1. Find four consecutive even integers such that four times the greatest is 36 more than the sum of the other three.

2. Find three consecutive integers such that the sum of the second and the third is half the first decreased by 15.


Please explain and show you solutions. ( Use X as your unknown number) Thanks in advance!


2

Answers

2014-03-05T21:35:42+08:00
1.
x
x+2
x+4
x+6

4(x+6) = 36 + 3x + 6
4x + 24 = 3x + 42
x = 18
x+2 = 18 + 2 = 20
x+4 = 18 + 4 = 22
x+6 = 18 + 6 = 24

2.
x
x+1
x+2

4x+6 = 30 - x
x = 24/5
x+1 = 24/5 + 1 = 29/5
x+2 = 24/5 + 2 = 34/5
0
2014-03-05T21:45:27+08:00
For #1
let x be the first number and the four consecutive numbers:
x+(x+2)+(x+4)+(x+6)

Four times the greatest is 36 more than the sum of the other three:
4(x+6)=36+(x+x+2+x+4)
distributive property------combine like terms:
                   4x+24=36+(3x+6)
                   4x+24=3x+42
                   4x-3x=42-24
                       x=18
therefore:
x=18
x+2=20
x+4=22
x+6=24

Try to check it yourself na lang :D

#2
x
x+1
x+2

the sum of the second and the third is half the first decreased by 15.
(x+1)+(x+2) = (x-15) / 2
2x+3=
 (x-15) / 2
remove the denominator by dividing both sides by 2
2[2x+3= (x-15) / 2]2
4x+6 = x-15
4x-x = -15 - 6
3x = - 21
x = -7

so the 3 consecutive integers are:
x= -7
x + 1 = -6
x + 2 = -5



 

0