algebra??????

Question

one number exceeds three times a second number by three. half of the larger number exceeds one-fifth of the smaller number by eight.find the number.

Answer 1

First, formulate the equations:
I: n1 = 3 n2 + 3;
II: 1/2 n1 = 1/5 n2 + 8;
combine and solve for n1:
II * -2:
II: -n1 = -2/5 n2 - 16;
Add I and (new) II;
0 = 13/5 n2 - 13
n2 = 5;
from I: n1 = 3 n2 + 3;
n1 = 3 (5) + 3 = 18!

Answer 2

Let the smaller number be 'x'
The larger number is '3x + 3'

(3x + 3)/2 = x/5 + 8
(3x + 3)/2 = (x + 40)/5
5(3x + 3) = 2(x + 40)
15x + 15 = 2x + 80
15x - 2x = 80 - 15
13x = 65
x = 65/13
x = 5
The smaller number is 5
The larger number is 3(5) + 3 = 15 + 3 = 18
The numbers are 5 and 18

Answer 3

lets take the 2nd number as x
thus, 1st number= 3x+3
1st number > 2nd number
according to the question,
1/2 of 3x+3 exceeds 1/5 of x by 8
therefore,
1/2(3x+3) - x/5 = 8
(3x/2+3/2)- x/5 = 8
(3x+3)/2 - x/5 = 8
(15x+15)/10 - 2x/10 = 8 (L.C.M)
(15x+15-2x)/10 = 8
13x+15 = 8*10
13x+15 = 80
13x = 80-15
13x = 65
x = 65/13=5
thus,
2nd number=x=5
1st number= 3x+3=3*5 +3=15+3=18

Answer 4

Let the two numbers by x and y.

y = 3x + 3
y/2 = x/5 + 8

y = 2x/5 + 16

3x + 3 = (2/5)x + 16
15x + 15 = 2x + 80
13x = 65
x = 5
y = 3x + 3 = 3(5) + 3 = 18

The two numbers are 5 and 18.

Answer 5

N1 = 3 X N2 + 3
1/2 X N1 = 1/5 X N2 + 8

N1 = 18, N2 = 5

Answer 6

let x , 3x + 3 be the numbers

given

(3x +3)/2 = x/5 + 8

(3x +3)/2 = ( x + 40) / 5

15x + 15 = 2x + 80

13x = 65

x = 5

put x=5 in 3x + 3 we get 18

Numbers 5,18

Answer 7

n+3=m
n/5+3=m/2

m-n=3
m/2-n/5=3

m-n=3
-m+2n/5=-6

n/5=3
n=15
m=18