Question 980167: Hello :)
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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.
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
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Found 2 solutions by onlinepsa, josgarithmetic:
Answer by onlinepsa(22)   (Show Source):
You can put this solution on YOUR website! 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
https://onlinepsa.wordpress.com
Answer by josgarithmetic(39616)  (Show Source):
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