Question 109125

find three consecutive even intergers such that the sum of the smallest and twice the second is 20 more than the third

{{{x}}},{{{x+2}}},{{{x+4}}}

{{{x}}} is smalest integer

{{{x+2}}} is the second integer, and twice the second is {{{2(x+2)}}}

{{{x+4}}} is the third integer

the sum of the smallest and twice the second is:

{{{x + 2(x+2)}}} 

the smallest and twice the second is {{{20}}} more than the third

{{{x + 2(x+2)= (x+4) + 20}}} 

now solve for {{{x}}}:


{{{x + 2x + 4= x+4 + 20}}}...move {{{x}}}to the left and {{{4}}} the the right

{{{x + 2x - x = 4 + 20 - 4}}}..add all terms on both sides

{{{2x = 20}}}.......divide both sides by {{{2}}}

{{{x = 10}}}.........SMALLEST integer

then
{{{x+2 = 12}}}.....second integer

and
{{{x+4= 14}}}...third integer


check:

the sum of smallest and second integer is: {{{10 + 2*12 = 34}}}

{{{34 = 14 + 20}}}........where {{{14}}} is the third integer