InterviewSolution
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InterviewSolution
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Find the solution of the pair of equations `(x)/(10) + (y)/(5) - 1 = 0` and `(x)/(8) + (y)/(6) = 15` and find `lambda`, if `y = lambda x + 5`. |
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Answer» Given pair of equations is `(x)/(10)+(y)/(5)-1=0 " " ...(i)` and `" " (x)/(8)+(y)/(6)=15 " " ...(ii)` Now, multiplying both sides of Eq. (i) by LCM (10,5)=10, we get `x+2y-10=0` `rArr " " x+2y=10 " " ...(iii)` Again, multiplying both sides of Eq. (iv) by LCM (8,6)=24, we get `3x+4y=360 " " ...(iv)` On, multiplying Eq. (iii) by 2 and then subtracting from Eq. (iv), we get `{:(3x + 4y = 360),( ul(underset(-)2x+underset(-)4y=underset(-)2 0)),(" "x=340):}` Put the value of x in Eq. (iii), we get 340+2y=10 `rArr " " 2y=10-340=-330` `rArr " " y=-165` Given that, the linear relation between x, y and `lambda` is `y=lambda x + 5` Now, put the values of x and y in above relation, we get `-165=lambda (340)+5` `rArr " " 340 lambda =-170` `rArr " " lambda=-(1)/(2)` Hence, the solution of the pair of equations is x=340, y=-165 and the required value of `lambda` is `-(1)/(2)`. |
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