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In the following question two equations are given. You have to solve the equations and Give answer
I. \( \Large x^{2} \) + x - 6 = 0
II. \( \Large 2y^{2} \) - 13y + 21 = 0
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A) x < y |
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B) x \( \Large \leq \) y |
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C) x = y |
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D) x \( \Large \geq \) y |
Correct Answer:
| A) x < y |
Description for Correct answer:
I. \( \Large x^{2} \) + x - 6 = 0
=> \( \Large x^{2} \) + 3x - 2x - 6 = 0
=> x ( x + 3 ) - 2 ( x + 3 ) = 0
=> ( x + 3) ( x - 2 ) = 0
\( \Large \therefore \) x = - 3 or 2
II. \( \Large 2y^{2} \) - 13y + 21 = 0
=> 2\( \Large y^{2} \) - 7y - 6y + 21 = 0
=> y ( 2y - 7 ) - 3 ( 2y - 7 ) = 0
=> ( 2y - 7 ) ( y -3 ) = 0
\( \Large \therefore \) y = \( \Large \frac{7}{2} \) or 3
I. \( \Large x^{2} \) + x - 6 = 0
=> \( \Large x^{2} \) + 3x - 2x - 6 = 0
=> x ( x + 3 ) - 2 ( x + 3 ) = 0
=> ( x + 3) ( x - 2 ) = 0
\( \Large \therefore \) x = - 3 or 2
II. \( \Large 2y^{2} \) - 13y + 21 = 0
=> 2\( \Large y^{2} \) - 7y - 6y + 21 = 0
=> y ( 2y - 7 ) - 3 ( 2y - 7 ) = 0
=> ( 2y - 7 ) ( y -3 ) = 0
\( \Large \therefore \) y = \( \Large \frac{7}{2} \) or 3
Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations
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