Question 980167
1.) The sum of two numbers is 15. The sum of their squares is 9 more than 13 times the larger number. Find the numbers.


Let the numbers be x and y, the latter being the larger of the two.


x+y=15  -----(A)
x^2 + y^2 = 13y ----(B)


Squaring equation (A), we get


x^2 + y^2 + 2xy = 225

=> 13y + 2xy = 225
=> y= 225/(13+2x)


13+2x must be a multiple of 5.

If x=1, y=15 . [Rejected as sum is 15]
If x=6, y= 225/25 = 9. [Accepted].


[Assumption, both x and y are natural numbers here].


Method 2:
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From equation (B) x^2 + y^2 = 13y

=> x^2 =13y-y^2
=> x^2 =y(13-y)

Now, minimum value of y can be 8 (as 7+8=15 and y has to be larger than x always).

By trial and error,

if y=8; y(13-y)= 8*5. Not a square. [Rejected as x^2 is in LHS]
if y=9; y(13-y)= 9*4. A square. [Accepted; and value of x=6].

(x,y)=(6,9)


2.) The product of two numbers is 10 less than 16 times the smaller number. If twice the smaller number is 5 more than the larger number, find the two numbers


Let the numbers be x and y, the latter being the larger of the two.


xy=16x-10 -----(A)


Also, 2x= y+5 ----(B)


Multiplying (A) by 2,

=>y*(2x) =16*(2x)-20
=>y(y+5) = 16(y+5) -20
=>y^2 + 5y = 16y + 80 -20
=>y^2 - 11y -60 = 0
=> y^2 - 15y +4y -60 =0
=> y=15, -4

(y,x)= (15,10) is the correct set.



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Thanks,
PRD
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