y - 3k = 0

So,

• $\sf a_{1} = 2 \:,\: a_{2} = k - 1$
• $\sf b_{1} = -3 \:,\: b_{2} = k + 2$
• $\sf c_{1} = -7 \:,\: c_{2} = -3k$

★ Putting all values :

$\qquad:\implies\:\sf \dfrac{2}{k - 1} = \dfrac{-3}{k + 2} = \dfrac{-7}{-3k}$

★ By cross multiplying :

$\qquad:\implies\:\sf 2(k + 2) = -3(k - 1)\:,\:-3(-3k) = -7(k + 2)$

$\qquad:\implies\:\sf 2k + 4 = -3k + 3\:,\:9k = -7k - 14$

$\qquad:\implies\:\sf 2k + 3k = 3 - 4\:,\:9k + 7k = -14$

$\qquad:\implies\:\sf 5k = -1\:,\:16k = -14$

$\qquad:\implies\:\sf k = \dfrac{-1}{5}\:,\:k = \dfrac{-14}{16}$

$\qquad:\implies\:\sf k = \dfrac{-1}{5}\:,\:k = \dfrac{\cancel{-14}}{\cancel{16}}$

$\qquad:\implies\:\bf{ k = \red{\dfrac{-1}{5}}\:,\:k = \red{\dfrac{-7}{8}}}$

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$\therefore\:{\underline{\frak {Value\: of\:k \:=\:\dfrac{-1}{5}\:or\:\dfrac{-7}{8}}}}$