SOLUTION: find three consecutive even integers, such that the sum of the second and third number is 20 more than the first.
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Question 517004: find three consecutive even integers, such that the sum of the second and third number is 20 more than the first.
Found 2 solutions by jessica43, JBarnum:
Answer by jessica43(140) (Show Source): You can put this solution on YOUR website!
To solve this equation you need to write equations using what you know. First, you know that the numbers are 3 consecutive even integers. So A is your first, B = A + 2 and C = B + 2.
Next, you know that the sum of the second and third numbers is 20 more than the first: B + C = 20 + A.
Now plug the first two equations into the second, for B and C:
B + C = 20 + A
(A + 2) + (B + 2) = 20 + A
A + 2 + B + 2 = 20 + A
A + B + 4 = 20 + A
B + 4 = 20
B = 16
So the three consecutive even numbers are 14, 16, 18.
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
x, x+2, x+4
add
subtract from both sides
subtract from both sides
, ,
check:
correct
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