For what value of k will the linear equation have infinitely many solutions.
2x - 3y = 7 ,
( k + 1 )x + ( 1 - 2k)y = 5k - 4
Class 10
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Answers
Given Equation is 2x - 3y = 7.
2x - 3y - 7 = 0
On comparing with a1x + b1y + c = 0, we get a1 = 2, b1 = -3, c1 = -7.
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Given Equation is (k + 1)x + (1 - 2k)y = 5k - 4
= > (k + 1)x + (1 - 2x)y - 5k + 4 = 0
On comparing with a2x + b2y + c2 = 0, we get a2 = (k + 1), b2 = (1 - 2k), c2 = (-5k + 4).
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Given that the equation have infinitely many solutions.
= > a1/a2 = b1/b2 = c1/c2
= > (2/k + 1) = (-3/1 - 2k) = (-7/-5k + 4)
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(1)
= > 2/k + 1 = -3/1 - 2k
on cross multiplication,we get
= > (1 - 2k)2 = -3(k + 1)
= > 2 - 4k = -3k - 3
= > -3k + 4k = 2 + 3
= > k = 5.
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(2)
= > (-3/1 - 2k) = (-7/-5k + 4)
= > -3(-5k + 4) = -7(1 - 2k)
= > 15k - 12 = -7 + 14k
= > k = 5.
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Therefore, value of k = 5.
Hope this helps!