Question

solve.
find three consecutive integers whose sum is the square of 6

Answer 1

so the equation you want is:
\[x + x + 1 + x +2 = 36\]

Answer 2

you combine like terms to get:
\[3x + 3 = 36\]

Answer 3

subtract 3 to get:

Answer 4

\[3x = 33\]

Answer 5

divide by 3 to get:
\[x = 11\]

Answer 6

so the three numbers are 11, 11+1 (12), and 11 + 2 (13)

Answer 7

and finally:
\[11 + 12 + 13 = 36 = 6^2\]

Answer 8

so it works

Answer 9

thank you

Answer 10

you are welcome

Answer 11

could you help with this one..the greater of two consecutive integers is 10 less than twice the smaller. find the integers.

Answer 12

so:
\[x + 1 = 2x - 10\]

Answer 13

add 10 to both sides:
\[x + 11 = 2x\]

Answer 14

subtract x:
\[x = 11\]

Answer 15

check:
\[2(11) = 22 = 12 + 10\]

Answer 16

so the integers are 11 and 12

Answer 17

i meant to put two consecutive even** integers

Answer 18

oh

Answer 19

o.k.

Answer 20

\[x + 2 = 2x - 10\]

Answer 21

because you skip a number

Answer 22

so you get x is 12 through the same process as above

Answer 23

check:
\[2(12) = 24 = 14 + 10\]

Answer 24

it works, so the numbers are 12 and 14

Answer 25

okay thanks

Answer 26

you are welcome