SOLUTION: find three consecutive integers such that 4 times the sum of the last two is 2 greater than 10 times the first. I don't understand how to find the sum of the last two.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find three consecutive integers such that 4 times the sum of the last two is 2 greater than 10 times the first. I don't understand how to find the sum of the last two.       Log On


   



Question 506079: find three consecutive integers such that 4 times the sum of the last two is 2 greater than 10 times the first. I don't understand how to find the sum of the last two.
Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!
Let
1st Integer = x
2nd Integer = x+1
3rd Integer = x+2
===============================================
Given
4((x+1)+(x+2))=10(x)+2
4(x+1+x+2)=10x+2
4(2x+3)=10x+2
8x+12=10x+2
8x-10x=+2-12
-2x=-10
-2x/-2=-10/-2
x=5
===============================================
1st Integer = x =5
2nd Integer = x+1 =5+1=6
3rd Integer = x+2=5+2=7
================================================
Check
4(6+7)=10(5)+2
4(13)=50+2
52=52