the difference bet. 2 nos. is 45 and their sum is 6 times the smaller no. decreased by 11. what are the nos?

Question

put the solution please....

Answer 1

Let the smaller number be x
The larger number will be x + 45

x + (x+45) = (6x-11)
2x + 45 = 6x - 11
4x = 56
x = 14
x + 45 = 59

The numbers are 14 and 59

Check:
59 - 14 = 45
59 + 14 = 73
6*14 - 11 = 84 - 11 = 73

Answer 2

let a > b and a -- b = 45 => a = b + 45
also (a + b) = 6b -- 11
=> b + 45 + b = 6(b -- 11)
=> 4b = 111 + b = 111/4
=> a = b + 45 = 111/4 + 45 = 291/4
numbers are 291/4 and 111/4
verify
291/4 -- 111/4 = 180/4 = 45
291/4 + 111/4 = 402/4 = 6(111/4 -- 11)

this also if a > b and a -- b =45 and
a + b = 6b -- 11
=> b + 45 + b = 6b -- 11
=> 4b = 56 => b = 14, a = 14 + 45 = 59
numbers are 59 and 14.

Answer 3

L - S = 45; L + S = 6(S - 11). Simplifying,
L + S = 6S -66; L = 5S -66; L = 45+S;
5S -66 = S + 45; 4S = 111; S = 27.75; L = 72.75. This differs from the previous answer in that there is a different interpretation of "6 times the smaller number decreased by 11." Is the 11 to be subtracted before or after the multiplication by 6?

Answer 4

The difference between two numbers is 45.

To avoid the complexities of double variable calculation, let the numbers be x and 45 + x.

x + x + 45 = 6x - 11
2x + 45 = 6x - 11
-4x = -56
x = 14

45 + x = 59

The numbers are 14 and 59

Answer 5

Let x = smaller number, y = greater number

y - x = 45
y = 45 + x

x + y = 6x - 11
y = 5x - 11

45 + x = 5x - 11
4x = 56
x = 14
y = 14 + 45 or 59

Answer: the smaller no. is 14; the greater no. is 59.

Proof:
59 - 14 = 45
45 = 45

14 + 59 = 6(14) - 11
73 = 84 - 11
73 = 73

Answer 6

sounds easy...
let the no.s be 'x' and 'y',and let y>x,
therefore,

y - x = 45, ---(1)

x + y =6x -11,
or, 5x - y = 11, --(2)
solving (1) and (2), we get,

x=14,
y=59...