find two numbers who sum is 66 and whose difference is 8?

Answer 1

let x be the bigger no. and y be the smaller no.
x+y=66 ---(1)
x-y=8 ---(2)
from (1):
x= 66-y
subst into (2):
66-y-y=8
66-2y=8
2y=58
y= 29
x=37

Answer 2

a million. you have 2 equations: x + y = a hundred (a million) x * y = f (2) resolve for y in (a million): y = a hundred - x Insert in (2): x * (a hundred - x) = f As you notice f is a function of x. that could be a quadratic function and so there exists a optimal (the term with x^2 is damaging) so which you would be able to the two differentiate the function and set f' = 0 and resolve for x or, in case you do no longer understand differential calculus, you ought to use right here formulation for the optimal of a quadratic funtion: (-b/(2*a) ; -(b^2 - 4*a*c)/(4*a)) the y-coordinate is the optimal fee of the made from the two numbers. you recognize the variety of x and so can discover y in equation (a million). 2. Set y = 0 and resolve for x. 3. 2 equations with 2 unknowns: a + b = sixty six (3) a - b = 8 (4) from (4): a = 8 + b. Insert in (3): 8 + b + b = sixty six <=> 2*b = fifty 8 <=> b = 29 so: a + 29 = sixty six <=> a = 37.

Answer 3

x = 1st number, y = 2nd number

Find values of y:
x + y = 66
y = 66 - x

x - y = 8
y = x - 8

Find value of x:
66 - x = x - 8
2x = 58
x = 29

Find value of y:
y = 66 - 29
y = 37

Answer: the numbers are 29 and 37

Proof: the sum of the two numbers is 66 (29 + 37); their difference is 8 (37 - 29).

Answer 4

x + (x - 8) = 66

2x = 74
x = 37

x - 8 = 29

Two numbers are -: 37 and 29

To check-:
37 + 29 = 66

Answer 5

let the biger: x => the other is : x-8
=> x+ (x-8) =66
=> 2x = 58
=> x =29

Answer 6

a+b=66 a-b=8
a=b+8
b+8+b=66
2b=66-8=58
b=29 a=37

Answer 7

Let the 2 nos. be x & y
sum of the nos. x + y =66........(1)
difference of the nos. x - y = 8 .......(2)
2x = 74........adding(1) & (2)
x =37
substituting x = 37 in (1)
y = 29
the 2 nos. are 29 & 37

Answer 8

29 37 ?

Answer 9

x + y = 66
x - y = 8-------ADD
2x = 74
x = 37
y = 29