help please

Question

find four consecutive odd integers such that 5 times the sum of the first and third integers exceeds four times the sum of the second and last integers by 14.

Answer 1

Hi Patryk

 

OK lets define some variables that will help us solve this thing

 

x = the first odd number

 

x + 2 = second odd #

 

x+ 4 = 3rd odd #

 

x + 6 = 4th odd #

 

now lets create and equation using the information in the problem

 

5(x + x +4)    this is five times the 1st plus 3rd

 

4(x+2 + x +6)  this is four times the 2nd plus 4th

 

now we we make the equation

 

5(x + x +4) = 4(x+2 + x +6) +14  (we add the 14 to make them equal)

 

now solve

 

10x + 20 = 8x + 32 + 14

 

10x + 20 = 8x + 46

 

2x = 26

 

x = 13

 

x+ 2 = 15

 

x + 4 = 17

 

x + 6 = 19

 

5(13+17) = 150

 

4(15+19)= 136  and if we added 14 we get 150

 

Hope this helps

Answer 2

So this is really a problem of one unknown, because the 2nd, 3rd, and 4th integers can be expressed in terms of the 1st, such as:

n, n+1, n+2, and n+3, where n is the unkown.

 

Now set up 1 equation to capture the conditions that were cited:

5(n+(n+2)) > 4( n+1 + n+2 ) +14

 

and then solve for 'n'. I think I did that right. Check my work, but you should get the idea.

 

- Ben