find five consecutive integers such that: The sum ofthe first and 3 times the third is equal to 54 less than twice the sum of the second, fourth, and fifth

Question

Find five consecutive integers such that:

"The sum of the first and 3 times the third is equal to 54 less than twice the sum of the second, fourth, and fifth."

I need help trying to set up this equations

 

Answer 1

Let's call our 5 integers a, b, c, d, and e.

 

a + 3c = 2(b+d+e) - 54

b = a+1

c = a+2

d = a+3

e=a+4

 

Now we can plug everything into the first and solve for a.

 

a + 3(a+2) = 2[ (a+1) + (a+3) + (a+4) ] -54

a + 3a + 6 = 2[3a+8] - 54

4a + 6 = 6a + 16 - 54

4a + 6 = 6a - 38

44 = 2a

22 = a

 

Which would mean b through e are 23, 24, 25, and 26 respectively. Now we can check that answer.

 

22 + 3(24) = 2(23+25+26) - 54

22 + 72 = 2(74) - 54

94 = 148 - 54

94=94